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melamori03 [73]

3 years ago

8

A fuel-air mixture is placed in a cylinder fitted with a piston. The original volume is 0.310-L. When the mixture is ignited, ga

ses are produced and 935 J of energy is released. To what volume will the gases expand against a constant pressure of 635 mmHg, if all the energy released is converted to work to push the piston? □ 12.8 L 114L 9.02L 10.2L

1 answer:

Soloha48 [4]3 years ago

3 0

Answer:

$V_2=11.35L$

Explanation:

Knowing that the system is at constant pressure, the energy balance is:

$\Delta H=Q-W$

If all the energy (enthalpy of combustion) is transformed into work:

$\Delta H=-W$

Work:

$W=-\int_{V1}^{V2}P*dV$

$W=-P*(V_2-V_1)$

$\Delta H=P*(V_2-V_1)$

Calculating:

$935J=84659 Pa*(V_2-3.1*10^{-4}m^3)$

$V_2=0.0113 m^3=11.35L$

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If an ocean current is moving northward from Antarctica and up the coast of Africa, the climate of the coastal land would be....

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Answer:

I would say a cooled

Explanation:

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Susie cooked sausages on a barbecue. Fat and water in the sausages changed state. which two changes of state occur during this p

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Answer:

The options a and b are correct

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Which change of state is shown in the model?

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3 years ago

Read 2 more answers

Each step in the following process has a yield of 90.0 %. CH 4 + 4 Cl 2 ⟶ CCl 4 + 4 HCl CCl 4 + 2 HF ⟶ CCl 2 F 2 + 2 HCl The CCl

Rainbow [258]

Answer:

4.86 moles of HCl

Explanation:

1. First write the balanced chemical equations involved in the process:

$CH_{4}+_4Cl_{2}=CCl_{4}+_4HCl$

$CCl_{4}+_2HF=CCl_{2}+F_{2}+_2HCl$

2. Calculate what amount of $CCl_{4}$ is formed in the first reaction.

$3.00molesCH_{4}*\frac{1molCCl_{4}}{1molCH_{4}}=3.00molesCCl_{4}$

As the yield of each reaction is 90.0%, the amount of $CCl_{4}$ produced is the following:

$3.00molesCCl_{4}*0.90=2.7molesCCl_{4}$

3. Calculate the amount of HCl produced.

$2.7molesCCl_{4}*\frac{2molHCl}{1molCCl_{4}}=5.4molesHCl$

The total amount of HCl produced with a 90.0% yield is:

$5.4molesHCl*0.90=4.86molesHCl$

4 0

3 years ago

At what time will the pressure of SO₂Cl₂ decline to 0.50 its initial value? Express your answer using two significant figures.

QveST [7]

Answer: The time is 0.69/k seconds

Explanation:

The following integrated first order rate law

ln[SO₂Cl₂] - ln[SO₂Cl₂]₀ = - k×t

where

[SO₂Cl₂] concentration at time t,

[SO₂Cl₂]₀ initial concentration,

k rate constant

Therefore, the time elapsed after a certain concentration variation is given by:

$t=\frac{ln[SO_{2}Cl_{2}]_{0} - ln[SO_{2}Cl_{2}]}{k}=\frac{ln\frac{[SO_{2}Cl_{2}]_{0}}{[SO_{2}Cl_{2}]} }{k}$

We could assume that SO₂Cl₂ behaves as a ideal gas mixture so partial pressure is proportional to concentration:

$p_{(SO_{2}Cl_{2})}V = n_{(SO_{2}Cl_{2})}RT$

$[SO_{2}Cl_{2}]= \frac{n_{(SO_{2}Cl_{2})}}{V}}=\frac{p_{(SO_{2}Cl_{2})}}{RT}}$

In conclusion,

t = ln( p(SO₂Cl₂)₀/p(SO₂Cl₂) )/k

$t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{p_{(SO_{2}Cl_{2})}} }{k}$

for

$p_{(SO_{2}Cl_{2})}=0.5p_{(SO_{2}Cl_{2})}_{0}$

$t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{0.5p_{(SO_{2}Cl_{2})_{0}}} }{k}$

$t=\frac{ln\frac{1}{0.5} }{k}$

$t=\frac{ln(2)}{k}$

$t=\frac{0.69}{k}}$

7 0

3 years ago