the masses of the reactants and products after the reaction are:
- Blank g Li (lithium is completely consumed)
- 26.24 g O2 (some oxygen remains)
- 5.12 g Li2O (this is the amount formed in the reaction)
To solve this problem, we first need to write a balanced chemical equation for the reaction between lithium and oxygen:
4Li + O2 → 2Li2O
a) To determine which reactant is present in excess, we need to calculate the amount of product that can be formed from each reactant. We can do this by assuming that one of the reactants is limiting and calculating the amount of product that would be formed based on that assumption. Then, we compare that amount to the amount of product that would be formed based on the other reactant being limiting. The reactant that produces less product is the limiting reactant, and the other reactant is present in excess.
Let's assume that lithium is the limiting reactant. To calculate the amount of product that can be formed from 1.54 g of lithium, we need to convert the mass of lithium to moles using its molar mass:
1.54 g Li × (1 mol Li/6.941 g Li) = 0.222 mol Li
From the balanced chemical equation, we see that 4 moles of lithium react with 1 mole of oxygen to produce 2 moles of Li2O. Therefore, the amount of product that can be formed from 0.222 mol of Li is:
0.222 mol Li × (2 mol Li2O/4 mol Li) = 0.111 mol Li2O
Now, let's assume that oxygen is the limiting reactant. To calculate the amount of product that can be formed from 6.56 g of oxygen, we need to convert the mass of oxygen to moles using its molar mass:
6.56 g O2 × (1 mol O2/32 g O2) = 0.205 mol O2
From the balanced chemical equation, we see that 1 mole of oxygen reacts with 4 moles of lithium to produce 2 moles of Li2O. Therefore, the amount of product that can be formed from 0.205 mol of O2 is:
0.205 mol O2 × (2 mol Li2O/1 mol O2) = 0.410 mol Li2O
Comparing the two amounts of product, we see that the amount of product that can be formed from lithium is smaller than the amount that can be formed from oxygen. Therefore, lithium is the limiting reactant and oxygen is present in excess.
b) To calculate the number of moles of Li2O formed
from the reaction, we can use the amount of limiting reactant (0.222 mol Li) and the mole ratio between the limiting reactant and the product (2 mol Li2O/4 mol Li) to find the amount of product produced:
0.222 mol Li × (2 mol Li2O/4 mol Li) = 0.111 mol Li2O
c) After the reaction, all of the limiting reactant (lithium) will be consumed, and some of the excess reactant (oxygen) will be left over. To calculate the amount of oxygen left over, we can use the amount of excess reactant and the mole ratio between the limiting reactant and the excess reactant (4 mol Li/1 mol O2):
0.205 mol O2 × (4 mol Li/1 mol O2) = 0.820 mol Li
Since we started with 6.56 g of oxygen, and oxygen has a molar mass of 32 g/mol, we can convert the amount of oxygen left over to grams:
(0.820 mol O2) × (32 g O2/mol) = 26.24 g O2 remaining
To calculate the mass of Li2O formed, we can use the amount of product we calculated in part (b) and the molar mass of Li2O (45.88 g/mol):
0.111 mol Li2O × (45.88 g Li2O/mol) = 5.12 g Li2O formed
Finally, to calculate the mass of lithium consumed in the reaction, we can use the mass of lithium we started with (1.54 g) and subtract the amount of lithium that was not consumed:
1.54 g Li - 0.222 mol Li × (6.941 g Li/mol) = 0.998 g Li consumed
Therefore, the masses of the reactants and products after the reaction are:
- Blank g Li (lithium is completely consumed)
- 26.24 g O2 (some oxygen remains)
- 5.12 g Li2O (this is the amount formed in the reaction)
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If 3grams of sodium reacts with 25 grams of sulfuric acid to form sodium sulfate and 1 gram of hydrogen and no sodium is left after the reaction but 9grams of acid remained unreacted how many grams of sodium sulfate were formed
The balanced chemical equation for the reaction between sodium and sulfuric acid to form sodium sulfate and hydrogen gas is:
2Na + H2SO4 -> Na2SO4 + 2H2
From the given information, we can see that the reaction is limited by the amount of sodium available, since all of the sodium is used up in the reaction.
Therefore, we can use the amount of sodium to determine the amount of sulfuric acid that reacted and the amount of sodium sulfate that was formed.
1. Calculate the amount of sulfuric acid that reacted:
m(Sulfuric acid) = 25 g - 9 g = 16 g
n(Sulfuric acid) = m(Sulfuric acid) / M(Sulfuric acid) = 16 g / 98.08 g/mol = 0.163 mol
2. Calculate the amount of sodium sulfate formed:
Since the mole ratio of Na to Na2SO4 is 2:1, the number of moles of sodium used is:
n(Na) = m(Na) / M(Na) = 3 g / 22.99 g/mol = 0.1305 mol
The amount of sodium sulfate formed is also 0.1305 mol, since the mole ratio of Na to Na2SO4 is 2:1.
m(Na2SO4) = n(Na2SO4) x M(Na2SO4) = 0.1305 mol x 142.04 g/mol = 18.54 g
Therefore, 18.54 grams of sodium sulfate were formed in the reaction.
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using wedge-dash notation to designate stereochemistry, draw (r)-3-aminobutan-1-ol.
To draw (R)-3-aminobutan-1-ol using wedge-dash notation, follow these steps: 1. Draw a four-carbon chain representing butan-1-ol. 2. Add an -OH group to the first carbon. 3. Add an -NH2 group to the third carbon.
To draw (R)-3-aminobutan-1-ol using wedge-dash notation to designate stereochemistry, we first need to determine the absolute configuration of the molecule. The priority of the substituents attached to the chiral center (the carbon with four different groups attached) must be determined according to the Cahn-Ingold-Prelog (CIP) rules. The highest priority group is given a number 1, the second-highest priority group is given a number 2, and so on. For (R)-3-aminobutan-1-ol, the substituents attached to the chiral center are: - NH2 (amino group) - highest priority - OH (hydroxy group) - second-highest priority - CH3 (methyl group) - third-highest priority - H (hydrogen) - lowest priority To determine the absolute configuration, we need to look at the orientation of the substituents in three-dimensional space. If the lowest priority group is pointing away from us (into the page), we use the right-hand rule to determine the orientation of the remaining three groups. If the sequence of priorities goes clockwise, the configuration is (R); if it goes counterclockwise, the configuration is (S). In the case of (R)-3-aminobutan-1-ol, we can assign the following orientations: - NH2 (highest priority) - wedge - OH - dash - CH3 - wedge - H (lowest priority) - into the page Based on this, we can see that the sequence of priorities goes clockwise, indicating that the configuration is (R). Therefore, the wedge-dash notation for (R)-3-aminobutan-1-ol is: H NH2 | | C---C | | CH3 OH The NH2 and CH3 groups are represented by wedges, indicating that they are coming out of the page towards the viewer. The OH group is represented by a dash, indicating that it is going into the page away from the viewer. The H group is represented by a thin line, indicating that it is behind the plane of the paper.
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calculate the molarity of 102.6 grams of sugar, C12H22O11 in 500. mL of solution
The molarity of the sugar solution is 0.5988 M (mol/L).
To calculate the molarity of a solution, we need to know the number of moles of solute (the substance being dissolved) and the volume of the solution in liters.
First, we need to determine the number of moles of sugar (C12H22O11) in the given mass of 102.6 grams:
The molar mass of C12H22O11 can be calculated as follows:
12(12.01 g/mol) + 22(1.01 g/mol) + 11(16.00 g/mol) = 342.3 g/mol
The number of moles of C12H22O11 in 102.6 grams can be calculated as:
102.6 g / 342.3 g/mol = 0.2994 mol
Next, we need to convert the volume of the solution from milliliters to liters:
mL = 0.5 L
Now we can calculate the molarity (M) of the solution:
M = moles of solute/liters of solution
M = 0.2994 mol / 0.5 L
M = 0.5988 M
Therefore, the molarity of the sugar solution is 0.5988 M (mol/L).
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1. How many liters of water will be produced if you have 17. 43 grams of ammonia (NH3)? *
(8 Points)
4 NH3 + 502 --> 4 NO + 6H2O
Enter your math answer
17.43 grams of NH₃ will produce 34.39 liters of water.
The balanced chemical equation is 4 NH₃ + 5O₂ → 4NO + 6H₂O. From the equation, we can see that for every 4 moles of NH₃ reacted, 6 moles of water are produced.
Therefore, to determine the number of moles of water produced, we need to convert the mass of NH₃ given to moles. The molar mass of NH₃ is 17.03 g/mol, so:
17.43 g NH₃ × (1 mol NH₃/17.03 g NH₃) = 1.023 mol NH₃
Using stoichiometry, we can calculate the number of moles of water produced:
1.023 mol NH₃ × (6 mol H₂O/4 mol NH₃) = 1.5345 mol H₂O
Finally, we can convert the number of moles of water to liters using the fact that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 L:
1.5345 mol H₂O × (22.4 L/mol) = 34.39 L
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At constant temperature and pressure, a system is most likely to undergo a reaction so that in its final state, as compared to its initial state, the system has:
A) lower energy and higher entropy
B) lower energy and lower entropy
C) higher energy and lower entropy
D) higher energy and higher entropy
In general, a system tends to favor a reaction that results in an increase in entropy, which is a measure of the number of possible arrangements of the system's particles. The answer is A) lower energy and higher entropy.
This is due to the fact that the increase in the number of particles in the system or the increase in the number of ways the particles can be arranged leads to an increase in entropy. On the other hand, a system also tends to favor a reaction that results in a decrease in energy, which is a measure of the system's ability to do work.
Therefore, when a system undergoes a reaction that decreases its energy while increasing its entropy, it is moving towards a more stable and disordered state.
This is because a lower energy state means that the system is releasing energy, while a higher entropy state means that the system is becoming more disordered and spread out. This tendency towards lower energy and higher entropy is known as the second law of thermodynamics, which governs the behavior of all physical systems.
The answer is A) lower energy and higher entropy.
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9. An unknown gas has a volume of 200L at 5 atm and -140°C. What is its volume at STP?
10. A Los Angeles class nuclear sub has an internal volume of eleven million liter at a
pressure of 1. 250 atm. If a crewman were to open one of the hatches to the outside
ocean while it was underwater (pressure of 15. 75 atm), what would be the new volume
of the air inside?
11. A man heats a balloon in the oven (Why?. Who knows?. It is a crazy world we live in).
If the balloon initially has a volume of 0. 40 L and a temperature of 20 °C, what is its
volume after he heats it to 250 °C?
Mixed Gas Laws
12. A gas has a pressure of 1. 26 atm and occupies a volume of 7. 40 L. If the gas is
compressed to a volume of 2. 93 L, what is its new pressure?
13. People who are angry sometimes say that they feel as if they'll explode. If a calm
person with a lung capacity of 3. 5 liters and a body temperature of 36 °C gets angry,
what is the volume of their lungs if their temperature rises to 39 °C. Do you think they
will really explode?
9. Using the combined gas law, the volume of the gas at STP can be calculated as 112.2 L. This equation takes into account the initial pressure, temperature, and volume, as well as the new pressure and temperature at STP.
10. Applying Boyle's law, the new volume of the air inside the submarine would be approximately 87,873.2 L. This is calculated by multiplying the initial volume and pressure, and dividing by the new pressure.
11. Using the combined gas law, the new volume of the balloon can be calculated as 0.98 L. This equation takes into account the initial temperature, volume, and pressure, as well as the new temperature.
12. Using Boyle's law, the new pressure of the gas can be calculated as 3.25 atm. This equation takes into account the initial pressure and volume, as well as the new volume.
13. Using Charles' law, the new volume of the person's lungs can be calculated as 3.8 L. This equation takes into account the initial lung capacity and temperature, as well as the new temperature.
It is highly unlikely that a person would actually explode from anger, as the body has mechanisms in place to regulate pressure and prevent such an event.
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Ite
a city is considering different water purification methods for the water supply. which fact would help inform the city if cost was
a significant constraint? (1 point)
iter
reverse osmosis systems are more expensive because of the chlorine treatments
iten
olon exchange systems are more expensive because of the chlorine treatments
iten
o reverse osmosis systems are more expensive because of the use of filters that need replacement
iten
olon exchange systems are more expensive because of the use of filters that need replacement
item
iten
If cost is a significant constraint for Itea city in choosing a water purification method for their water supply, then the fact that ion exchange systems are more expensive because of the use of filters that need replacement would be important to consider.
While reverse osmosis systems are also more expensive, it is primarily due to the chlorine treatments, which may not be a significant factor for the city. Therefore, ion exchange systems may not be a cost-effective option in the long run due to the ongoing expenses of replacing filters.
This information can help inform the city's decision-making process and ensure that they choose a water purification method that meets their needs while also being financially feasible.
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Chemical equilibrium is a dynamic process. What does this mean?
1. Nothing is changing.
2. There are multiple reactants and products involved in the chemical reaction.
3. It appears as though nothing is happening, but there is constant change occurring.
4.The reaction has reached completion and stopped reacting.
Answer: 3. It appears as though nothing is happening, but there is constant change occurring.
Explanation:
equilibrium is the state when the changes cancel each other, and the net change is 0.
think of it like a stalemate in tug of war; both people are pulling, but you wont see anything change, because their forces are equal and in opposite direction :)
Determine which of the substrates will and will not react with naome in an sn2 reaction to form an appreciable amount of product.
The substrates that will react are CH₃CH₂CH₂Br and CH₃CH₂CH₂CH₂Br and (CH₃)₃CNH₂ and CH₃CH₂OH will not react with naome in an sn2 reaction to form an appreciable amount of product.
Based on the Sn2 reaction mechanism, substrates with good leaving groups and low steric hindrance are more likely to react with nucleophiles like NaOMe.
Therefore, the substrates CH₃CH₂Br, (CH₃)₂CHBr, CH₃CH₂I, and (CH₃)₃CBr are expected to react with NaOMe to form appreciable amounts of product. On the other hand, substrates with poor leaving groups or high steric hindrance are less likely to undergo Sn2 reactions.
Therefore, the substrates (CH₃)₃CNH₂ and CH₃CH₂OH are not expected to react with NaOMe to form appreciable amounts of product. Finally, CH₃CH₂CH₂Br and CH₃CH₂CH₂CH₂Br may react with NaOMe, but to a lesser extent due to their higher steric hindrance.
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Complete question :
Determine which of the substrates will and will not react with NaOMe in an Sy2 reaction to form an appreciable amount of product. Substrate will react Substrate will NOT react Answer Bank CH,CH.CH,BE (CH),CBE (CH), CHRE CH, CH,CH,NH, (CH),CCH,BE CH,CH.CH, OH
A solution contains 1.49×10-2 M potassium chromate and 1.04×10-2 M ammonium phosphate.
Solid barium acetate is added slowly to this mixture.
A. What is the formula of the substance that precipitates first?
formula =______ B. What is the concentration of barium ion when this precipitation first begins?
[Ba2+] =__________ M
the concentration of barium ion when precipitation begins is approximately 3x =
3(7.93 × 10^-9 M) = 2.38 × 10^-8 M.
Hence, the concentration of barium ion when precipitation begins is approximately 2.38 × 10^-8 M.
To determine which substance precipitates first and the concentration of barium ion when precipitation begins, we need to consider the solubility product (Ksp) of the possible precipitation reactions.
The possible precipitation reactions are:
Ba(CrO4)2(s) ⇌ Ba2+(aq) + CrO42-(aq) Ksp1 = [Ba2+][CrO42-]^2
Ba3(PO4)2(s) ⇌ 3Ba2+(aq) + 2PO43-(aq) Ksp2 = [Ba2+]^3[PO43-]^2
The substance that precipitates first is the one with the lower solubility product (Ksp) value. To determine the Ksp values, we need to look up the relevant values of the solubility products.
From the solubility product table, we find:
- Ksp1 for Ba(CrO4)2 is 1.17 × 10^-10
- Ksp2 for Ba3(PO4)2 is 1.34 × 10^-23
Comparing the Ksp values, we see that Ksp1 is much larger than Ksp2, indicating that Ba(CrO4)2 is more soluble than Ba3(PO4)2.
Therefore, the precipitate that forms first is Ba3(PO4)2(s).
To determine the concentration of barium ion when precipitation begins, we can use the Ksp2 expression and assume that x mol/L of Ba3(PO4)2(s) dissolves, forming 3x mol/L of Ba2+ and 2x mol/L of PO43-. Since the initial concentration of ammonium phosphate is 1.04×10^-2 M, which is much less than the initial concentration of potassium chromate (1.49×10^-2 M), we can assume that all of the phosphate ions come from the ammonium phosphate and ignore the small contribution from the autoionization of water.
Using the Ksp2 expression and the concentrations of PO43- and Ba2+, we get:
Ksp2 = [Ba2+]^3[PO43-]^2
1.34 × 10^-23 = (3x)^3(2x)^2
Solving for x, we get:
x = 7.93 × 10^-9 M
Therefore, the concentration of barium ion when precipitation begins is approximately 3x =
3(7.93 × 10^-9 M) = 2.38 × 10^-8 M.
Hence, the concentration of barium ion when precipitation begins is approximately 2.38 × 10^-8 M.
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For photosynthesis to occur, 2801 kJ/mole of energy is required. Add the ΔH to the correct side of the equation below:
6 CO2 (g) + H2O (l) → C6H12O6 (aq) + 6 O2 (g)
The correct presentation is;
6 CO2 (g) + H2O (l) → C6H12O6 (aq) + 6 O2 (g) ΔH = 801 kJ/mole
What is the energy that is required?A chemical reaction known as an endothermic reaction draws energy from its surroundings, causing the temperature of those surroundings to drop. This indicates that energy must be added to the system in order for the reaction to take place because the reactants of the reaction have a lower enthalpy (energy content) than the products.
Because the absorbed energy during an endothermic reaction is typically in the form of heat, the reaction feels cold to the touch.
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Determine the mass of ammonium chloride, NH4Cl, required to prepare 0. 250 L of a 0. 35 M solution of ammonium chloride.
We need 4.68 g of ammonium chloride (NH₄Cl) to prepare 0.250 L of a 0.35 M solution.
To determine the mass of ammonium chloride (NH₄Cl) required to prepare a 0.250 L (liters) of a 0.35 M (molar) solution, follow these steps:
1. Recall the formula for molarity: M = moles of solute / volume of solution in liters.
2. Rearrange the formula to solve for moles of solute: moles of solute = M x volume of solution in liters.
3. Calculate the moles of NH₄Cl needed: moles of NH₄Cl = 0.35 M x 0.250 L = 0.0875 moles.
4. Determine the molar mass of NH₄Cl by adding the molar masses of its constituent elements: (N = 14.01 g/mol, H = 1.01 g/mol, Cl = 35.45 g/mol): 14.01 + (4 x 1.01) + 35.45 = 53.49 g/mol.
5. Calculate the mass of NH₄Cl required: mass = moles x molar mass = 0.0875 moles x 53.49 g/mol = 4.680125 g.
So, you need 4.68 g of ammonium chloride (NH₄Cl) to prepare 0.250 L of a 0.35 M solution.
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1: calculate the ph of a 0.25m solution of h3o+
2: calculate the ph of a 6.3x10-8m solution of h3o+
3: look at your answer for 4 and 5 which one is a base?
4: look at 4 and 5 which one is a strong acid
please show your work
The pH of a 6.3 x [tex]10^{-8[/tex]M solution of H₃O+ is approximately 7.20.
A 0.25 M solution of H₃O+ is not a strong acid, since it is not a single acid that completely dissociates in water.
A 6.3 x [tex]10^{-8[/tex] M solution of H₃O+ is not a strong acid, since it is a very weak acid with a very low concentration of H₃O+ ions.
The pH of a 0.25 M solution of H₃O+ can be calculated using the formula:
pH = -log[H₃O+]
where [H₃O+] is the concentration of H₃O+ ions in moles per liter (M).
In this case, [H3O+] = 0.25 M,
pH = -log(0.25) = 0.602
Therefore, the pH of a 0.25 M solution of H₃O+ is approximately 0.602.
The pH of a 6.3 x [tex]10^{-8[/tex] M solution of H₃O+ can be calculated using the same formula:
pH = -log[H₃O+]
In this case, [H₃O+] = 6.3 x [tex]10^{-8[/tex]M, so we have:
pH = -log(6.3 x [tex]10^{-8[/tex]) = 7.20
Therefore, the pH of a 6.3 x [tex]10^{-8[/tex] M solution of H₃O+ is approximately 7.20.
There is no information given for question 3.
A strong acid is an acid that completely dissociates in water to produce H₃O+ ions. The most common example of a strong acid is hydrochloric acid (HCl).
Looking at the given solutions:
A 0.25 M solution of H₃O+ is not a strong acid, since it is not a single acid that completely dissociates in water.
A 6.3 x [tex]10^{-8[/tex] M solution of H₃O+ is not a strong acid, since it is a very weak acid with a very low concentration of H₃O+ ions.
Therefore, neither of the given solutions is a strong acid.
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16. Lab Analysis: You forgot to label your chemicals and do not know whether your unknown solution is strontium nitrate or magnesium nitrate. You use the solutions potassium carbonate and potassium sulfate in order to determine your mistake. unknown + potassium carbonate & unknown + potassium sulfate . Justify your unknown solution in complete sentences, using your observations and the solubility rules as evidence in your explanation.
Based on the lab analysis, we used potassium carbonate and potassium sulfate to determine whether our unknown solution was strontium nitrate or magnesium nitrate.
When we mixed the unknown solution with potassium carbonate, we observed a white precipitate forming, indicating that the unknown solution contained a carbonate ion. When we mixed the unknown solution with potassium sulfate, we observed no change, indicating that the unknown solution did not contain a sulfate ion.
Using the solubility rules, we know that strontium carbonate is insoluble, while magnesium carbonate is soluble. Therefore, since we observed a white precipitate forming, we can conclude that our unknown solution was strontium nitrate.
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do avalanchers play a large part in shaping the Earth's surface?
Answer:
yes
Explanation:
yes, avalanches a big part in the shaping of the earths surface.
Yes, avalanches can play a significant role in shaping the Earth's surface, particularly in mountainous areas.
The movements of snow, ice, and debris down a slope known as avalanches can significantly impact the Earth's surface, especially in mountainous regions. These natural occurrences can cause various landscape changes, erosion, and deposition.
A 4 L sample of gas at 30 degrees celcius and 1 atm is changed to 0 degrees celcius and 800torr. What is its new volume?
A 4 L sample of gas at 30 degrees celcius and 1 atm is changed to 0 degrees celcius and 800torr. 4.51 L is its new volume.
To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas.
[tex]P1V1/T1 = P2V2/T2[/tex]
where P1, V1, and T1 are the initial conditions, and P2, V2, and T2 are the final conditions.
Substituting the given values, we get:
[tex]\left(\frac{{1 , \text{atm} \cdot 4 , \text{L}}}{{303 , \text{K}}}\right) = \left(\frac{{0.8 , \text{atm} \cdot V2}}{{273 , \text{K}}}\right)[/tex]
Solving for V2, we get:
[tex]V2 = \frac{{1 , \text{atm} \cdot 4 , \text{L} \cdot 273 , \text{K}}}{{303 , \text{K} \cdot 0.8 , \text{atm}}} = 4.51 , \text{L}[/tex]
Therefore, the new volume of the gas is 4.51 L when the temperature is changed from 30 degrees Celsius to 0 degrees Celsius and the pressure is changed from 1 atm to 800 torr.
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Solve the following problems using the chemical formulas as a conversion factor.
1. How many grams of Lead (Pb) contain 1. 25x104 grams of PbCO3?
2. Determine the number of moles of Hydrogen (H) in 0. 0737 mol of N2H4
3. How many grams of Iron (Fe) contain 6. 45x10-3 grams of Fe3O4?
4. Determine the number of moles of Sodium (Na) in 4. 2 mol of NaClO3
There are 0.1474 moles of hydrogen atoms in 0.0737 mol of N2H4.
What is Mole?
In chemistry, a mole is a unit used to express the amount of a substance. One mole of a substance is defined as the amount of that substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in 12 grams of carbon-12.
To determine the mass of lead in PbCO3, we need to use the molar mass of PbCO3 and the stoichiometric relationship between Pb and PbCO3. The molar mass of PbCO3 is 267.21 g/mol, and the stoichiometric relationship between Pb and PbCO3 is 1:1.
Thus, the mass of Pb in 1.25x10^4 g of PbCO3 can be calculated as follows:
Mass of Pb = (1.25x10^4 g PbCO3) x (1 mol PbCO3/267.21 g PbCO3) x (1 mol Pb/1 mol PbCO3) x (207.2 g Pb/mol Pb)
= 1.02x10^4 g Pb
Therefore, 1.02x10^4 g of Pb is contained in 1.25x10^4 g of PbCO3.
The formula for N2H4 indicates that there are two hydrogen atoms for every molecule of N2H4. Therefore, we can calculate the number of moles of hydrogen atoms in 0.0737 mol of N2H4 as follows:
Number of moles of H atoms = (0.0737 mol N2H4) x (2 mol H atoms/1 mol N2H4)
= 0.1474 mol H
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2. Dragonflies can travel at speeds up to 35 miles perhour. How many meters per second is that? (1 mile = 1609 meters)
3. The Hyperion is the tallest redwood tree in the worldat 379. 7 feet. How many centimeters is that? (1 inch = 2. 54 cm)
4. How many atoms are in 2. 35 moles sulfur?
5. How many molecules are in 3. 45 moles sucrose?
Pls Help ASAP!
2. To convert miles per hour to meters per second, we need to divide by 2.237.
Thus, 35 miles per hour is equal to (35/2.237) meters per second.
Simplifying, we get:
= 15.646 m/s
3. To convert feet to centimeters, we need to multiply by 30.48.
Thus, 379.7 feet is equal to (379.7 x 30.48) centimeters.
Simplifying, we get:
= 1158.754 centimeters
4. To calculate the number of atoms in 2.35 moles of sulfur, we need to use Avogadro's number, which is 6.022 x 10^23 atoms per mole.
Therefore, the number of atoms in 2.35 moles of sulfur is:
2.35 moles x 6.022 x 10^23 atoms/mole = 1.41 x 10^24 atoms
5. To calculate the number of molecules in 3.45 moles of sucrose, we need to use Avogadro's number, which is 6.022 x 10^23 molecules per mole.
Therefore, the number of molecules in 3.45 moles of sucrose is:
3.45 moles x 6.022 x 10^23 molecules/mole = 2.08 x 10^24 molecules
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H2 + Br2 → 2HBr. How many liters of hydrogen gas are needed to react with 9.0 g of bromine?
We need 1.26 liters of hydrogen gas to react with 9.0 g of bromine.
To solve this problem, we need to use the balanced chemical equation for the reaction between hydrogen gas (H₂) and bromine (Br₂):
[tex]H_2 + Br_2 - > 2HBr[/tex]
According to the stoichiometry of this equation, one mole of Br₂ reacts with one mole of H₂ to produce two moles of HBr. Therefore, we need to determine the number of moles of Br₂ in 9.0 g, and then use the mole ratio to find the number of moles of H₂ required.
Finally, we can convert the number of moles of H₂ to liters using the ideal gas law.
First, we need to calculate the number of moles of Br₂ in 9.0 g:
The molar mass of Br₂ is 2(79.90 g/mol) = 159.80 g/mol
The number of moles of Br₂ in 9.0 g is:
9.0 g / 159.80 g/mol = 0.0563 mol Br₂
Next, we use the mole ratio from the balanced equation to find the number of moles of H₂ required:
According to the balanced equation, one mole of Br₂ reacts with one mole of H₂, so we need 0.0563 moles of H₂.
Finally, we can use the ideal gas law to convert the number of moles of H₂ to liters:
The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
We can assume standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm.
At STP, one mole of an ideal gas occupies 22.4 L.
Therefore, the volume of H2 required is:
V = (0.0563 mol) x (22.4 L/mol) = 1.26 L
Therefore, we need 1.26 liters of hydrogen gas to react with 9.0 g of bromine.
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A student made the claim that a 4 gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy as a 1 gram bb pellet fired from a bb gun at 180 m/s do you agree or disagree with the student's claim?
I agree with the student's claim that a 4-gram paintball fired from a paintball gun at 90 m/s could have about the same kinetic energy as a 1-gram bb pellet fired from a bb gun at 180 m/s.
To answer this question, we need to compare the kinetic energy of the paintball and the bb pellet. The formula for kinetic energy is 1/2mv^2, where m is the mass of the object and v is its velocity.
For the paintball, with a mass of 4 grams and a velocity of 90 m/s, the kinetic energy is:
1/2 * 0.004 kg * (90 m/s)^2 = 18.18 joules
For the bb pellet, with a mass of 1 gram and a velocity of 180 m/s, the kinetic energy is:
1/2 * 0.001 kg * (180 m/s)^2 = 16.2 joules
So, the student's claim is actually true - the 4-gram paintball fired at 90 m/s has slightly more kinetic energy than the 1-gram bb pellet fired at 180 m/s. However, it's worth noting that the two projectiles have different sizes and shapes, and would behave differently upon impact.
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When ammonium is added to water the temperature of the water decreases. Ammonium nitrates can be recovered by evaporating the water added Which explains those observations A the ammonium nitrates dissolved in water and process is endothermic B the ammonium nitrate reacts with the water and process is endothermic C the ammonium nitrates dissolved in water and process is exothermic D the ammonium nitrate reacts with the water and process is exothermic
Ammonium nitrates can be recovered by evaporating the water added explains that ammonium nitrates dissolved in water and process is endothermic. Thus, option A is correct.
When ammonium is added to water, the temperature of the water decreases. This is because the dissolution of ammonium in water is an endothermic process, meaning it requires energy in the form of heat to take place. When ammonium dissolves in water, it absorbs heat from the surroundings, which causes the temperature of the water to decrease.
Furthermore, ammonium nitrates can be recovered by evaporating the water that was added. This indicates that the ammonium nitrates dissolved in water and the process is endothermic. If the ammonium nitrate had reacted with the water, it would not be possible to recover it by evaporation.
Therefore, option A, "the ammonium nitrates dissolved in water and process is endothermic," is the correct explanation for the observations that when ammonium is added to water, the temperature decreases, and ammonium nitrates can be recovered by evaporating the water added.
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How many grams of steam are produced when 675 grams of oxygen gas combust?
2c8h18 (1) + 2502 (g) --> 16co2 (g) + 18h20 (g) (balanced)
Based on the balanced chemical equation provided, the combustion of 675 grams of oxygen gas (O₂) will produce 275.1 grams of water (H₂O) in the form of steam. Therefore, 275.1 grams of steam are produced when 675 grams of oxygen gas combust.
To determine how many grams of steam are produced when 675 grams of oxygen gas combust, we'll use the balanced equation you provided: 2C₈H₁₈ (l) + 25O₂ (g) --> 16CO₂ (g) + 18H₂O (g).
Step 1: Calculate the molar mass of O₂ and H₂O.
O₂: 16.00 g/mol * 2 = 32.00 g/mol
H₂O: (1.01 g/mol * 2) + 16.00 g/mol = 18.02 g/mol
Step 2: Calculate the moles of oxygen (O₂) in the 675 grams of oxygen gas.
moles of O₂ = 675 g / 32.00 g/mol = 21.09375 mol
Step 3: Use the stoichiometry from the balanced equation to find the moles of H₂O (steam) produced.
(18 mol H₂O / 25 mol O2) * 21.09375 mol O₂ = 15.271125 mol H2O
Step 4: Convert moles of H₂O to grams.
grams of H₂O = 15.271125 mol * 18.02 g/mol = 275.097895 g
So, approximately 275.1 grams of steam are produced when 675 grams of oxygen gas combust.
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A container has 0.182 mol of CO₂ gas at STP. How many liters does the gas take up?
Answer:
4.08 L
Explanation:
At standard temperature and pressure, a mole of any gas equals 22.4 L.
We have 0.182 mol of CO₂ gas. We know that every mole of gas is 22.4 L, so
[tex]0.182mol*\frac{22.4L}{1mol} =4.08L[/tex]
⇒ 4.08 L of CO₂ is the answer
SI Unit: Volume = 4.133 L of carbon dioxide
Non-SI Unit: Volume = 4.079 L carbon dioxide
Molar Volume of Gases:At STP conditions (Standard Temperature and Pressure), which is conditions at 100 kPa and at 0°C or 273.15 K, it is a given that the volume of 1 mole of ideal gas is 22.71 L.
[tex]\large \textsf{$\therefore$ if 1 mol of CO$_2$ = 22.71 L}\\\\\large \textsf{hence, 0.182 $\times$ 1 mol of CO$_2$ = 22.71 $\times$ 0.182}\\\\\large \textsf{$\implies$ \boxed{\boxed{$volume = 4.133 L of CO$_2}}}[/tex]
Note: The value used for pressure above, 100 kPa (kilopascals), is a standard SI unit (International System of Units), used by most countries around the world.
However, another commonly used value for pressure (though not the preferred SI unit), is 1 atm (atmospheric pressure), which is equivalent to 101.325 kPa.
Using this value, the volume of 1 mole of ideal gas at STP is then 22.41 L. Solving this:
[tex]\large \textsf{if 1 mol of CO$_2$ = 22.41 L}\\\\\large \textsf{$\therefore$ 0.182 $\times$ 1 mol of CO$_2$ = 22.41 $\times$ 0.182}\\\\\large \textsf{$\implies$ \boxed{\boxed{$volume = 4.079 L CO$_2}}}[/tex]
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Consider these two entries from a fictional table of standard reduction potentials.
X3+ + 3e—>
X(s)
E° = -2. 43 V
Y3+ + 3e—>
Y(S)
E° = -0. 44 V
What is the standard potential of a galvanic (voltaic) cell where X is the anode and Y is the cathode?
Edell
=
V
The standard potential of the galvanic cell where X is the anode and Y is the cathode is 1.99 V.
The standard potential of a galvanic cell can be calculated by subtracting the reduction potential of the anode (X) from the reduction potential of the cathode (Y).
E°cell = E°cathode - E°anode
In this case, Y has a higher reduction potential than X, so Y will be the cathode and X will be the anode.
E°cell = E°Y - E°X
E°cell = (-0.44 V) - (-2.43 V)
E°cell = 1.99 V
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B) Express the answer to this multistep calculation using the appropriate number of significant figures: 87. 95 feet x 0. 277 feet +5. 02 feet - 1. 348 feet + 10. 0 feet.
The answer to the multistep calculation, expressed using the appropriate number of significant figures, is 24.3 feet.
In order to determine the appropriate number of significant figures in the answer, we need to follow the rules of significant figures for addition and subtraction.
When adding or subtracting numbers, the answer should be rounded to the same number of decimal places as the measurement with the least number of decimal places.
Here, the measurement with the least number of decimal places is 10.0 feet, which has one decimal place. Therefore, we should round the final answer to one decimal place as well.
Now, let's perform the calculation:
87.95 feet x 0.277 feet + 5.02 feet - 1.348 feet + 10.0 feet = 24.3108725 feet
Rounding to one decimal place, the final answer is:
24.3 feet
Therefore, the answer to the multistep calculation, expressed using the appropriate number of significant figures, is 24.3 feet.
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If 450 ml of water are added to 550 ml of a 0.75 m k2so4 solution, what will the molarity of the diluted solution be?
To determine the molarity of the diluted solution, we need to use the equation:
M1V1 = M2V2
where M1 is the initial molarity of the solution, V1 is the initial volume of the solution, M2 is the final molarity of the solution, and V2 is the final volume of the solution.
In this case, the initial solution is a 0.75 M K2SO4 solution with a volume of 550 mL, and water is added to make a final volume of 450 mL. We can write:
M1 = 0.75 M
V1 = 550 mL
V2 = 450 mL
We can solve for M2:
M1V1 = M2V2
0.75 M × 550 mL = M2 × 450 mL
M2 = (0.75 M × 550 mL) / 450 mL
M2 = 0.92 M
Therefore, the molarity of the diluted solution is 0.92 M.
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What concentration of ethylene glycol is needed to raise the boiling point
of water to 105°C? (K⬇️b = 0. 51°C/m)
The concentration of ethylene glycol needed to raise the boiling point of water to 105°C is 9.8 mol/kg or 9.80 molal concentration.
To calculate the concentration of ethylene glycol needed to raise the boiling point of water to 105°C, we can use the following formula:
ΔTb = Kb x molality
Where ΔTb is the change in boiling point, Kb is the boiling point elevation constant for water (0.51°C/m), and molality is the number of moles of solute per kilogram of solvent.
First, we need to calculate the ΔTb, which is the difference between the boiling point of the solution (105°C) and the boiling point of pure water (100°C):
ΔTb = 105°C - 100°C = 5°C
Next, we can plug in the values and solve for the molality:
5°C = 0.51°C/m x molality
Therefore;
molality = 5°C / 0.51°C/m
= 9.8 mol/kg
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When 200. Ml of 2. 0 m naoh(aq) is added to 500. Ml of 1. 0 m hcl(aq), the ph of the resulting mixture is closest to
The pH of the resulting mixture is closest to 2.48, which is in the acidic range.
The reaction between HCl and NaOH produces water and NaCl:
HCl + NaOH → NaCl + H₂O
Moles of HCl = 1.0 mol/L × 0.5 L = 0.5 moles
Moles of NaOH = 2.0 mol/L × 0.2 L = 0.4 moles
NaOH is a limiting factor since it has fewer moles than HCl.
Excess H⁺ ions = 0.5 moles - 0.4 moles = 0.1 moles
Excess OH⁻ ions = 0.4 moles
To calculate the pH of the solution, we need to know the concentration of excess H⁺ or OH⁻ ions. Since we know the amount of excess H⁺ and OH⁻ ions, we can calculate their concentrations using the volume of the solution.
The total volume of the solution is 200 mL + 500 mL = 0.7 L
The concentration of excess H+ ions is:
[H⁺] = 0.1 moles ÷ 0.7 L = 0.143 mol/L
The concentration of excess OH- ions is:
[OH⁻] = 0.4 moles ÷ 0.7 L = 0.571 mol/L
Since the concentration of OH⁻ ions is higher than the concentration of H⁺ ions, the solution is basic. The pH can be calculated using the equation:
pH = 14 - pOH
pOH = -log[OH⁻]
pOH = -log(0.571)
pOH = 0.242
Thus, pH = 14 - 0.242 = 13.76
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During this reaction, water is evaporating from the solution at the same time some of the co2 is dissolving into the water. How might these factors affect the results of the experiment? explain each effect and the overall effect.
The evaporation of water and dissolution of CO2 can affect the results of the experiment in several ways:
Concentration changes: As water evaporates, the concentration of the solute in the remaining solution increases. This can affect the rate of reaction, as the concentration of the reactants is a key factor in determining the rate. Similarly, as CO2 dissolves in the water, the concentration of dissolved CO2 increases, which can affect the pH of the solution.
Mass changes: As water evaporates, the mass of the solution decreases. This can affect the accuracy of the results, as the mass is often used to calculate the amount of product formed.
Temperature changes: Evaporation is an endothermic process, meaning that it requires energy in the form of heat. As a result, the temperature of the solution may decrease during the reaction, which can affect the rate of the reaction.
Overall, the effects of water evaporation and CO2 dissolution will depend on the specific conditions of the experiment, including the starting concentrations of the reactants and the rate of evaporation. In general, these factors can affect the accuracy and precision of the results, and must be carefully controlled or accounted for in order to obtain reliable data.
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A student finds the mass and volume of four mystery liquids. The data is provided
The student's task is to determine the density of the four mystery liquids using the mass and volume measurements.
Density is a physical property that describes the amount of mass per unit volume.
The formula for density is density = mass/volume. Once the density of each liquid is determined, the student can compare it to known densities of different substances to identify the liquid.
This information can be useful in various fields such as chemistry, pharmacology, and environmental science.
The student may also use this data to calculate other properties of the liquids such as viscosity, surface tension, and boiling point. Overall, measuring mass and volume is a fundamental method in scientific research and analysis.
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