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pantera1 [17]
3 years ago
11

A 9.0 mL sample of oxygen gas (Q) is stored at a pressure of 6.3 atm. If the pressure of the sample

Chemistry
1 answer:
Julli [10]3 years ago
6 0

Answer:

<h2>The answer is 23.63 mL</h2>

Explanation:

In order to find the new volume we use the formula for Boyle's law which is

P_1V_1 = P_2V_2 \\

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question

P1 = 6.3 atm

V1 = 9 mL

P2 = 2.4 atm

So we have

V_2 =  \frac{6.3 \times 9}{2.4}  \\  = 23.625

We have the final answer as

<h3>23.63 mL</h3>

Hope this helps you

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What is the percent composition of NaHCO3?
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Answer:

                Option-C (27.36% Na, 1.20% H, 14.30% C, and 57.14% O)

Explanation:

<em>Percent Composition</em> is defined as the <u><em>%age by mass of each element present in a compound</em></u>. Therefore, it is a relative amount of each element present in a compound.

Calculating Percent Composition of NaHCO₃:

1: Calculating Molar Masses of all elements present in NaHCO₃:

              a) Na  =  22.99 g/mol

             b) H  =  1.01 g/mol

              c) C  =  12.01 g/mol

              d) O₃  =  16.0 × 3 =  48 g/mol

2: Calculating Molecular Mass of NaHCO₃:

              Na  =  22.99 g/mol

             H    =  1.01 g/mol

              C    =  12.01 g/mol

              O₃  =  48 g/mol

                       ----------------------------------  

Total                  84.01 g/mol

3: Divide each element's molar mass by molar mass of NaHCO₃ and multiply it by 100:

For Na:

                 =  22.99 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  27.36 %

For H:

                 =  1.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  1.20 %

For C:

                 =  12.01 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  14.29 % ≈ 14.30 %

For O:

                 =  48.0 g.mol⁻¹ ÷ 84.01 g.mol⁻¹ × 100

                 =  57.13 % ≈ 57.14 %

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