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denis-greek [22]
3 years ago
8

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) A reaction mixture is made containing an initial [SO2Cl2] of 2.2×10−2M

. At equilibrium, [Cl2]= 1.3×10−2M .
Chemistry
1 answer:
lukranit [14]3 years ago
5 0
I think you want to ask about Keq. At equilibrium, we can know [SO2Cl2] is 2.2*10-2 M -1.3*10-2M=9*10^-3 M. And [SO2]=[Cl2]. So the Keq=1.88*10^-2.
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A 5.00 L sample of helium expands to 12.0 L at which point the
mina [271]

Answer:

1.73 atm

Explanation:

Given data:

Initial volume of helium = 5.00 L

Final volume of helium = 12.0 L

Final pressure = 0.720 atm

Initial pressure = ?

Solution:

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

P₁ × 5.00 L = 0.720 atm × 12.0 L

P₁ = 8.64 atm. L/5 L

P₁ = 1.73 atm

7 0
2 years ago
2. Convert 340,000 to scientific notation
arlik [135]

Answer:

3.4 x 10^5

Explanation:

8 0
3 years ago
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What is the concentration of a solution with a volume of 660L that contains 33.4g of AlCO3?
lana [24]

Answer:

The concentration of the solution is 5.8168 × 10^{-4} mol.dm^{-3}

Explanation:

Here, we want to calculate the concentration of the solution.

The unit of this is mol/dm^3

So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3

The number of moles = mass/molar mass

molar mass of AlCO3 = 27 + 12 + 3(16)  = 27 + 12 + 48 = 87g/mol

Number of moles = 33.4/87 = 0.384 moles

This 0.384 moles is present in 660 L

x moles will be present in 1 dm^3

Recall 1 dm^3 = 1L

x * 660 = 0.384 * 1

x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3

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2 years ago
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Answer:

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Explanation:

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