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sladkih [1.3K]
3 years ago
12

A 10.0 L balloon contains helium gas at a pressure of 655 mmHg. What is the final pressure, in millimeters of mercury, if the fi

nal volume is 13.8 L?
Chemistry
1 answer:
joja [24]3 years ago
7 0

Answer:

474.64mmHg

Explanation:

From the question given, we obtained the following:

V1 = 10L

P1 = 655mmHg

V2 = 13.8L

P2 =?

P1V1 = P2V2

10 x 655 = P2 x 13.8

Divide both side by the coefficient P2 i.e 13.8

P2 = (10 x 655) / 13.8

P2 = 474.64mmHg

Therefore, the new pressure will be 474.64mmHg

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Following are the answer to this question:

Explanation:

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calculating the new concentrated value of NH_4Cl= \frac{0.050\times 0.10}{0.120}= 0.04166 \ Mcalculating the new concentrated value of H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M when 1 mol H_2So_4 produced 2 mols H^{+} so, 0.0125 in H_2So_4produced:

=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}

create the ICE table:    

NH_3    \ \ \ \ \ \ \ \     + H^{+}  \ \ \ \ \ \ \longrightarrow NH_4^{+}                    

I (m)       0.033(m)            0.025                       0.04166

C            -0.025                 -0.025                       + 0.025  

E            8.3\times 10^{-3}     0                    0.0667

now calculating pH:

when ph= 8.83:

P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769

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