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docker41 [41]
3 years ago
14

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade​

Chemistry
1 answer:
egoroff_w [7]3 years ago
5 0

Answer:

Volume occupied by oxygen gas at 15 degree centigrade​ is equal to 316.5 centimeter cube

Explanation:

Assuming Pressure is constant.

\frac{V_1}{T_1} = \frac{V_2}{T_2}

where T1 and T2 are temperature in Kelvin

Substituting the give values we get-

\frac{300}{273} = \frac{V_2}{288}

V_2 = \frac{288*300}{273} \\V_2 = 316.5

Volume occupied by oxygen gas at 15 degree centigrade​ is equal to 316.5 centimeter cube

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3 years ago
What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso
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The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247

The pH of the buffer can be known as

pH = pK_{a} + log[\frac{[A-]}{[HA]}}]

The concentration of [A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\

Similarly, the concentration of [HA] = \frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236

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4 years ago
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4 0
3 years ago
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8 0
3 years ago
at what temperature will a fixed amount of gas with a volume of 175 L at 15 degrees celsius and 760mmHg occupy a volume of 198L
Jlenok [28]

<u>Answer:</u> The temperature when the volume and pressure has changed is 274 K

<u>Explanation:</u>

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=760mmHg\\V_1=175L\\T_1=15^oC=[15+273]K=288K\\P_2=640mmHg\\V_2=198L\\T_2=?K

Putting values in above equation, we get:

\frac{760mmHg\times 175L}{288K}=\frac{640mmHg\times 198L}{T_2}\\\\T_2=274K

Hence, the temperature when the volume and pressure has changed is 274 K

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3 years ago
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