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77julia77 [94]
1 year ago
9

Consider 5.00 L of a gas at 365 mmHg and 20. ∘C . If the container is compressed to 2.30 L and the temperature is increased to 4

0. ∘C , what is the new pressure, P2, inside the container? Assume no change in the amount of gas inside the cylinder.
Chemistry
1 answer:
lakkis [162]1 year ago
6 0

Answer:

P₂ = 1.12 atm

Explanation:

To find the new pressure, you need to use the Combined Gas Law:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

In this equation, "P₁", "V₁", and "T₁" represent the initial pressure, volume, and temperature. "P₂", "V₂", and "T₂" represent the new pressure, volume, and temperature. Before plugging the values into the equation, you need to

(1) convert the pressure from mmHg to atm (760 mmHg = 1 atm)

(2) convert the temperatures from Celsius to Kelvin (°C + 273)

The final answer should have 3 sig figs like the given values.

P₁ = 365 mmHg / 760 = 0.480 atm           P₂ = ? atm

V₁ = 5.00 L                                                   V₂ = 2.30 L

T₁ = 20°C + 273 = 293 K                             T₂ = 40°C + 273 = 313 K

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}                                              <----- Combined Gas Law

\frac{(0.480 atm)(5.00 L)}{293 K}=\frac{P_2(2.30 L)}{313 K}                       <----- Insert values

0.00819=\frac{P_2(2.30 L)}{313 K}                                     <----- Simplify left side

2.56 = P_2(2.30L)                                      <----- Multiply both sides by 313

1.12 = P_2                                                  <----- Divide both sides by 2.30

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How many moles of CO2 are produced from the combustion of 5.25 moles of CH3OH?
iVinArrow [24]

First, we write the reaction for CH3OH combustion

CH3OH+3/2O2--->CO2+2H2O

for 1 mole of methanol, we get 1 mole of CO2, therefore for 5,25 moles of methanol we will get 5,25 moles of CO2

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rusak2 [61]
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In a process for producing acetic acid, oxygen gas is bubbled into acetaldehyde, CH3CHO, containing manganese(II) acetate (catal
grigory [225]

<u>Answer:</u> The mass of acetic acid that can be produced is 30.24 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For acetaldehyde:</u>

Given mass of acetaldehyde = 22.2 g

Molar mass of acetaldehyde = 44 g/mol

Putting values in equation 1, we get:

\text{Moles of acetaldehyde}=\frac{22.2g}{44g/mol}=0.504mol

  • <u>For oxygen gas:</u>

Given mass of oxygen  gas = 12.6 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{12.6g}{32g/mol}=0.394mol

The given chemical equation follows:

2CH_3CHO(l)+O_2(g)\rightarrow 2CH_3COOH(l)

By Stoichiometry of the reaction:

2 moles of acetaldehyde reacts with 1 mole of oxygen gas

So, 0.504 moles of acetaldehyde will react with = \frac{1}{2}\times 0.504=0.252mol of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.

Thus, acetaldehyde is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of acetaldehyde produces 2 moles of acetic acid

So, 0.504 moles of acetaldehyde will produce = \frac{2}{2}\times 0.504=0.504moles of acetic acid

Now, calculating the mass of acetic acid from equation 1, we get:

Molar mass of acetic acid = 60 g/mol

Moles of acetic acid = 0.504 moles

Putting values in equation 1, we get:

0.504mol=\frac{\text{Mass of acetic acid}}{60g/mol}\\\\\text{Mass of acetic acid}=(0.504mol\times 60g/mol)=30.24g

Hence, the mass of acetic acid that can be produced is 30.24 grams

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