The net ionic equation of the reaction is as follows:
4 Al3+(aq) + 12 NO3-(aq) + 3 Ag(s) = 4 Al(s) + 12 Ag+(aq) + 12 NO3-(aq)
Ions which remain in their ground state and do not take part in the reaction are called spectator ions. The net ionic equation cancels out these ions, which are present on both the reactant and product sides of the equation.
Spectator ions, which can be found on both the reactant and product sides, but are not included in the finished reaction from the net ionic equation. The [tex]NO^3^-[/tex] ions are spectator ions in this example, thus taking them out of the equation. The net ionic equation makes up the rest of the species.
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diffrences in water temperature in the ocean create movement because-
Diffrences in water temperature in the ocean create movement because bodies of water at different temperatures have different densities.
How can the differences be explained?Water that is colder is generally denser than water that is warmer, so when a body of water with colder, denser water is next to a body of water with warmer, less dense water, a density gradient is established. This gradient creates a difference in pressure between the two bodies of water, with the colder, denser water being at a higher pressure than the warmer, less dense water.
This difference in pressure creates a force that drives the movement of water from the denser, colder region to the less dense, warmer region. This movement of water is known as convection, and it can occur both vertically and horizontally in the ocean. Vertical convection occurs when differences in temperature cause water to rise or sink, while horizontal convection occurs when water moves laterally due to differences in temperature.
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missing options:
1. as water heats up, the atoms of water more faster.
2. warm water is pulled more by gravity than cold water.
3. warm and cold water mix and reach the same temperature.
4. bodies of water at different temperatures have different densities.
7 Suppose you weighed a different sample, of 2.500-g, which consisted of a mixture of CuO and potassium chloride and dissolved it in 25.00 mL of 0.437 M H₂SO4 solution. Some acid remains after treatment of the sample. Determine: a) If 35.4-mL of 0.108 M NaOH were required to titrate the excess sulfuric acid, how (6) many moles of CuO were present in the original sample?
The initial sample had 0.010925 mol of Copper(II) oxide, or one mole.
What exactly is kinetic-molecular theory?The kinetic-molecular theory, which describes the states of matter, is based on the presumption that matter is composed of minuscule particles that are constantly in motion. This theory explains the observable properties and behaviours of solids, liquids, and gases. The container's walls and the quickly moving particles' collisions with one another are constant.
Copper(II) oxide + Sulfuric acid → Cupric sulfate + Water
One mole of Copper(II) oxide interacts with one mole of Sulfuric acid, as shown by the equation. The amount of Sulfuric acid that reacted with the Copper(II) oxide in the sample is therefore equal to the amount of Copper(II) oxide in the sample.
We must first determine how many moles of Sulfuric acid interacted with the sample:
moles Sulfuric acid = concentration × volume
moles Sulfuric acid = 0.437 mol/L × 0.025 L
moles Sulfuric acid = 0.010925 mol
Since the acid is in excess, the moles of Sulfuric acid remaining after treatment of the sample is:
moles Sulfuric acid remaining = moles Sulfuric acid added – moles Sulfuric acid reacted
moles Sulfuric acid remaining = 0.437 mol/L × 0.0354 L – 0.010925 mol
moles Sulfuric acid remaining = 0.007571 mol
To determine the number of moles of Copper(II) oxide in the original sample, we can use the following equation:
moles Copper(II) oxide = moles Sulfuric acid reacted = 0.010925 mol
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need help with this problem
Answer:
Na < Al < Mg < S < Cl
Explanation:
Sodium has the smallest ionization energy because it wants to lose an electron as an alkali metal.
Aluminum has the second smallest because losing an electron would leave it with just a full s orbital.
Magnesium has the third smallest because although it's removing an electron from a full s orbital, it has less protons than sulfur and chlorine to keep the electron in the shell.
Sulfur has the second largest because it has more protons to pull at the electrons.
Chlorine has the largest ionization energy because it really wants an electron to fill the p orbital. Due to its number of protons, the element is also very small and it will be difficult to remove an electron.
If an ideal gas has a pressure of 1.71 atm, a temperature of 68.16 ∘C, and a volume of 12.85 L how many moles of gas are in the sample?
Answer:
0.745 moles
Explanation:
We can use the ideal gas law, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas:
P V = n R T
where R is the gas constant.
We can rearrange this equation to solve for n:
n = (P V) / (R T)
We can look up the value of the gas constant for units of atm L / (mol K). The value is approximately 0.08206 (atm L) / (mol K).
Substituting the given values, we get:
n = (1.71 atm) * (12.85 L) / (0.08206 (atm L) / (mol K) * (68.16 + 273.15) K)
where we have converted the temperature from Celsius to Kelvin by adding 273.15.
Evaluating this expression gives us:
n ≈ 0.745 mol
Therefore, there are approximately 0.745 moles of gas in the sample.
3. In a lab, students mixed HCI acid with a Mg strip. The Mg started to bubble and dissolved within a few seconds. The rate at which the reaction occurs is determined by the A. number of effective collisions B. large AH C. the stabilization of the reactants D. mass of the products after the reaction
Answer:It might exposed
Explanation: or a spayed H2O might change because different water change over time
A 10 g piece of metal at 50°C absorbs 900 G of energy after which the temperature of the metal is 350°C what is the specific heat of the metal
A 10 g piece of metal at 50°C absorbs 900 G of energy after which the temperature of the metal is 350°C. 0.35J/g°C is the specific heat of the metal.
The amount of heat needed to raise a substance's temperature by one degree Celsius in one gramme, also known as specific heat. Typically, calories and joules per gramme per degree Celsius are used as the measurement units of specific heat.
For instance, water has a specific heat of 1 calorie per gramme per degree Celsius. The notion of specific heat was developed by the Scottish scientist Joseph Black in the 18th century as a result of his discovery that equal masses of different substances required varying quantities.
q = m×c×ΔT
900= 10×c×( 350-50)
c=0.35J/g°C
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Propane, C3H8 (approximate molar mass = 44 g/mol) is used in gas barbeques and burns according to the thermochemical equation: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH = –2046 kJ. If it takes 1.7 x 103 kJ to fully cook a pork roast on a gas barbeque, how many grams of propane will be required, assuming all the heat from the combustion reaction is absorbed by the pork?
The mass (in grams) of propane that will be required, assuming all the heat from the combustion reaction is absorbed by the pork is 36.56 grams
How do i determine the mass propane required?The mass of propane that will be required can be obtain as illustrated below:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g) ΔH = –2046 KJ
Molar mass of C₃H₈ = 44 g/molMass of C₃H₈ from the balanced equation = 1 × 44 = 44 gFrom the balanced equation above,
2046 KJ of heat energy required 44 g of propane, C₃H₈
Therefore,
1.7×10³ KJ of heat energy will require = (1.7×10³ KJ × 44 g) / 2046 KJ = 36.56 g of propane, C₃H₈
Thus, we can conclude that the mass of propane, C₃H₈ required is 36.56 grams
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The question is in the picture
Charles's law of gases states that the density of an ideal gas is inversely proportional to its temperature at constant pressure.
The equation is as follows;
Va/Ta = Vb/Tb
Where;
Va and Ta = initial volume and temperature respectivelyVb and Tb = final volume and temperature respectively0.67/362 = 1.12/Tb
0.00185Tb = 1.12
Tb = 605.41K
This temperature in °C is 605.41 - 273 = 332°C
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What is the S-P difference (sec)?
What is the amplitude (mm)?
What is the distance (km)?
What is the magnitude (M)?
The S-P difference (sec) is used to calculate the distance (km) between an earthquake epicenter and a seismic station, while the magnitude (M) is a measure of the energy released during the earthquake.
These parameters are important for understanding the severity and impact of an earthquake, as well as for predicting future seismic activity.
The S-P difference (sec) refers to the time difference between the arrival of the primary (P) waves and the secondary (S) waves at a seismic station. This time difference is used to calculate the distance (km) between the earthquake epicenter and the seismic station, using the equation: distance (km) = S-P difference (sec) x 8 km/sec. This calculation assumes that the waves travel at a constant speed through the Earth's interior.
The magnitude (M) of an earthquake is a measure of the energy released during the earthquake, and is usually determined using a seismometer. The magnitude scale is logarithmic, meaning that each increase of one unit represents a tenfold increase in seismic energy. For example, an earthquake with a magnitude of 5.0 is ten times more powerful than one with a magnitude of 4.0, and 100 times more powerful than one with a magnitude of 3.0.
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What concentration results from the dilution of 500.0 mL of 4.267 M to a volume of 1.85 L?
To calculate the concentration resulting from the dilution of 500.0 mL of 4.267 M to a volume of 1.85 L, we can use the equation:
M1V1 = M2V2
where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
Plugging in the given values, we get:
4.267 M)(500.0 mL) = M2(1.85 L)
Simplifying this equation, we get:
M2 = (4.267 M)(500.0 mL) / (1.85 L)
M2 = 1.153 M
Therefore, the concentration resulting from the dilution is 1.153 M.
To calculate the concentration after dilution, you can use the dilution formula: C1V1 = C2V2, where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume.
Given:
C1 = 4.267 M
V1 = 500.0 mL = 0.5 L (converted to liters)
V2 = 1.85 L
Now, find C2:
C2 = (C1 * V1) / V2
C2 = (4.267 M * 0.5 L) / 1.85 L
C2 ≈ 1.153 M
The concentration after dilution is approximately 1.153 M.
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To find the concentration resulting from the dilution, we can use the equation:
M1V1 = M2V2
where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
Plugging in the given values, we get:
(4.267 M)(500.0 mL) = M2(1.85 L)
Simplifying and converting units, we get:
M2 = (4.267 M)(500.0 mL) / (1.85 L)
M2 = 1.16 M
Therefore, the concentration resulting from the dilution is 1.16 M.
To find the concentration after dilution, you can use the dilution formula:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Given:
C1 = 4.267 M
V1 = 500.0 mL (0.5 L)
V2 = 1.85 L
Rearrange the formula to solve for C2:
C2 = (C1V1) / V2
Now, plug in the given values:
C2 = (4.267 M * 0.5 L) / 1.85 L
C2 ≈ 1.154 M
So, the resulting concentration after dilution is approximately 1.154 M.
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A 210.00 g sample of water with an initial temperature of 29.0°C absorbs 7,000.0 J of heat. What is the final temperature of the water?
Note: Use C (capital C) for degrees Celsius when typing units. So it might look like 35C or 2.03 J/gC. Give your answer in 3 sig figs.
The 210.00 g sample of the water with the initial temperature of the 29.0°C absorbs the 7,000.0 J of heat. The final temperature of the water is the 36.9 °C .
The mass of the water = 210 g
The initial temperature = 29.0 °C
The final temperature = ?
The heat energy = 7000 J
The specific heat capacity = 4.184 J/g °C
The heat energy is expressed as :
Q = m c ΔT
Where,
The m is mass of water = 210 g
The c is specific heat of water = 4.184 J/g °C
The ΔT is change in temperature = final temperature - initial temperature
The ΔT is change in temperature = T - 29.0 °C
7000 = 210 × 4.184 ( T - 29.0 )
7000 = 878.64 ( T - 29.0 )
( T - 29.0 ) = 7.966
T = 36.9 °C
The final temperature is 36.9 °C .
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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)
, as described by the chemical equation
MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)
How much MnO2(s)
should be added to excess HCl(aq)
to obtain 105 mL Cl2(g)
at 25 °C and 765 Torr
?
Chemistry Table balance A+B→C
Table 1 attached
The reaction A + B → C has the following rate expression is 197.62 [A][B] M/s
How to determine rate expression?Using the experimental data to determine the order of the reaction with respect to A and B, assume that the rate of the reaction is given by:
rate = [tex]k[A]^x[B]^y[/tex]
where k = rate constant and
x and y = orders of the reaction with respect to A and B, respectively.
Compare the rates of the reaction in trials 1 and 2 while keeping the concentration of A constant:
rate1/rate2 = [tex]\frac{k[A]^x[B]^y}{k[A]^x[B]^y} = \frac{[B]^y}{[B]^y} = 1[/tex]
Conclude that the reaction is first-order with respect to B.
Similarly, compare the rates of the reaction in trials 1 and 3 while keeping the concentration of B constant:
rate1/rate3 =[tex]\frac{k[A]^x[B]^y}{k[A]^x[B]^y} = \frac{[A]^x}{[A]^x} = 1[/tex]
Therefore, the reaction is first-order with respect to A.
The rate expression for the reaction A + B → C is:
rate = k[A][B]
Using any of the experimental trials to determine the value of the rate constant k, use trial 1:
rate1 =[tex]k[A]^1[B]^1[/tex]
k = [tex]\frac{rate1}{[A]^1[B]^1}[/tex] = (3.30 E-3)/(0.012 M x 0.014 M) = 197.62 M⁻² s⁻¹
Therefore, the rate expression for the reaction A + B → C is:
rate = 197.62 [A][B] M/s
In this case, the units of k are M⁻¹ s⁻¹ because the reaction is first-order with respect to both A and B.
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Answer the following questions in complete sentences, and justify your responses.
After how many time intervals (shakes) did one-half of your atoms (candies) decay?
What is the half-life of your substance?
If the half-life model decayed perfectly, how many atoms would be remaining (not decayed) after 12 seconds?
If you increased the initial number of atoms (candies) to 300, would the overall shape of the graph be altered? Explain your answer.
Go back to your data table and for each three-second interval, divide the number of candies decayed by the number previously remaining and multiply by 100. Show your work.
The above percentage calculation will help you compare the decay modeled in this experiment to the half-life decay of a radioactive element. Did this activity perfectly model the concept of half-life? If not, was it close?
Compare how well this activity modeled the half-life of a radioactive element. Did the activity model half-life better over the first 12 seconds (four decays) or during the last 12 seconds of the experiment? If you see any difference in the effectiveness of this half-life model over time, what do you think is the reason for it?
To answer these questions, we need to know what substance you are referring to, as well as the data from the experiment.
1. After a certain number of time intervals (shakes), one-half of your atoms (candies) would decay. This number would depend on the specific substance and its half-life.
2. The half-life of a substance is the time it takes for half of its atoms to decay.
3. If the half-life model decayed perfectly, the number of remaining atoms after 12 seconds would depend on the initial number of atoms and the half-life of the substance.
4. If you increased the initial number of atoms (candies) to 300, the overall shape of the graph would not be altered. This is because the half-life decay is a percentage-based process, meaning it would still follow an exponential decay pattern.
5. To calculate the percentage of decay for each three-second interval, you would divide the number of candies decayed by the number previously remaining and multiply by 100. This would show the percentage of decay for each interval.
6. This activity may not perfectly model the concept of half-life, but it can provide a close approximation. Any discrepancies may be due to experimental errors or limitations.
7. To compare how well this activity modeled the half-life of a radioactive element, you would need to analyze the decay percentages over time. If there are differences in the effectiveness of the half-life model, it could be due to the limitations of the experimental setup, such as using candies as a representation of atoms.
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Which of the following represents an exothermic reaction?
Question 5 options:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + energy
2H2O(l) + energy → 2H2(g) + O2(g)
6CO2(g) + 6H2O(l) + energy → C6H12O6(aq) + 6O2(g)
Answer:
exothermic reaction is: Reactants → Products + Energy.
Explanation:
Note: ΔH represents the change in energy. If the energy produced in an exothermic reaction is released as heat, it results in a rise in temperature.
A balloon vendor at a street fair is using a tank of helium to fill her balloons. The tank has a volume of 109.0 L and a pressure of 107.0 atm at 25.0 °C. After a while she notices that the valve has not been closed properly. The pressure had dropped to 97.0 atm. (The tank is still at 25.0 °C.) How many moles of gas has she lost?
The number of mole of the gas lost, given that the pressure had dropped to 97.0 atm is 44.5 moles
How do i determine the number of mole lost?First, we shall determine the initial mole of the gas. Details below:
Initial volume (V₁) = 109.0 LInitial temperature (T₁) = 25 °C = 25 + 273 = 298 KInitial pressure (P₁) = 107.0 atmGas constant (R) = 0.0821 atm.L/mol KInitial mole (n₁) =?P₁V₁ = n₁RT₁
107 × 109 = n₁ × 0.0821 × 298
Divide both sides by (0.0821 × 298)
n₁ = (107 × 109) / (0.0821 × 298)
n₁ = 476.7 mole
Next, w shall determine the final mole of the gas. Details below
Final volume (V₂) = 109.0 LFinal temperature (T₂) = 25 °C = 25 + 273 = 298 KFinal pressure (P₂) = 97.0 atmGas constant (R) = 0.0821 atm.L/mol KFinal mole (n₂) =?P₂V₂ = n₂RT₂
97 × 109 = n₂ × 0.0821 × 298
Divide both sides by (0.0821 × 298)
n₂ = (97 × 109) / (0.0821 × 298)
n₂ = 432.2 mole
Finally, we shall determine the mole of the gas that was lost. Details below:
Initial mole (n₁) = 476.7 molesFinal mole (n₂) = 432.2 molesMole lost =?Mole lost = n₁ - n₂
Mole lost = 476.7 - 432.2
Mole of gas lost = 44.5 moles
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What two statements are true about a system?A) systems are a group of objects analyzed as one unit? B) energy that moves across a system boundaries is covered? C) only one way to define the boundary of a system? D) systems are made by humans?
The two true statements about a system are:
A) Systems are a group of objects analyzed as one unit.
B) Energy that moves across system boundaries is covered.
In general, a system can be defined as a group of objects or components that are connected or related to one another in some way, and that can be analyzed as a single unit. The components within a system can interact with each other, and with the environment outside of the system, in various ways. One of the key characteristics of a system is that it has a boundary or interface that separates it from the surrounding environment.
Energy, matter, or other quantities may flow across this boundary, and the interactions between the system and its environment can affect the behavior and properties of the system as a whole.
Overall, systems are a fundamental concept in many fields of science and engineering, and they can be used to model and analyze a wide range of phenomena, from physical systems like engines and circuits, to social and ecological systems like cities and ecosystems.
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The reactant concentration in a zero-order reaction was 6.00×10−2 M
after 175 s
and 3.50×10−2 M
after 315 s
. What is the rate constant for this reaction?
How much heat, in joules, would be required to raise the temperature of 450 g of
Aluminum (c Al = 0.21 cal/g o C) from 19.5 o C to 31.2 o C?
Answer:
[tex]\huge\boxed{\sf Q = 1105.65\ cal}[/tex]
Explanation:
Given data:Mass = m = 450 g
T₁ = 19.5 °C
T₂ = 31.2 °C
Change in Temperature = ΔT = 31.2 - 19.5 = 11.7 °C
c = 0.21 cal/g °C
Required:Heat = Q = ?
Formula:Q = mcΔT
Solution:Put the given data in the above formula.
Q = (450)(0.21)(11.7)
Q = 1105.65 cal
[tex]\rule[225]{225}{2}[/tex]
Suppose 10.0 g of ice at -10.0C is placed into 300.0 g of water in a 200.0-g copper calorimeter. The final temperature of the water and copper calorimeter is 18.0C.
1) What was the initial common temperature of the water and copper? (Express your answer to three significant figures.)
The intital common temperature of copper and water is 9.5°C, under the condition that 10.0 g of ice at -10.0C is placed into 300.0 g of water in a 200.0-g copper calorimeter.
Now to evaluate the initial common temperature of the water and copper calorimeter, we have to apply the formula
m1c1(Tk - Ti) + m2c2(Tk - Ti)
= mcopperccopper(Tk - Ti)
Here,
m1 = mass of water,
c1 =specific heat capacity of water,
m2 = mass of copper calorimeter,
c2 = specific heat capacity of copper calorimeter, mcopper = mass of copper block
ccopper =specific heat capacity of copper.
Here, this equation to evaluate Ti
Ti = (m1c1Tk + m2c2Tk - mcopperccopperTk - m1c1Ti - m2c2Ti) / (m1c1 + m2c2 - mcopperccopper)
Staging the given values into this equation
Ti = (-300.0 g)(4.18 J/g°C)(18.0°C) + (200.0 g)(0.385 J/g°C)(18.0°C) + (10.0 g)(0.385 J/g°C)(18.0°C) / [(300.0 g)(4.18 J/g°C) + (200.0 g)(0.385 J/g°C) - (10.0 g)(0.385 J/g°C)]
Ti = 9.5°C
Hence, the initial common temperature of the water and copper calorimeter was 9.5°C.
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What is the molar mass of a compound if a gaseous sample has a density of 0.978 g/L at 30 °C and 615 torr? The temperature in Celsius is known to two significant figures.
If a gaseous sample has a density of 0.978 g/L at 30 °C and 615 torr, the molar mass of the compound is 24.8 g/mol.
To calculate the molar mass of the compound, we first need to calculate the number of moles present in the gaseous sample using the ideal gas law:
PV = nRT
Where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
Converting the given pressure of 615 torr to atm:
615 torr = 0.811 atm
Converting the given temperature of 30°C to Kelvin:
30°C + 273.15 = 303.15 K
Rounding off to two significant figures, we get:
P = 0.81 atm
T = 303 K
Now, rearranging the ideal gas law equation to solve for n:
n = PV/RT
Substituting the given values:
n = (0.978 g/L) x (1 L) / (0.081 atm x 0.0821 L atm/mol K x 303 K)
n = 0.0394 mol
Next, we can calculate the molar mass of the compound using the formula:
molar mass = mass / mole
molar mass = (0.978 g/L) x (1 L) / 0.0394 mol
molar mass = 24.8 g/mol
Therefore, 24.8 g/mol is the molar mass of the compound.
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Arrange the following ions in order of increasing ionic radius: selenide ion, rubidium ion, bromide ion, strontium ion.
Answer:
Br, Se, Sr, Rb
Explanation:
Atomic radius increases as you move to the left and down the periodic table. The increase in radius as you move left is due to decreasing effective nuclear charge (the pull an electron feels from the nucleus) since the number of protons decrease. The increase in radius as you move down is due to a higher number of principle energy levels (orbital in which the electron is located relative to the atom's nucleus), causing the electrons to be farther from the nucleus.
If heat is going INTO the system, that means that energy must have come OUT FROM the ____________
If heat is going into a system, it means that energy must have come out from the surroundings.
How is energy/heat transferred?Heat is a form of energy transfer from a hotter object to a cooler one, and the direction of heat flow is always from the hotter object to the cooler one.
Therefore, if heat is entering a system, it must be gaining energy from its surroundings, which are at a lower temperature and therefore have less thermal energy.
Conversely, if heat is leaving a system, it means that energy is being transferred from the system to its surroundings, which are at a higher temperature and therefore have more thermal energy.
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What is volume of 12.0 g of carbon dioxide at stp?
Answer: 6.11 L
Explanation:
STP= 1atm, 273.15K
Molar mass of CO2=44.01g/mol so n= (12.0/44.01)
PV=nRT
V=(nRT)/P
V=((12.0/44.01)(0.0821)(273.15))/1
V=6.11L
Calculate the volume of hydrogen produced at s.t.p. When 25g of zinc are added to excess dilute hydrochloride acid at 31°c and 778mm Hg pressure. (H=1, Zn=65, Cl=35.5, molar volume of a gas at s.t.p = 22.4 dm3
To solve this problem, we need to use the balanced chemical equation for the reaction between zinc (Zn) and hydrochloric acid (HCl):
[tex]Zn + 2HCl - > ZnCl_2 + H_2[/tex]
According to the stoichiometry of this equation, one mole of Zn reacts with two moles of HCl to produce one mole of H2. Therefore, we need to determine the number of moles of Zn in 25 g, and then use the mole ratio to find the number of moles of H2 produced.
Finally, we can convert the number of moles of H2 to volume at STP using the molar volume of a gas.
First, we need to calculate the number of moles of Zn in 25 g:
The molar mass of Zn is 65.38 g/mol
The number of moles of Zn in 25 g is:
25 g / 65.38 g/mol = 0.383 mol Zn
Next, we use the mole ratio from the balanced equation to find the number of moles of H2 produced:
According to the balanced equation, one mole of Zn reacts with one-half mole of H2, so we produce 0.5 x 0.383 = 0.192 mol H2.
Finally, we can use the molar volume of a gas at STP to convert the number of moles of H2 to volume:
The molar volume of a gas at STP is 22.4 dm3/mol
Therefore, the volume of H2 produced is:
V = (0.192 mol) x (22.4 dm3/mol) = 4.30 dm3 or 4,300 ml
Therefore, the volume of hydrogen gas produced at STP is 4.30 dm3 or 4,300 ml when 25 g of zinc is added to excess dilute hydrochloric acid at 31°C and 778 mm Hg pressure.
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a student mixed 20 grams of salt into a beaker with 200 milliliters of warm water. then, the student set the cup of saltwater on a windowsill undisturbed for one week. what changes did the student observe? include what happened when salt was mixed with warm water and what most likely happened to the saltwater after one week.
Answer:
Water molecules pull the sodium and chloride ions apart, breaking the ionic bond that held them together. After the salt compounds are pulled apart, the sodium and chloride atoms are surrounded by water molecules, as this diagram shows. Once this happens, the salt is dissolved, resulting in a homogeneous solution.
Explanation:
Ethane burns in oxygen according to the following equation: 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
(a) How many liters of O2 at 41 °C and 0.307 atm will be needed to burn 8.57 L of C2H6 at 41 °C and 0.307 atm?
(b) How many liters of CO2 at 41 °C and 0.307 atm will be produced? Report your answers to parts (a) and (b) to 3 significant figures.
a) We need 32.6 liters of [tex]O_2[/tex] at 41 °C and 0.307 atm to burn 8.57 L of [tex]C_2H_6[/tex] at 41 °C and 0.307 atm
b) 18.5 liters of [tex]CO_2[/tex] will be produced at 41 °C and 0.307 atm.
To answer this question, we will use the ideal gas law, which relates pressure, volume, temperature, and number of moles of a gas. We will also use stoichiometry to relate the amount of ethane and oxygen consumed and the amount of carbon dioxide and water produced.
(a) To determine how many liters of [tex]O_2[/tex] are needed to burn 8.57 L of [tex]C_2H_6[/tex] , we first need to convert the volume of ethane to moles using the ideal gas law:
n([tex]C_2H_6[/tex] ) = PV/RT = (0.307 atm)(8.57 L)/(0.0821 L·atm/mol·K)(314 K) = 0.342 mol
From the balanced equation, we see that 2 moles of [tex]C_2H_6[/tex] react with 7 moles of [tex]O_2[/tex] . Therefore, the amount of [tex]O_2[/tex] needed is:
n([tex]O_2[/tex]) = (7/2) n([tex]C_2H_6[/tex]) = (7/2)(0.342 mol) = 1.20 mol
Now we can use the ideal gas law again to calculate the volume of [tex]O_2[/tex] needed:
V([tex]O_2[/tex] ) = n([tex]O_2[/tex])RT/P = (1.20 mol)(0.0821 L·atm/mol·K)(314 K)/(0.307 atm) = 32.6 L
Therefore, 32.6 liters of [tex]O_2[/tex] are needed to burn 8.57 L of [tex]C_2H_6[/tex] at at 41 °C and 0.307 atm
(b) From the balanced equation, we see that 2 moles of [tex]C_2H_6[/tex] produce 4 moles of [tex]CO_2[/tex] . Therefore, the amount of [tex]CO_2[/tex] produced is:
n([tex]CO_2[/tex]) = 2 n([tex]C_2H_6[/tex]) = 2(0.342 mol) = 0.684 mol
V([tex]CO_2[/tex]) = n([tex]CO_2[/tex])RT/P = (0.684 mol)(0.0821 L·atm/mol·K)(314 K)/(0.307 atm) = 18.5 L
Therefore, 18.5 liters of [tex]CO_2[/tex] at 41 °C and 0.307 atm will be produced.
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Calculate the mass of Kr
in a 9.95 L
cylinder at 91.2 ∘C
and 4.50 bar
.
What is heredity worksheet answer?
1. The long-haired cat in the P generation is a purebred. This is because it has two copies of the recessive allele (hh) responsible for long hair.
2. The short-haired cat in the P generation is a hybrid. We know this because the offspring include both short-haired (Hh) and long-haired (hh) cats, indicating that the short-haired parent must have one dominant (H) and one recessive (h) allele (Hh).
3. If the short-haired cat in the P generation were purebred (HH), all offspring would have short hair, as they would inherit one dominant allele (H) from the short-haired parent and one recessive allele (h) from the long-haired parent, resulting in Hh offspring.
4. The black horse is a hybrid. Since the cross between a black horse (B...) and a brown horse (bb) produced a brown foal (bb), the black horse must carry one dominant allele (B) and one recessive allele (b) - making it a hybrid (Bb).
5. To determine whether a guinea pig with a smooth coat (S...) is a hybrid or a purebred, perform a test cross by mating it with a guinea pig with a rough coat (ss). If all offspring have smooth coats (Ss), the smooth-coated guinea pig is likely purebred (SS). If any offspring have a rough coat (ss), the smooth-coated guinea pig is a hybrid (Ss).
What is a dominant allele and recessive allele?
A dominant allele is a variant of a gene that expresses its trait even when only one copy is present in an individual's genotype. In other words, it masks the effect of another variant (allele) of the same gene when they are together.
A recessive allele is a variant of a gene that only expresses its trait when two copies are present in an individual's genotype. The trait associated with the recessive allele is "masked" by the presence of a dominant allele, and it will only be expressed if both copies of the gene are recessive.
The above answer is based on the question below;
In a test cross, the organism with the trait controlled by a dominant allele is crossed with an organism with a trait controlled by a recessive allele. If all offspring have the trait controlled by the dominant allele, then the parent is probably a purebred. If any offspring has the recessive strait, then the dominant parent is a hybrid.
1. Is the long-haired cat in the P generation a hybrid or a purebred? Explain your answer.
2. Is the short-haired cat in the P generation a hybrid or a purebred? Explain your answer.
3. If the short-haired cat in the P generation were purebred, what would you expect the offspring to look like?
4. In horses, the allele for a black coat (B) is dominant over the allele for a brown coat (b). A cross between a black horse and a brown horse produces a brown foal. Is the black horse a hybrid or a purebred? Explain.
5. In guinea pigs, the allele for a smooth coat (S) is dominant over the allele for a rough coat (s). Explain how you could find out whether a
guinea pig with a smooth coat is a hybrid or a purebred.
H= Short hair
h = Long hair
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A helium-filled balloon of the type used in long-distance flying contains 1.5 ✕ 107 L of helium. Let us say you fill the balloon with helium on the ground where the pressure is 837 mm Hg and the temperature is 18.4°C. When the balloon ascends to a height of 6 miles where the pressure is only 707. mm Hg and the temperature is -31°C, what volume is occupied by the helium gas? Assume the pressure inside the balloon matches the external pressure.
We can use the combined gas law to solve this problem:
(P1V1/T1) = (P2V2/T2)
where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.
We are given that the initial pressure is P1 = 837 mm Hg and the initial volume is V1 = 1.5 × 10^7 L. The initial temperature is T1 = 18.4°C, which we need to convert to Kelvin by adding 273.15:
T1 = 18.4°C + 273.15 = 291.55 K
We are also given that the final pressure is P2 = 707 mm Hg and the final temperature is T2 = -31°C, which we need to convert to Kelvin:
T2 = -31°C + 273.15 = 242.15 K
Now we can solve for the final volume, V2:
(P1V1/T1) = (P2V2/T2)
V2 = (P1V1T2) / (P2T1)
V2 = (837 mm Hg * 1.5 × 10^7 L * 242.15 K) / (707 mm Hg * 291.55 K)
V2 = 5.26 × 10^6 L
Therefore, the volume occupied by the helium gas at the higher altitude is 5.26 × 10^6 L.
