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2 years ago

7) Muscles work to lift a bone in your arm by contracts and relaxes tohelp please

Chemistry

1 answer:

Andru [333]2 years ago

3 0

Answer:

By contracting, muscles pull on bones and allow the body to move. When you want to bend your elbow, your biceps muscle contracts, and, at the same time, the triceps muscle relaxes. The biceps is the flexor, and the triceps is the extensor of your elbow joint.

Explanation:

yeah, hope it helps a little. :)

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If a gas has a proportionality constant of 4.32 x 10-4 mol at room temperature for a particular solvent, what will the

inysia [295]

$0.0467 X 10^{-4} M/kPa$ is the solubility of the gas when it exerts a partial pressure of 92.4kPa.

<h3>What is Henry's law?</h3>

Mathematically, we can get this from Henry's law

From Henry law;

Concentration = Henry constant × partial pressure

Thus Henry constant = $\frac{Concentration}{partial \;pressure}$

Henry constant = $\frac{4.32 \;X \;10^{-4} mol}{92.4kPa}$

$= 0.0467 X 10^{-4} M/kPa$

Hence, $0.0467 X 10^{-4} M/kPa$ is the solubility of the gas when it exerts a partial pressure of 92.4kPa.

Learn more about the Henry's law here:

brainly.com/question/16222358

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6 0

1 year ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating 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 $H_2So_4$ produced:

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

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:

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