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Phoenix [80]
3 years ago
10

A flexible container at an initial volume of 3.10 L contains 9.51 mol of gas. More gas is then added to the container until it r

eaches a final volume of 19.1 L . Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Chemistry
1 answer:
Schach [20]3 years ago
7 0

Answer:

49.09 moles of gas are added to the container

Explanation:

Step 1: Data given

Initial volume = 3.10 L

Number of moles gas = 9.51 moles

The final volume = 19.1 L

The pressure, temperature remain constant

Step 2: Calculate number of moles gas

V1/n1 = V2/n2

⇒with V1 = the initial volume of the gas = 3.10 L

⇒with n1 = the initial number of moles = 9.51 moles

⇒with V2 = the increased volume = 19.1 L

⇒with n2 = the final number of moles gas

3.10L / 9.51 moles = 19.1 L / n2

n2 = 58.6 moles

The new number of moles is 58.6

Step 3: calculate the number of moles gas added

Δn = 58.6 - 9.51 = 49.09 moles

49.09 moles of gas are added to the container

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Consider the following reaction, equilibrium concentrations, and equilibrium constant at
REY [17]

Answer:

Equilibrium concentration of H_{2}O is 12.5 M

Explanation:

Given reaction: C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH

Here, K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}

where K_{c} represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations

Here, [C_{2}H_{4}]=0.015M , [C_{2}H_{5}OH]=1.69M and K_{c}=9.0

So, [H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M

Hence equilibrium concentration of H_{2}O is 12.5 M

5 0
3 years ago
Please help with questions 7-12
Rufina [12.5K]

Answer:

negitive 5 or -5

Explanation

usally 12-7 = 5 but 7 is first so it you turn 5 to negitive 5

or -5

3 0
3 years ago
What does it mean to dilute a solution?
LenKa [72]

Answer:

Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to the solution

7 0
3 years ago
20.
sergiy2304 [10]
I think the anwser is A
8 0
3 years ago
Read 2 more answers
What is the molality of impurities in thesolvent? If the impurity is largely hexachloroethane, C2Cl6, how many grams of this imp
Contact [7]

Answer:

a) grams of this impurity per kg of CCl4 = 3.416 g/kg of solvent.

b) mass purity % = 99.66%

Explanation:

Given, the freezing point of pure CCl₄ = - 23°C

Presence of impurities lowers the freezing point to - 23.43°C

The freezing point depression constant, Kբ = 29.8°C/m

The lowered freezing point is related to all the parameters through the relation

ΔT = i Kբ × m

where ΔT is the lowered freezing point, that is, the difference between freezing point of pure substance (T⁰) and freezing point of substance with impurities (T).

i = Van't Hoff factor which measures how much the impurities influence/affect colligative properties (such as freezing point depression) and for most non-electrolytes like this one, it is = 1

Kբ = The freezing point depression constant = 29.8°C/m

m = Molality = ?

T⁰ - T = i Kբ m

- 23 - (-23.43) = 1 × 29.8 × m

m = 0.43/29.8 = 0.0144 mol/kg

Then, we're told to calculate impurity of the CCl₄

we convert the Molality to (gram of solute)/(kg of solvent) first

Solute = C₂Cl₆

Molar mass = 236.74 g/mol

So, (molality × molar mass) = (gram of solute)/(kg of solvent)

(gram of solute)/(kg of solvent) = 0.0144 × 236.74 = 3.416 (gram of solute)/(kg of solvent)

Mass purity % = (1000 g of pure substance)/(1000 g of pure substance + mass of impurity in 1000 g of pure substance)

1000 g of solvent contains 3.416 grams of impurities

Mass purity % =100% × 1000/(1003.416)

Mass purity % = 99.66 %

5 0
4 years ago
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