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iren2701 [21]
3 years ago
8

A solution with a ph of 4 has _________ the concentration of h+ present compared to a solution with a ph of 5.

Chemistry
1 answer:
Marina CMI [18]3 years ago
7 0
<span>A solution with a pH of 4 has ten times the concentration of H</span>⁺<span> present compared to a solution with a pH of 5.
</span>pH <span>is a numeric scale for the acidity or basicity of an aqueous solution. It is  the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
</span>[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.
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2 years ago
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3 years ago
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Infrared (IR) spectroscopy is used to identify functional groups within a molecule. Click on the peak that corresponds to the st
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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]

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

<u>Explanation:</u>

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

PV=nRT

where,

P = pressure of the gas = 755 mmHg

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

6 0
3 years ago
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