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VikaD [51]
3 years ago
15

H2S(g) is an endothermic process. A 6.1589−g sample of the solid is placed in an evacuated 4.000−L vessel at exactly 24°C. After

equilibrium has been established, the total pressure inside is 0.709 atm. Some solid NH4HS remains in the vessel. (a) What is the KP for the reaction? (b) What percentage of the solid has decomposed? % (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel? More than 50% would remain. Between 20% and 50% would remain. Between 10% and 20% would remain. Less than 10% would remain.
Chemistry
1 answer:
rewona [7]3 years ago
7 0

s an endothermic process. A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at exactly 24°C. After equilibrium has been established, the total pressure inside is 0.709 atm hope this helps

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The gas in an aerosol can is at a pressure of 3.16 atm at 32.2°C. What would the gas pressure in the can be at 22.9°C?
olasank [31]

Answer:

The pressure of the gas would be 3.06 atm

Explanation:

Amonton's law states that the pressure is directly proportional to the absolute temperature of a gas under constant volume. The equation is:

P1 / T1 = P2 / T2

<em>Where P1 is the initial pressure = 3.16atm</em>

<em>T1 is initial absolute temperature = 273.15 + 32.2°C = 305.35K</em>

<em>P2 is our incognite</em>

<em>And T2 is = 273.15 + 22.9°C = 296.05K</em>

<em />

Replacing:

3.16atm / 305.35K = P2 / 296.05K

3.06 atm = P2

<h3>The pressure of the gas would be 3.06 atm</h3>
5 0
3 years ago
What evidence supports a conservation law? 6 CO2 → C6H12O6 6 O2 6 CO2 6 H2O light → C6H12O6 6 O2 6 H2O light → C6H12O6 6 O2 6 CO
docker41 [41]

The law of conservation has been stated that the mass and energy has neither be created nor destroyed in a chemical reaction.

The law of conservation has been evident when there has been an equal number of atoms of each element in the chemical reaction.

<h3>Conservation law</h3><h3 />

The given equation has been assessed as follows:

  • \rm 6\;CO_2\;\rightarrow\;C_6H_1_2O_6

The reactant has absence of hydrogen, while hydrogen has been present in the product. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;+\;6\;CO_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The number of atoms of each reactant has been different on the product and the reactant side. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;H_2O\;+\;Light\;\rightarrow\;C_6H_1_2O_6

The reactant has the presence of carbon, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

  • \rm 6\;O_2\;+\;6\;CO_2\;\rightarrow\;3\;C_6H_1_2O_6\;+\;3\;O_2

The product has the presence of hydrogen, while it has been absent in the reactant. Thus, the reaction will not follow the law of conservation.

Learn more about conservation law, here:

brainly.com/question/2175724

4 0
2 years ago
I need HELP!!!<br><br> Do the benefits of nuclear power outweigh the risks?
Georgia [21]
While there is no such thing as 100 percent safe, having nuclear energy is much safer than you think. It's thousands of times safer than conventional coal and other fossil-fuel-derived energy, not to mention the specter of environmental disaster from continued use of carbon-based energy sources.
3 0
3 years ago
ObIel WiLll unt COl.. USSMS A certain chemical reaction releases 31.2 kJ/g of heat for each gram of reactant consumed. How can y
aleksandr82 [10.1K]

Answer:

The expression to calculate the mass of the reactant is m = \frac{1.080kJ}{31.2kJ/g}

Explanation:

<em>The amount of heat released is equal to the amount of heat released per gram of reactant times the mass of the reactant.</em> To keep to coherence between units we need to transform 1,080 J to kJ. We do so with proportions:

1,080J.\frac{1kJ}{10^{3}J } =1.080kJ

Then,

1.080kJ=31.2\frac{kJ}{g} .m\\m = \frac{1.080kJ}{31.2kJ/g}

5 0
3 years ago
Read 2 more answers
What does D represent?
Vitek1552 [10]

Answer:

C. Reactants

I think soo

5 0
3 years ago
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