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11111nata11111 [884]

3 years ago

9

Type the correct answer in the box. Express your answer to three significant figures. A balloon is filled with 0.250 mole of air

at 35°C. If the volume of the balloon is 6.23 liters, what is the absolute pressure of the air in the balloon? The absolute pressure of the air in the balloon is kilopascals.

1 answer:

Setler79 [48]3 years ago

7 0

Answer:

102.8 kPa.

Explanation:

We can use the general law of ideal gas: <em>PV = nRT.</em>

where, P is the pressure of the gas in atm (P = ??? atm).

V is the volume of the gas in L (V = 4000 mL = 6.23 L).

n is the no. of moles of the gas in mol (n = 0.25 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 35ºC + 273 = 308 K).

∴ P = nRT/V = (0.25 mol)(0.0821 L.atm/mol.K)(308 K)/(6.23 L) = 1.015 atm.

To convert to kPa:

<em><u>multiply the pressure value by 101.325 </u></em>

∴ P = (1.0 atm)(101.325 kPa/1.0 atm) = 102.8 kPa.

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${\qquad\qquad\huge\underline{{\sf Answer}}}$

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Calculate how many grams of sodium azide (NaN3) are needed to inflate a 25.0 × 25.0 × 20.0 cm bag to a pressure of 1.35 atm at a

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Answer : The mass of $NaN_3$ at temperature $20^oC$ 28.47 g.

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Solution : Given,

Pressure of gas = 1.35 atm

Temperature of gas = $20^oC=273+20=293K$ $(0^oC=273K)$

Volume of gas = $25\times 25\times 20cm=12500cm^3=12.5L$ $(1L=1000cm^3)$

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Using ideal gas equation,

$PV=nRT$

where,

P = pressure of the gas

V = volume of the gas

T = temperature of the gas

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in this formula, we get

$(1.35atm)\times (12.5L)=n\times (0.0821Latm/moleK)\times (293K)$

By rearranging the terms, we get the value of 'n'

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$20NaN_3(s)+6SiO_2(s)+4KNO_3(s)\rightarrow 32N_2(g)+5Na_4SiO_4(s)+K_4SiO_4(s)$

As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

So, 0.7015 moles of $N_2$ produced from $\frac{20}{32}\times 0.7015=0.438$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

$\text{ Mass of }NaN_3=(0.438moles)\times (65g/mole)=28.47g$

Therefore, the mass of $NaN_3$ needed are 28.47 g.

Part 2 : We have to calculate the moles of gas at temperature $10^oC$ and same volume & pressure.

Using ideal gas equation,

$PV=nRT$

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As, 32 moles of $N_2$ produced from 20 moles of $NaN_3$

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Now we have to calculate the mass of $NaN_3$ .

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

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Therefore, the mass of $NaN_3$ needed are 29.51 g.

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