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Pachacha [2.7K]
3 years ago
3

How many moles of CO2 are in 210 liters at STP of the compound​

Chemistry
1 answer:
Zina [86]3 years ago
8 0

Answer:

9.4 mol CO₂

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

  • Using Dimensional Analysis
  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Explanation:

<u>Step 1: Define</u>

210 L CO₂ at STP

<u>Step 2: Identify Conversions</u>

22.4 L = 1 mol at STP

<u>Step 3: Convert</u>

<u />210 \ L \ CO_2(\frac{1 \ mol \ CO_2}{22.4 \ L \ CO_2} ) = 9.375 mol CO₂

<u>Step 4: Check</u>

<em>We are given 2 sig figs. Follow sig fig rules and round.</em>

9.375 mol CO₂ ≈ 9.4 mol CO₂

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The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

\text{NH}_4\text{Cl} is a salt soluble in water. \text{NH}_4\text{Cl} dissociates into ions completely when dissolved.

\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq).

The first test tube used to contain \text{NH}_4\text{OH}. \text{NH}_4\text{OH} is a weak base that dissociates partially in water.

\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+}  \;(aq)+ {\text{OH}}^{-} \; (aq).

There's also an equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l).

\text{OH}^{-} ions from \text{NH}_4\text{OH} will shift the equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} to the right and reduce the amount of {\text{H}_3\text{O}}^{+} in the solution.

The indicator equilibrium will shift to the right to produce more {\text{H}_3\text{O}}^{+} ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding \text{NH}_4\text{Cl} will add to the concentration of {\text{NH}_4}^{+} ions in the solution. Some of the {\text{NH}_4}^{+} ions will combine with \text{OH}^{-} ions to produce \text{NH}_4\text{OH}.

The equilibrium between  \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions will shift to the left to produce more of both ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)

The indicator equilibrium will shift to the left as the concentration of {\text{H}_3\text{O}}^{+} increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

7 0
3 years ago
According to the kinetic-molecular theory, how do gas particles behave?
mezya [45]

Answer:

They are in constant motion.

Explanation:

More energy\heat= more kinetic energy=more motion\movement

7 0
3 years ago
At a certain temperature the vapor pressure of pure acetic acid HCH3CO2 is measured to be 226.torr. Suppose a solution is prepar
Rudiy27

Answer:

The partial pressure of acetic acid is 73.5 torr

Explanation:

Step 1: Data given

Total pressure is 226 torr

mass of acetic acid = 126 grams

mass of methanol = 141 grams

 

Step 2: Calculate moles of acetic acid

moles acetic acid = mass acetic acid / molar mass acetic acid

moles acetic acid = 127 grams / 60.05 g/mol

moles acetic acid = 2.115 moles

Step 3: Calculate moles of methanol

moles methanol = 141 grams / 32.04 g/mol

moles methanol = 4.40 moles

Step 4: Calculate total moles

Total moles = moles of acetic acid + moles methanol

Total moles = 2.115 moles + 4.40 moles

Total moles = 6.515 moles

Step 5: Calculate mole fraction of acetic acid

2.115 moles / 6.515 moles = 0.325

Step 6: Calculate partial pressure of acetic acid

P(acetic acid) = 0.325 * 226

P(acetic acid) = 73.45 torr ≈73.5

 

We can control this by calculating the partial pressure of methanol

mole fraction of methanol = (6.515-2.115)/6.515 = 0.675

P(methanol) = 0.675 * 226 = 152.55

226 - 152.55 = 73.45 torr  

The partial pressure of acetic acid is 73.5 torr

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3 years ago
2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to
evablogger [386]

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

3 0
3 years ago
Select the correct answer.
Svetlanka [38]

Answer:

alcohol

I hope this helps you

7 0
3 years ago
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