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2 years ago

6. A balloon filled with air has a volume of 3.25 L at 30°C. It is placed in a freezer at

Chemistry

1 answer:

Viefleur [7K]2 years ago

7 0

Answer:

2.82 L

T₁ = 303 K

T₂ = 263 K

The final volume is smaller.

Explanation:

Step 1: Given data

Initial temperature (T₁): 30 °C

Initial volume (V₁): 3.25 L

Final temperature (T₂): -10 °C

Final volume (V₂): ?

Step 2: Convert the temperatures to Kelvin

We will use the following expression.

K = °C + 273.15

T₁: K = 30°C + 273.15 = 303 K

T₂: K = -10°C + 273.15 = 263 K

Step 3: Calculate the final volume of the balloon

Assuming constant pressure and ideal behavior, we can calculate the final volume using Charles' law. Since the temperature is smaller, the volume must be smaller as well.

V₁/T₁ = V₂/T₂

V₂ = V₁ × T₂/T₁

V₂ = 3.25 L × 263 K/303 K = 2.82 L

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Vedmedyk [2.9K]

Answer:

$pH=13$

Explanation:

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In this case, given the acid, we can suppose a simple dissociation as:

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Which occurs in aqueous phase, therefore, the law of mass action is written by:

$Ka=\frac{[H^+][A^-]}{[HA]}$

That in terms of the change $x$ due to the reaction's extent we can write:

$1x10^{-20}=\frac{x*x}{0.1M-x}$

But we prefer to compute the Kb due to its exceptional weakness:

$Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}} =1x10^{-6}$

Next, the acid dissociation in the presence of the base we have:

$Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}$

Whose solution is $x=0.0999M$ which equals the concentration of hydroxyl in the solution, thus we compute the pOH:

$pOH=-log([OH^-])=-log(0.0999)=1$

Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

$pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13$

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3 years ago

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The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
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3 years ago

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Ivahew [28]

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