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BigorU [14]
3 years ago
11

A sample of gas has a volume of 20.0 liters at 22.0° C. If the pressure remains constant, what is the volume at 100.0° C?

Chemistry
2 answers:
miv72 [106K]3 years ago
6 0
The problem applies Charles' law since constant pressure with varying volume and temperature are given. Assuming ideal gas law, the equation to be used is \frac{ V_{1} }{ T_{1} }=\frac{ V_{2} }{ T_{2} }. We make sure the temperatures are expressed in Kelvin, hence the given added with 273. The volume 2 is equal to 25.2881 liters.
padilas [110]3 years ago
4 0

Answer: The volume at 100.0^0C is 25.3 L

Explanation:

To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

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

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=20.0L\\T_1=22.0^oC=(22.0+273)K=295.0K\\V_2=?\\T_2=100.0^oC=(100.0+273)K=373.0K

Putting values in above equation, we get:

\frac{20.0}{295.0K}=\frac{V_2}{373.0}\\\\V_2=25.3L

The volume at 100.0^0C is 25.3 L

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Ionic equation:

\text{H}_2\text{PO}_4^{-} + \text{HAsO}_4^{2-} \to \text{HPO}_4^{2-} + \text{H}_2\text{AsO}_4^{-}

The acid and base in a conjugate pair differ by only one proton \text{H}^{+}. The acid loses one proton to produce a conjugate base, whereas the base gains a proton to produce its conjugate acid.

\text{H}_2\text{PO}_4^{-} loses one proton to produce \text{HPO}_4^{2-} in this reaction.

\text{H}_2\text{PO}_4^{-} \to \text{H}^{+} + \text{HPO}_4^{2-}

Meanwhile, \text{HAsO}_4^{2-} gains one proton to form \text{H}_2\text{AsO}_4^{-}.

\text{HAsO}_4^{2-} + \text{H}^{+} \to \text{H}_2\text{AsO}_4^{-}

Therefore

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8 0
3 years ago
Read 2 more answers
In a titration of 0.35 M HCl and 0.35 M NaOH, how much NaOH should be added to 45.0 ml of HCl to completely neutralize the acid?
Bezzdna [24]
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Firstly, here is your formula: M(HCI) x V(HCI) = M(NaOh) x V(NaOH) 
With the values put in: 0.35 x 45 = 0.35 x V(NaOH) 
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There is 45 ml of V(NaOH)

Let me know if you need anything else. :)

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