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3 years ago

What is the OH- ion concentration, in moles per liter, of an aqueous solution with a pH of 7?

Chemistry

1 answer:

topjm [15]3 years ago

3 0

Answer:

Explanation:

the pH scale is based on the function

1x10-14

So since pH is neutral the concentration of OH and H must be equal hence the only option in which the concetration would [OH]{H}= 1x10-14 would be when both are equal so it cannot be a or d and when they are expressed in concetration so the only option is B.

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Which of the following objects will have the greatest change in motion? here's the image

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Predict the effect of adding a non competitive inhibitor to the reaction mixture on the rate of reaction at a high substrate con

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A noncompetitive inhibitor can only bind to an enzyme with or without a substrate at several places at a particular point in time

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3 years ago

A sample of nitrogen gas occupies a volume of 2.55 L when it is at 755 mm Hg and 23 degrees Celsius. Use this information to det

ivolga24 [154]

Answer: The number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

Explanation:

To calculate the amount of nitrogen gas, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

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V = Volume of the gas = 2.55 L

T = Temperature of the gas = $23^oC=[23+273]K=296K$

R = Gas constant = $62.364\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = ?

Putting values in above equation, we get:

$755mmHg\times 2.55L=n\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 296K\\\\n=\frac{755\times 2.55}{62.364\times 296}=0.1043mol$

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=755mmHg\\V_1=2.55mL\\T_1=23^oC=[23+273]K=296K\\P_2=?\\V_2=4.10L\\T_2=18^oC=[18+273]K=291K$

Putting values in above equation, we get:

$\frac{755mmHg\times 2.55L}{296K}=\frac{P_2\times 4.10L}{291K}\\\\P_2=\frac{755\times 2.55\times 291}{4.10\times 296}=461.6mmHg$

Hence, the number of moles of nitrogen gas are 0.1043 moles and the pressure when volume and temperature has changed is 461.6 mmHg

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shutvik [7]

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