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devlian [24]
3 years ago
10

A container holds 3.41×10−3mol of carbon dioxide (CO2). After the addition of 8.41×10−4mol of carbon dioxide, the volume of the

container increases to 95.2mL, with the temperature and pressure remaining constant. What was the initial volume of the container?
Chemistry
2 answers:
mel-nik [20]3 years ago
3 0

Answer:

76.366

Explanation:

Knewton answer is 76.4

poizon [28]3 years ago
3 0

Answer:

76.4mL

Explanation:

First, calculate the final number of moles of carbon dioxide (n2) in the container.

n2 = n1+n added = 3.41×(10^-3) mol+8.41×(10^-4) mol = 4.25×(10^-3) mol

Rearrange Avogadro's law to solve for V1.

V1 =(V2×n1)/n2

Substitute in the known values for n1, V2, and n2.

V1 = (95.2mL×(3.41×10^-3 mol))/(4.25×(10^-3) mol)=76.4mL

So the initial volume is 76.4mL.

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