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Slav-nsk [51]
3 years ago
15

A sample of ammonia gas has a volume of 3213 mL at 11°C and a pressure of 822 torr. What will the volume of the gas be in liters

if the
moles of gas and the temperature do not change but the pressure changes to 2.33 atm?
Chemistry
1 answer:
EleoNora [17]3 years ago
6 0

Answer:

<u>We are given:</u>

v1 = 3.2 L                   v2 = x L

p1 = 822 torr OR 1.08 atm        p2 = 2.33 atm

t1 = t2 = 284 k

<u></u>

<u>From the Ideal gas equation:</u>

PV = nRT

Since the number of moles (n), universal gas constant (R) and Temperature (T) are constant

PV = k      (where k is a constant)

it can also be written as:

P1V1 = k --------------------------(1)

Similarly,

P2V2 = k (where k is the same constant as before)----------(2)

<u>Solving for V2:</u>

From (1) and (2):

P1V1 = P2V2

Replacing the variables

1.08 * 3.2 = 2.33 * x

x = (1.08 * 3.2)/2.33

x = 1.5 L (approx)

Therefore, the final volume of the solution will be 1.5L

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2C3H7OH + 9O2 --&gt; 6CO2 + 8H2O
Ne4ueva [31]
<h3>Answer:</h3>

733 g CO₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
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  7. Subtraction
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<u>Atomic Structure</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced]   2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O

[Given]   5.55 mol C₃H₇OH

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol C₃H₇OH → 6 CO₂

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up conversion:                    \displaystyle 5.55 \ mol \ C_3H_7OH(\frac{6 \ mol \ CO_2}{2 \ mol \ C_3H_7OH})(\frac{44.01 \ g \ CO_2}{1 \ mol \ CO_2})
  2. Multiply/Divide:                                                                                               \displaystyle 732.767 \ g \ CO_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

732.767 g CO₂ ≈ 733 g CO₂

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