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julia-pushkina [17]
3 years ago
14

Assuming all gas is removed from the tank when filling balloons, how many 0.75 L balloons can be filled from a tank that contain

s 375 g CO2 at 1.0 atm of pressure at a temperature of 298 K
Chemistry
2 answers:
Marrrta [24]3 years ago
8 0

Answer:

278 balloons can be aired.

Explanation:

We apply the Ideal Gases Law to determine the total available volume for each baloon of 0.75L. So we need to divide the volume we obtain by the volume of each baloon.

We determine the moles of CO₂

375 g. 1 mol / 44g = 8.52 moles

V = (n . R . T) / P → (8.52 mol . 0.082 . 298K) / 1 atm = 208.3 L

That's the total volume from the tank, so we can inflate (208.3 / 0.75) = at least 278 balloons

lana [24]3 years ago
6 0

Answer:

277 balloons can be filled with CO2

Explanation:

Step 1: Data given

Mass of CO2 = 375 grams

Pressure of CO2 gas = 1.0 atm

Temperature of the gas = 298 K

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 375 grams / 44.01 g/mol

Moles CO2 = 8.52 moles

Step 3: Calculate volume of the gas

p*V= n*R*T

V = (n*R*T)/p

⇒with V = the volume of CO2 = TO BE DETERMINED

⇒with n = the number of moles CO2 = 8.52 moles CO2

⇒with R = the gas constant = 0.08206 L*atm / mol*K

⇒with T = the temperature = 298 K

⇒with p= the pressure : 1.0 atm

V = (8.52 * 0.08206* 298)/1.0

V = 208.35 L

Step 4: Calculate number of ballons that can be filled

Number of ballons = 208.35 L / 0.75 L

Number of balloons = 277.8

277 balloons can be filled with CO2

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Answer:

Hope this helps!

Explanation:

Ans: 15.1 grams

Given reaction:

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Mass of Na2CO3 = 20.0 g

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# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

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4 0
3 years ago
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Explanation:

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Explanation:

From the given information:

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= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

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\Delta \ T_f = 2.9 ^0 \ C

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= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

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