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IgorLugansk [536]
3 years ago
5

A tank of nitrogen has a volume of 14.0 L and a pressure of 1 atm. Find the volume of the nitrogen when its pressure is changed

to 0.8 atm while the temperature is held constant .
Chemistry
1 answer:
ANEK [815]3 years ago
7 0

Answer:

17.5 L

Explanation:

Step 1: Given data

  • Initial volume of the tank of nitrogen (V₁): 14.0 L
  • Initial pressure of nitrogen (P₁): 1 atm
  • Final volume of the tank of nitrogen (V₂): ?
  • Final pressure of nitrogen (P₂): 0.8 atm

Step 2: Calculate the final volume of the nitrogen

We have a gas that undergoes a change at a constant temperature. If we assume an ideal behavior, we can calculate the final volume of the nitrogen using  Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 1 atm × 14.0 L/0.8 atm = 17.5 L

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Draw the condensed structural formulas for all the possible haloalkane isomers that have four carbon atoms and a bromine.
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<u>Answer:</u> The isomers are shown in the image below.

<u>Explanation:</u>

Isomers are defined as the chemical compounds having the same number and kinds of atoms but arrangement are different.

For the alkane having four carbon atoms and 1 bromine atom, the IUPAC name of the haloalkane is bromobutane

There are 4 possible isomers for the given haloalkane compound:

  1. 1-bromobutane
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The isomers of the given organic compound is shown in the image below.

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Calculate the number of moles of solute present in 185.0 mL of a 1.50 mol/L solution of HCl.
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Answer:

0.278 mol HCl

Explanation:

We currently have 185.0 mL of a 1.50 mol/L solution of HCl. We want to find the number of moles there are.

Based on the given information, our volume is 185.0 mL and our molarity is 1.50. Because molarity is defined as moles / Litre, we can easily find the moles given volume by multiplying molarity by volume.

First, though, we must convert millilitres to litres. There are 1000 mL in 1 L, so divide 185.0 by 1000:

185.0 / 1000 = 0.185 L

Now, multiply 0.185  by 1.50:

0.185 L * 1.50 mol/L = 0.278 mol HCl

Thus the answer is 0.278 mol HCl.

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