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liberstina [14]
3 years ago
10

A SCUBA tank can normally hold 18 liters of gas at 3000 psi. How many liters of gas would this be equivalent to if the gas were

at 14.7 psi?
Chemistry
1 answer:
NikAS [45]3 years ago
5 0

Liters of gas at 14.7 psi: 3674 L

Explanation:

We can solve this problem by using Boyle's law, which states that:

"For an ideal gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume"

Mathematically:

pV=const.

where

p is the gas pressure

V is its volume

This law can also be rewritten as

p_1 V_1 = p_2 V_2

Where:

p_1 = 3000 psi is the initial pressure of the gas

V_1 = 18 L is the initial volume of the gas

p_2 = 14.7 psi is the final pressure of the gas

V_2 is the final volume

Therefore, we can calculate the volume of gas corresponding to 14.7 psi of pressure:

V_2=\frac{p_1 V_1}{p_2}=\frac{(3000)(18)}{14.7}=3674 L

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

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Answer:

3.01 × 10²⁴ atoms S

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Using Dimensional Analysis
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Explanation:

<u>Step 1: Define</u>

5.00 mol S

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

<u />5.00 \ mol \ S(\frac{6.022 \cdot 10^{23} \ atoms \ S}{1 \ mol \ S} ) = 3.011 × 10²⁴ atoms S

<u />

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

3.011 × 10²⁴ atoms S ≈ 3.01 × 10²⁴ atoms S

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Boron has two naturally-ocurring isotopes. Boron-10 has an abundance of 19.8% and actual mass of 10.013 amu, and boron-11 has an
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Answer:

Average atomic mass = 10.812 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

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% = 19.8 %

Mass = 10.013 amu

<u>For second isotope, Boron-11: </u>

% = 80.2 %

Mass = 11.009 amu

Thus,  

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