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ArbitrLikvidat [17]
3 years ago
6

Now, find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 11. Then div

ide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C. What is the concentration of H+ ions at a pH = 11? mol/L What is the concentration of OH– ions at a pH = 11? mol/L What is the ratio of H+ ions to OH– ions at a pH = 11? :1, OR 1:
Physics
2 answers:
Hunter-Best [27]3 years ago
8 0

Answer

1.0/5

4

IlaMends

Ambitious

2.1K answers

12.9M people helped

Explanation:

When pH of the solution is 11.

..(1)

At pH = 11, the concentration of ions is .

When the pH of the solution is 6.

..(2)

At pH = 6, the concentration of ions is .

On dividing (1) by (2).

The ratio of hydrogen ions in solution of pH equal to 11 to the solution of pH equal to 6 is .

Difference between the ions at both pH:

This means that Hydrogen ions in a solution at pH = 7 has ions fewer than in a solution at a pH = 6

natima [27]3 years ago
5 0

1.) 0.00000000001

2.) 0.001

3.) 0.00000001: 1, OR  1: 100,000,000

- Chilio

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Answer:

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Explanation:

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\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

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\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

5 0
3 years ago
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Answer:

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Answer:

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Stating this mathematically. this implies that:

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Answer:

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