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3 years ago

What pressure is exerted by 0.625 moles of a gas in a 45.4 L container at -24.0 °C? Find pressure in units of atm.

Chemistry

1 answer:

scoundrel [369]3 years ago

3 0

Answer:

Ok to solve this you will need to use the Ideal Gas Law Formula which is as follows:

PV = nRT

P= pressure

V= volume

n= # of moles

R= Universal Gas Constant (0.0821 L x atm/mol x K)

T= Kelvin temperature

1.Simplify the Ideal Gas Law formula to what you need to solve for:

P = (nRT)/ V

2. List all you components as follows (this makes the process easier):

P = ?

V = 45.4 L

n = 0.625 mol

R = 0.0821 L x atm/ mol x K

T = 249 K

To find the Kelvin temperature K = C + 273

3. Plug in all your components in your set up formula:

P = [(0.625 mol)(0.0821 L x atm/ mol x K)(249 K)] / (45.4 L)

4. Cross out all similar units so the only thing left is atm because you are trying to find pressure.

P = [(0.625)(0.0821atm)(249)] / (45.4)

5. Multiply through and simplify

P = 0.28 atm

B. is the correct answer.

Glad I could help!! If you have any other questions just message me. Hopefully this was helpful.

Explanation:

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sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4 × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$ $\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$

= 21284.56606

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2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

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2 years ago

Which elements did you include in your question?

SIZIF [17.4K]

Answer:

It should be all of them

Explanation:

It’s because it was in your question you wrote out

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2 years ago

Consider this equilibrium reaction at 400 K. Br2(g)+Cl2(g)↽−−⇀2BrCl(g)Kc=7.0 If the composition of the reaction mixture at 400 K

Genrish500 [490]

Answer:

Q = 7.0

Q = kc. The reaction is in equilibrium

Explanation:

Based on the reaction:

Br₂ + Cl₂ ⇄ 2BrCl

Equilibrium constant of the reaction, kc, is the ratio of <em>equilibrium concentrations</em> products over reactants powered to its reaction coefficient:

Kc = [BrCl]² / [Br₂] [Cl₂] = 7.0

Now, reaction quotient, Q, is write as the same Kc but the concentrations are actual concentrations:

Q = [BrCl]² / [Br₂] [Cl₂]

Replacing:

Q = [0.00415M]² / [0.00366M] [0.000672M]

Q = 7.0

Now, as Q = Kc = 7.0, the reaction mixture is in equilibrium

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3 years ago

PLEASE HURRY. Identify the following as

strojnjashka [21]

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6 0

2 years ago

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In the unbalanced chemical reaction for the combustion of propane determine at standard temperature and pressure how many liters

Semmy [17]

Answer:

9 liters of CO₂ are produced by this combustion

Explanation:

In order to determine the volume of produced CO₂, we start with the reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

We need, O₂ density to find out the mass, that has reacted.

δ O₂ = O₂ mass / O₂ volume → δ O₂ . O₂ volume = O₂ mass

δ O₂ = 1.429 g /dm₃ (1dm³ = 1L) 1.429 g/L . 15L = 21.4 g of O₂

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By stoichiometry, 5 moles of O₂ can produce 3 moles of CO₂

Then, 0.670 moles of O₂ will produce (0.670 . 3) /5 = 0.402 moles of dioxide.

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3 years ago