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2 years ago

Please help with #2 and #3

Chemistry

1 answer:

Vaselesa [24]2 years ago

6 0

Answer:

2. $V_2=17L$

3. $V=82.9L$

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

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

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

$V_2=\frac{V_1T_2}{T_1}$

Therefore, we plug in the given data to obtain:

$V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L$

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

$V=\frac{nRT}{P}$

Therefore, we plug in the data to obtain:

$V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L$

Best regards!

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Answer:

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Given that;

acid dissociation constant ( $K_a$ ) = 0.42

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$0.42 = \frac{[x][x]}{0.52-x}}$

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0.42(0.52-x) = x²

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Using the quadratic formula;

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= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

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x = 0.30225

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