Iodine-131 is a radioactive isotope of iodine that has a half-life of 8.0 days. This means that after 8.0 days, half of the original amount of iodine-131 will have decayed, and after another 8.0 days (a total of 16 days), half of the remaining iodine-131 will have decayed again.
The activity of a radioactive sample is a measure of the number of radioactive decays that occur in a given time period. It is measured in units of becquerels (Bq) or curies (Ci). One curie is equal to 3.7 x 10^10 becquerels.
In this case, we are given that the initial activity of the sample is 200 mCi (milliCuries). To find the activity after 16 days, we can use the following equation:
Activity = Initial activity x (1/2)^(t/half-life)
where t is the time elapsed and half-life is the half-life of the isotope.
Substituting the given values, we get:
Activity = 200 mCi x (1/2)^(16/8)
Activity = 200 mCi x (1/2)^2
Activity = 200 mCi x 0.25
Activity = 50 mCi
Therefore, the activity observed after 16 days is 50 mCi. This means that half of the original iodine-131 has decayed in that time period. It is important to note that the actual number of atoms remaining in the sample will also be halved after 16 days, but the activity will be reduced by a factor of four (since activity is proportional to the number of decays per unit time).
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What is the energy of a photon that emits a light of frequency 6. 42 x 1014 Hz?
A. 3. 10 x 10-19 J
B. 4. 25 x 10-19 J
C. 9. 69 x 10-19 J
D. 4. 67 x 10-19 J
The energy of a photon that emits a light of frequency 6. 42 x 1014 Hz is 4.25 x 10^-19 J.
The energy of a photon can be calculated using the equation:
E=hf,
where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J.s), and f is the frequency of the light emitted by the photon.
Plugging in the given frequency of 6.42 x 10^14 Hz into the equation, we get
E=(6.626 x 10^-34 J.s)(6.42 x 10^14 Hz) = 4.25 x 10^-19 J.
Therefore, the correct answer is B i.e, 4.25 x 10^-19 J.
It should be emphasized that a photon's energy is directly linked to its frequency and inversely related to its wavelength. Therefore, light with higher frequency, such as blue light, contains more energy than light with lower frequency, such as red light.
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which of the following options correctly describe the oxidation of primary alcohols? select all that apply. multiple select question. primary alcohols require different oxidizing conditions than secondary alcohols. a carboxylic acid can be produced by oxidation of a primary alcohol. during oxidation, a primary alcohol will rearrange to produce a more substituted oxidation product. mild oxidizing conditions will result in an aldehyde product. harsher oxidizing conditions will produce a ketone from a primary alcohol.
The options that describe the oxidation of the primary alcohols is a carboxylic acid can be produced by oxidation of a primary alcohol. Mild oxidizing conditions will result in an aldehyde product.
The Primary alcohols will be oxidized to form the aldehydes and the carboxylic acids. The secondary alcohols will be oxidized to give the ketones. The Tertiary alcohols, in the contrast, cannot be oxidized by without breaking the molecules of the C–C bonds.
The Primary alcohols and the aldehydes will be normally oxidized to the carboxylic acids using the potassium dichromate solution in the presence of the dilute sulfuric acid that is H₂SO₄.
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A 30g piece of metal absorbs 1,200 joules of heat energy, and its
temperature changes from 25°C to 175°C. Calculate the specific capacity of
the metal. What is the likely metal?
Answer:
Niobium (Columbium)
Explanation:
Specific heat capacity has the units J/(kg °C). To find the heat capacity, all we need to do is organize the values so the units match up.
1200 J / (0.03 kg * 150°C) = 266.67 or 267 J/(kg °C)
The closest metal to a 267 heat capacity is Niobium I believe.
12. What is the weight/volume percent concentration of 100. ML of a 30. 0% (w/v) solution of
vitamin C after diluting to 200. ML?
The weight/volume percent concentration of the diluted solution is 15%.
The initial solution is a 30.0% (w/v) solution, which means that 30.0 grams of vitamin C is dissolved in 100 mL of the solution. Therefore, the amount of vitamin C in the initial solution is:
30.0% (w/v) = 30.0 g / 100 mL = 0.3 g/mL
The initial solution is then diluted to a final volume of 200 mL. Since the amount of vitamin C in the solution remains constant, we can use the following equation to calculate the final concentration:
CiVi = CfVf
where Ci and Vi are the initial concentration and volume, and Cf and Vf are the final concentration and volume.
We can rearrange the equation to solve for the final concentration:
Cf = (CiVi) / Vf
Substituting the values, we get:
Cf = (0.3 g/mL x 100 mL) / 200 mL
Cf = 0.15 g/mL
Finally, we can convert the concentration to weight/volume percent by multiplying by 100:
weight/volume percent = Cf x 100%
weight/volume percent = 0.15 g/mL x 100%
weight/volume percent = 15%
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During the combustion of propane(C3H8), 197. 4 grams of oxygen gas is consumed. How much water vapor is produced as a result?
197.4 grams of oxygen gas is consumed during the combustion of propane. Using stoichiometry, it is calculated that 88.43 grams of water vapor is produced as a result.
The balanced chemical equation for the combustion of propane is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
From the equation, we can see that for every mole of propane (C₃H₈) consumed, 4 moles of water (H₂O) are produced.
To solve the problem, we need to first find the number of moles of oxygen (O₂) consumed:
Moles of O₂ = Mass of O₂ / Molar mass of O₂
Molar mass of O₂ = 32 g/mol (from the periodic table)
Moles of O₂ = 197.4 g / 32 g/mol
Moles of O₂ = 6.16875 mol
Since the balanced chemical equation shows that 5 moles of O₂ are required for every mole of C₃H₈, we can find the number of moles of C₃H₈ consumed:
Moles of C₃H₈ = Moles of O₂ / 5
Moles of C₃H₈ = 6.16875 mol / 5
Moles of C₃H₈ = 1.23375 mol
Now, we can find the number of moles of H₂O produced:
Moles of H₂O = Moles of C₃H₈ x 4
Moles of H₂O = 1.23375 mol x 4
Moles of H₂O = 4.935 mol
Finally, we can find the mass of H₂O produced:
Mass of H₂O = Moles of H₂O x Molar mass of H₂O
Molar mass of H₂O = 18 g/mol (from the periodic table)
Mass of H₂O = 4.935 mol x 18 g/mol
Mass of H₂O = 88.43 g
Therefore, 88.43 grams of water vapor is produced as a result of the combustion of propane with 197.4 grams of oxygen gas.
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1.85 l of a gas is collected over water at 98.0 kpa and 22.0 °c. what is the volume of the dry gas at stp?
In this problem, we are given the volume of a gas collected over water at a certain temperature and pressure. We need to determine the volume of the dry gas at STP (standard temperature and pressure).
First, we need to understand why the presence of water is important in this problem. When a gas is collected over water, some of the water vapor dissolves in the gas, which affects the volume of the gas we measure. In order to account for this, we need to use the concept of vapor pressure.
The vapor pressure of water at 22.0°C is 2.64 kPa. This means that at 22.0°C and 98.0 kPa, the total pressure is the sum of the pressure due to the gas and the pressure due to the water vapor. We can use Dalton's Law of Partial Pressures to calculate the pressure due to the gas alone:
P_gas = P_total - P_water vapor
P_gas = 98.0 kPa - 2.64 kPa
P_gas = 95.36 kPa
Now we can use the Ideal Gas Law to calculate the volume of the dry gas at STP:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. At STP, P = 101.3 kPa and T = 273.15 K.
We can rearrange the Ideal Gas Law to solve for the volume of the dry gas:
V_dry gas = (V_collected gas * P_gas * T_STP) / (P_STP * T_collected gas)
where V_collected gas is the volume of the gas collected over water, T_collected gas is the temperature of the gas collected over water, and T_STP is the temperature at STP.
Plugging in the numbers, we get:
V_dry gas = (1.85 L * 95.36 kPa * 273.15 K) / (101.3 kPa * 295.15 K)
V_dry gas = 1.60 L
Therefore, the volume of the dry gas at STP is 1.60 L. It's important to note that the volume of the dry gas is smaller than the volume of the gas collected over water, because some of the volume was occupied by water vapor.
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h2o is a molecular compound that is a liquid at room temperature (22 degrees celsius). this is primarily due to the fact that it has relatively what strength of intermolecular forces?
H2O, or water, is a molecular compound that is a liquid at room temperature (22 degrees Celsius). This state is primarily due to the fact that it has relatively strong intermolecular forces.
These forces are the attractive forces between the molecules of the compound, and in the case of water, these forces are called hydrogen bonds.
Hydrogen bonds are a type of dipole-dipole interaction that occurs between molecules containing a hydrogen atom bonded to a highly electronegative element, such as oxygen in water. The oxygen atom attracts the electrons in the bond, creating a partial negative charge on the oxygen and a partial positive charge on the hydrogen.
This causes an electrostatic attraction between the partially positive hydrogen atom and the partially negative oxygen atom of a neighboring water molecule.
These hydrogen bonds give water its unique properties, such as its relatively high boiling and melting points compared to other molecular compounds with similar molecular weights.
The strong intermolecular forces provided by hydrogen bonding are what make water a liquid at room temperature, as they are strong enough to hold the molecules together, but not so strong that they form a solid at this temperature.
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Assuming pressure is constant. There are 12. 75 mL of chemical product associated with a temperature reading of 68 degrees Celsius. What will the final temperature be if the volume increased to 5. 25 mL
The final temperature will be approximately -55.6 degrees Celsius when the volume is reduced to 5.25 mL.
According to Charles' Law, when pressure is constant, the volume of a gas is directly proportional to its temperature (in Kelvin). The formula for Charles' Law is V1/T1 = V2/T2.
First, convert the initial temperature from Celsius to Kelvin (68 + 273.15 = 341.15 K). Then, plug in the values: (12.75 mL / 341.15 K) = (5.25 mL / T2).
To solve for T2, multiply both sides by T2 and divide by 5.25 mL: T2 = (341.15 K * 5.25 mL) / 12.75 mL ≈ 139.6 K. Finally, convert back to Celsius: 139.6 K - 273.15 ≈ -55.6 degrees Celsius.
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Use the VSEPR Theory to predict the molecular geometry of the following molecules:
(Remember, you may need to draw the lewis structure before making a prediction. )
HI
CBr4
CH2Cl2
SF2
PCl3
To predict the molecular geometry of these molecules using the VSEPR theory, we first need to draw the Lewis structure for each molecule:
1. HI
Lewis structure: H-I (single bond)
The central atom (Iodine) has 7 valence electrons, and each hydrogen atom contributes 1 valence electron. Therefore, the total valence electrons in the molecule is 9.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 0 + 1 = 1
Molecular geometry: linear
2. CBr4
Lewis structure:
:Br-C-Br:
: | :
:Br-C-Br:
The central atom (Carbon) has 4 valence electrons, and each Bromine atom contributes 7 valence electrons. Therefore, the total valence electrons in the molecule is 32.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 0 + 4 = 4
Molecular geometry: tetrahedral
3. CH2Cl2
Lewis structure:
H : Cl
| :
H-C-H
| :
Cl:
The central atom (Carbon) has 4 valence electrons, each hydrogen atom contributes 1 valence electron, and each chlorine atom contributes 7 valence electrons. Therefore, the total valence electrons in the molecule is 20.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 2 + 4 = 6
Molecular geometry: octahedral
However, the two lone pairs of electrons on the central atom will repel the bonded pairs more than the bonded pairs will repel each other. Therefore, the shape will be bent or V-shaped.
4. SF2
Lewis structure:
F : S : F
\ /
F
The central atom (Sulfur) has 6 valence electrons, each Fluorine atom contributes 7 valence electrons. Therefore, the total valence electrons in the molecule is 20.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 1 + 2 = 3
Molecular geometry: trigonal planar
However, the lone pair of electrons on the central atom will repel the bonded pairs more than the bonded pairs will repel each other. Therefore, the shape will be bent or V-shaped.
5. PCl3
Lewis structure:
Cl : P : Cl
:
Cl
The central atom (Phosphorus) has 5 valence electrons, each Chlorine atom contributes 7 valence electrons. Therefore, the total valence electrons in the molecule is 26.
Steric number = number of lone pairs of electrons + number of atoms bonded to central atom = 0 + 3 = 3
Molecular geometry: trigonal planar
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A 0.205g sample of caco3 is added to a flask with 7.50ml of 2.00 m hcl.
caco3(aq)+2hcl(aq)-cacl2(aq) + h2o(l) + co2
enough water is added to make a 125.0ml solution.a 10.00ml aliquot of this solution is taken and titred with 0.058 naoh
naoh (aq) + hcl - h2o + nacl
how many ml of naoh are used?
The volume of [tex]NaOH[/tex] used to titrate the[tex]HCl[/tex] is 5.80 mL
First, we need to find the number of moles of [tex]HCl[/tex] that reacted with the [tex]CaCO3[/tex].
2 mol [tex]HCl[/tex] react with 1 mol [tex]CaCO3[/tex]
Moles of [tex]HCl[/tex] = (7.50 mL) x (2.00 mol/L) = 0.015 mol [tex]HCl[/tex]
From the balanced equation, we see that 1 mol of [tex]CaCO3[/tex] reacts with 2 mol of [tex]HCl[/tex]. Therefore, the number of moles of [tex]CaCO3[/tex] in the original 0.205 g sample is:
Moles of[tex]CaCO3[/tex] = 0.205 g / 100.09 g/mol = 0.002049 mol [tex]CaCO3[/tex]
Since 1 mol of [tex]CaCO3[/tex] produces 1 mol of [tex]CO2[/tex], we have:
Moles of[tex]CO2[/tex]produced = 0.002049 mol [tex]CaCO3[/tex]
Now we need to calculate the concentration of [tex]CO2[/tex] in the final 125.0 mL solution.
Concentration of [tex]CO2[/tex] = Moles of [tex]CO2[/tex] produced / Volume of solution
Concentration of [tex]CO2[/tex] = 0.002049 mol / 0.125 L = 0.0164 mol/L
Finally, we can use the balanced equation for the titration reaction to calculate the number of moles of [tex]NaOH[/tex]used.
1 mol [tex]NaOH[/tex] reacts with 1 mol [tex]HCl[/tex]
Moles of [tex]NaOH[/tex] used = (0.058 L) x (0.1000 mol/L) = 0.0058 mol [tex]NaOH[/tex]
Since the volume of the aliquot is 10.00 mL or 0.0100 L, the concentration of [tex]HCl[/tex] is:
Concentration of [tex]HCl[/tex] = Moles of NaOH used / Volume of [tex]HCl[/tex]
Concentration of [tex]HCl[/tex] = 0.0058 mol / 0.0100 L = 0.580 M
Therefore, the volume of [tex]NaOH[/tex] used to titrate the [tex]HCl[/tex]is:
Volume of [tex]NaOH[/tex] = (0.580 M) x (0.0100 L) = 0.00580 L or 5.80 mL
So, the answer is 5.80 mL.
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2. A Snickers bar is burned in a bomb calorimeter containing 3500 grams of water causing a
72°C temperature change. How many joules are there in a bar?
The Snickers bar released 1,077,280 joules of energy when burned.
To calculate the energy released by burning a Snickers bar, we can use the formula:
q = mcΔT
where q is the heat energy released, m is the mass of water, c is the specific heat of water, and ΔT is the temperature change.
We know the mass of water is 3500 g, and the temperature change is 72°C. The specific heat of water is 4.184 J/g°C.
Therefore:
q = (3500 g) x (4.184 J/g°C) x (72°C) = 1077280 J
So, the Snickers bar released 1,077,280 joules of energy when burned.
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A solution is 5 mM in each of the following ions:
number ion Ksp of M(OH)2
1 Mg2+ 1. 8e-11
2 Cd2+ 2. 5e-14
3 Co2+ 1. 6e-15
4 Zn2+ 4. 4e-17
5 Cu2+ 2. 2e-20
Indicate which of the metal ions would precipitate (or start to precipitate) at each of the following pH values. Indicate your answer with the number of the ion. Use 0 to indicate no precipitate. If more than one precipitate is expected, list the numbers in increasing order and separate them with commas. For example, 3,4,5 is ok but 5,4,3 is not.
pH = 6. 00: _______________? (1,2,3,4,5 list all that apply?)
pH = 8. 00: __________? (1,2,3,4,5 list all that apply?)
What is the pH to the nearest 0. 1 pH unit at which Cu(OH)2 begins to precipitate? pH = ______?
pH = 6.00: 0, 1, 2, 3, 4, 5 will not precipitate.
pH = 8.00: 0, 1, 2, 3, 4, 5 will not precipitate.
To determine the pH at which Cu(OH)₂ begins to precipitate, we need to calculate the hydroxide ion concentration at which the product of [Cu²⁺] and [OH⁻]² reaches the Ksp value of Cu(OH)₂ (2.2e⁻²⁰). At this point, Cu(OH)₂ will begin to precipitate. Thus, we have:
Ksp = [Cu²⁺][OH⁻]²2.2e⁻²⁰ = (5e⁻³ M)[OH⁻]²[OH⁻]² = 4.4e⁻¹⁷[OH⁻] = 2.1e⁻⁸ MpOH = -log[OH⁻] = -log(2.1e⁻⁸) = 7.68pH = 14 - pOH = 6.32 (rounded to the nearest 0.1 pH unit)Therefore, Cu(OH)₂ begins to precipitate at a pH of 6.3.
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Which of the following is equal to 2?
O A. 6+4 ÷ (2+1) × 3
O B. (6+4 ÷ 2) - 1×3
O
C. 6+ (4÷ 2) + 1 × 3
O D. (6 + 4)÷2-1×3
O D. (6 + 4)÷2-1×3
the cacuclator gives u the answer to this
A 58. 3g sample of NH3 is reacted with 126g O2, according to this reaction what is the limiting reagent? 4NH3 + 7O2 --> 4NO + 6H2O
The ratio of NH₃ to O₂ is less than 4:7, it means that NH₃ is the limiting reagent. Therefore, NH₃ will be completely consumed before O₂ and the amount of product formed will be determined by the amount of NH₃ available.
To determine the limiting reagent, we need to compare the amount of each reactant with their respective stoichiometric coefficients in the balanced chemical equation.
First, we convert the given masses of NH₃ and O₂ to moles using their molar masses:
58.3 g NH₃ × (1 mol NH₃ ÷ 17.03 g NH₃) = 3.42 mol NH₃
126 g O₂ × (1 mol O₂ ÷ 32 g O₂) = 3.94 mol O₂
Next, we compare the number of moles of NH₃ and O₂ to the stoichiometric coefficients in the balanced equation:
NH₃ : O₂ ratio = 4:7
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Which one? Please help I don't understand
Based on the rate law, the equivalent expression to d[NO₂]/dt is -2k[O₃][NO₂]; option B.
What is the rate law of a chemical reaction?A rate law gives a mathematical explanation of how variations in a substance's amount affect the rate of a chemical reaction.
To determine the equivalent expression to d[NO₂]/dt, differentiate the rate law with respect to [NO₂].
d/dt[k[O₃][NO₂]] = k[d[O₃]/dt][NO₂] + k[O₃][d[NO₂]/dt]
We assume d[O₃]/dt is a constant = k1 (since it is not given in the rate law)
The coefficient for NO₂ is -2,
Substituting in the equation above:
d[NO₂]/dt = (-2k/k1)[O₃][NO₂]
d[NO₂]/dt = -2k[O₃][NO₂]/k1
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4. 3 moles of a gas are at a temperature of 28°C with a pressure of 1. 631 atm. What volume does the gas occupy?
The gas occupies a volume of approximately 28.18 liters at a temperature of 28°C and a pressure of 1.631 atm
To determine the volume the gas occupies at a temperature of 28°C and a pressure of 1.631 atm, we will use the Ideal Gas Law, which is defined as PV = nRT. In this equation, P represents pressure, V represents volume, n represents the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin: T(K) = T(°C) + 273.15. In this case, T(K) = 28 + 273.15 = 301.15 K.
Now, we can use the Ideal Gas Law to find the volume of the gas. The ideal gas constant (R) is 0.0821 L atm/mol K. Therefore, we have:
1.631 atm (V) = 3 moles (0.0821 L atm/mol K) (301.15 K)
To find the volume (V), we can rearrange the equation and isolate V:
V = (3 moles * 0.0821 L atm/mol K * 301.15 K) / 1.631 atm
V = 45.98271 L/mol / 1.631 atm
V ≈ 28.18 L
So, the gas occupies a volume of approximately 28.18 liters at a temperature of 28°C and a pressure of 1.631 atm.
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Part B
One of the main components of an airbag is the gas that fills it. As part of the design process, you need to determine the exact amount of nitrogen that should be produced. Calculate the number of moles of nitrogen required to fill the airbag. Show your work. Assume that the nitrogen produced by the chemical reaction is at a temperature of 495°C and that nitrogen gas behaves like an ideal gas. Use this fact sheet to review the ideal gas law.
Part C
Recall the balanced chemical equation from part B of task 1:
2NaN3 → 2Na + 3N2.
Calculate the mass of sodium azide required to decompose and produce the number of moles of nitrogen you calculated in part B of this task. Refer to the periodic table to get the atomic weights
To calculate the number of moles of nitrogen required to fill the airbag, we need to use the ideal gas law.
We know the temperature of the nitrogen gas produced by the chemical reaction, which is 495°C, and we assume that it behaves like an ideal gas.
We also know the volume of the airbag, which we can use to calculate the number of moles of nitrogen using the ideal gas law equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
Once we have calculated the number of moles of nitrogen required, we can move on to part C of the question, which asks us to calculate the mass of sodium azide required to produce that amount of nitrogen.
To do this, we need to refer to the balanced chemical equation given in part B and use the atomic weights from the periodic table to calculate the mass of sodium azide needed.
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How is entropy related to the spontaneity of a reaction?
O A. AS<0 contributes to spontaneity.
O B. AS does not affect spontaneity.
C. AS = 0 contributes to spontaneity.
D. AS> O contributes to spontaneity.
ΔS> O contributes to spontaneity. This is the relationship between entropy and spontaneity. Therefore, the correct option is option D.
Entropy is a measureable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty. From classical thermodynamics, where it was originally recognised, through the microscopic description of nature in statistical physics, to the fundamentals of information theory, the phrase and concept are employed in a variety of disciplines. It has numerous applications in physics and chemistry, biological systems and how they relate to life, cosmology, economics, sociology, weather science, and information systems, especially the exchange of information. ΔS> O contributes to spontaneity.
Therefore, the correct option is option D.
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Can someone answer the questions in the image?.
“Balancing equations”
Ans.1
blank 1 =1
blank 2 = 3
blank 3 = 2
Ans.2
blank 1 = 6
blank 2 = 4
blank 3 = 5
Ans.
blank 1 = 11
blank 2 = 7
blank 3 = 8
15 moles of NaOH are dissolved in 2. 0 L of solution. What is the molarity of the solution?
The molarity of a solution is defined as the number of moles of solute dissolved in one liter of solution. To calculate the molarity of the NaOH solution, we need to divide the number of moles of NaOH by the volume of the solution in liters.
Given that 15 moles of NaOH are dissolved in 2.0 L of solution, the molarity (M) of the solution can be calculated as:
M = number of moles of solute / volume of solution in liters
M = 15 moles / 2.0 L
M = 7.5 M
Therefore, the molarity of the NaOH solution is 7.5 M.
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Calculate the volume of 3. 00 M H2SO4 required to prepare 200. ML of 0. 200 N H2SO4. (Assume the acid is to be completely neutralized. )
Approximately 13.3 mL of 3.00 M H₂SO₄ is required to prepare 200. mL of 0.200 N H₂SO₄.
To calculate the volume of 3.00 M H₂SO₄ required to prepare 200. mL of 0.200 N H₂SO₄, we can use the formula for molarity:
Molarity (M) = moles of solute / volume of solution in liters
We can rearrange this formula to solve for volume:
Volume (in liters) = moles of solute / molarity
First, let's calculate the moles of H₂SO₄ in 200. mL of 0.200 N solution:
0.200 N = 0.200 mol/L
Moles of H₂SO₄ = 0.200 mol/L x 0.200 L = 0.0400 mol
Next, we can use this value and the concentration of the 3.00 M H₂SO₄ to calculate the volume of the concentrated acid needed:
Volume = moles of solute / molarity
Volume = 0.0400 mol / 3.00 mol/L
Volume = 0.0133 L or 13.3 mL
So, to make 200 mL of 0.200 N H₂SO₄ , roughly 13.3 mL of 3.00 M H₂SO₄ is required.
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How many grams of iron are produced from 300. moles of carbon monoxide reacting with 15,000. grams of ferric oxide? 3CO + Fe2O3 →2Fe + 3C02
11,169 grams of iron is produced from 300 moles of carbon monoxide reacting with 15,000 grams of ferric oxide.
The balanced chemical equation shows that 3 moles of CO react with 1 mole of [tex]Fe_2O_3[/tex] to produce 2 moles of Fe. Therefore, we can calculate the number of moles of Fe produced from 300 moles of CO reacting with [tex]Fe_2O_3[/tex] as follows:
1 mole [tex]Fe_2O_3[/tex] produces 2 moles Fe
300 moles CO produces (2/3) x 300 = 200 moles Fe (by stoichiometry)
Next, we can use the molar mass of Fe to convert moles to grams:
1 mole Fe = 55.845 g Fe
200 moles Fe = 200 x 55.845 = 11,169 g Fe
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Question 3 & what is the hydronium concentration for a solution with a poh = 12.04 o -1.08 m o.98 m 0.011 m p o 1.96 m question 4 a solution is made by combining 2.5 moles of hf (ka 3,5 x 19 and 3.5 mol click save and submit to save and submit chick save asters to small ans
For question 3, we can use the relationship pH + pOH = 14 to solve for the pH, which is 1.96.
Then, we can use the equation Kw = [[H₃O⁺][OH⁻] = 1.0 x 10⁻¹⁴ to solve for the hydronium concentration, which is 5.01 x 10⁻¹³ M.
For question 4, we can use the equation for the acid dissociation constant (Ka) to solve for the concentration of the conjugate base, F-. Ka = [H₃O⁺][F⁻]/[HF].
We know the concentration of HF is 2.5 moles, so we can convert this to molarity using the volume of the solution. Then, we can plug in the values we have and solve for [F-], which is 2.77 M. This solution will be acidic, as the Ka value is less than 1.
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During a period of discharge of a lead-acid battery, 378 grams of Pb from the anode is converted into PbSO (s). What mass of PbO,(s) in grams is reduced at the cathode during this same period?
During the discharge of a lead-acid battery, the oxidation reaction occurs at the anode where lead (Pb) is converted into lead sulfate (PbSO4) and electrons are released:
Pb(s) → PbSO4(s) + 2e-
Meanwhile, reduction occurs at the cathode where lead dioxide (PbO2) is reduced to lead sulfate (PbSO4) by gaining those electrons released at the anode:
PbO2(s) + 4H+(aq) + 2e- → PbSO4(s) + 2H2O(l)
The balanced chemical equation shows that for every two electrons transferred at the anode, one molecule of PbSO4 is formed. Therefore, the 378 grams of Pb from the anode would produce 378/207 = 1.82 moles of PbSO4.
Since the reaction at the cathode involves the reduction of PbO2 to PbSO4, the same number of moles of PbSO4 should be formed at the cathode. The molar mass of PbO2 is 239.2 g/mol, so the mass of PbO2 that is reduced at the cathode would be:
1.82 moles x 239.2 g/mol = 435.8 g
Therefore, during the same period of discharge, 435.8 grams of PbO2 would be reduced at the cathode.
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A solution contains 55 grams of potassium iodide, KI, dissolved in 100 grams of water at 15 °C. How many more grams of KI would have to be added to make it a saturated solution?
In order to respond to this query, it is necessary to first define a saturated solution. When a solution reaches its maximal solubility, no more solute can dissolve in the solvent, creating a saturated solution.
Potassium iodide is the solute in this scenario, while water is the solvent. Potassium iodide is most soluble in water at a temperature of around 74.2 grammes per 100 grammes of water.
We must thus add 19.2 additional grammes of KI to the solution in order to make it saturated. This implies that there would be 74.2 grammes of KI in the entire solution.
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What is the ability to do work or produce heat?
Answer: Energy
Explanation:
Energy is the ability to do work or produce heat.
Find the balance and net ionic equation for the statements below. Answer what you can.
1. Calcium + bromine —>
2. Aqueous nitric acid, HNO3, is mixed with aqueous barium chloride
3. Heptane, C7H16, reacts with oxygen
4. Chlorine gas reacts is bubbles through aqueous potassium iodide (write both the balanced and net ionic equation)
5. Zn (s) + Ca (NO3)2 (aq) —>
6. Aqueous sodium phosphate mixes with aqueous magnesium nitrate (write both the balanced and net ionic equation)
7. Aluminum metal is placed in aqueous zinc chloride
8. Iron (III) oxide breaks down
9. Li(OH) (ag) + HCI (aq) —>
(write both the balanced and net ionic equation)
10A. Solid sodium in water. Hint: Think water, H2O, as H(OH)
10B. What would happen if you bring a burning splint to the previous reaction?
A- The burning splint continues to burn.
B - The burning splint would make a "pop" sound.
C - The burning splint would go out.
Ca +Br2 ---> CaBr2
2HNO3 + BaCl2 --->Ba(NO3)2 +2HCl
C7H16 + 11O2 → 7CO2 + 8H2O
Cl2 + 2KI --->2KCl + I2
No reaction
2Na3PO4 + 3Mg(NO3)2 → Mg3(PO4)2 + 6NaNO3
2Al + 3ZnCl2 → 3Zn + 2AlCl3
Li(OH) (ag) + HCI (aq) —>LiCl + H2O
2Na + 2H2O → 2NaOH + H2
The burning splint would make a "pop" sound.
What is the balanced equation?A balanced equation is a chemical equation that has an equal number of atoms of each element on both the reactant and product sides.
In other words, a balanced equation follows the law of conservation of mass, which states that the total mass of the reactants must equal the total mass of the products in a chemical reaction.
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What are two results of the uneven heating of Earth's surface?
A. Ocean currents
B. Earth's axis tilt
c. Global winds
D. Coriolis effect
SUBMIT
Pls tell me the answe
According to the question the two results of the uneven heating of Earth's surface are ocean currents and global winds.
What is currents?Currents are electrical energy that flows through a circuit. They are typically measured in amperes (amps), and they result from the flow of electrons through the circuit. Currents can be either direct or alternating, and they are used to power many electrical appliances and power systems. Direct currents are generated from sources such as batteries, while alternating currents are produced by generators and power plants. Currents can also be generated artificially, using devices such as transformers, or naturally, through processes such as lightning. The magnitude of currents depends on the voltage and resistance of the circuit. Currents can be used to control the operation of many electrical circuits and components, such as motors, relays, and switches. They can also be used to provide power for many electrical devices, including lights, computers, and other electronic equipment.
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consider 5 sequential reactions where the product of each reaction is the reactant of the next and the 5 percent yields are 80%, 90%, 65%, 76% and 30%. if you begin with 100 molecules of the first limiting reagent, what is the maximum number of product molecules you can form at the end of the final reaction? \textbf{hint:} remember that you cannot have parts of a molecule!
Starting with 100 molecules of the first limiting reagent, the maximum number of product molecules that can be formed at the end of the final reaction, given the yields of each reaction, is 11 molecules.
Let's call the starting number of molecules of the first limiting reagent "A". Then, the number of molecules of each reactant and product after each reaction can be represented as follows,
Reaction 1: A → B (80% yield)
Starting molecules of A = 100
Molecules of B produced = 80
Reaction 2: B → C (90% yield)
Starting molecules of B = 80
Molecules of C produced = 72
Reaction 3: C → D (65% yield)
Starting molecules of C = 72
Molecules of D produced = 46.8 (rounded to 47)
Reaction 4: D → E (76% yield)
Starting molecules of D = 47
Molecules of E produced = 35.72 (rounded to 36)
Reaction 5: E → F (30% yield)
Starting molecules of E = 36
Molecules of F produced = 10.8 (rounded to 11)
Therefore, the maximum number of product molecules that can be formed at the end of the final reaction is 11, rounded to the nearest whole number.
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At what condition do magnesium chloride and silver nitrate react?
Magnesium chloride and silver nitrate react in aqueous solution when they come into contact with each other. In other words, they need to be dissolved in water for the reaction to occur. This is because both compounds are ionic and require a medium for their ions to interact and exchange. Therefore, the condition for the reaction between magnesium chloride and silver nitrate is an aqueous solution.
Magnesium chloride (MgCl₂) and silver nitrate (AgNO₃) react in an aqueous solution. The condition required for the reaction to occur is that both substances are dissolved in water. When this condition is met, a double displacement reaction takes place, leading to the formation of silver chloride (AgCl) precipitate and magnesium nitrate (Mg(NO₃)₂) in the solution. The reaction can be represented by the following balanced equation:
MgCl₂ (aq) + 2AgNO₃ (aq) → 2AgCl (s) + Mg(NO₃)₂ (aq)
1. Dissolve magnesium chloride (MgCl₂) and silver nitrate (AgNO₃) in water to create aqueous solutions.
2. Mix the two aqueous solutions together.
3. Observe the formation of silver chloride (AgCl) precipitate and magnesium nitrate (Mg(NO₃)₂) in solution as a result of the double displacement reaction.
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