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Karo-lina-s [1.5K]
3 years ago
12

What is the pressure of 0.540 mol of an ideal gas at 35.5 L and 223 K? Use mc031-1.jpg and mc031-2.jpg.

Chemistry
2 answers:
Reil [10]3 years ago
8 0
28.2 kPa -----------
Orlov [11]3 years ago
7 0

Answer: The pressure of an ideal gas 0.278 atm.

Explanation:

Volume of the ideal gas ,V = 35.5 L

Pressure of an ideal gas,P = 0.138 atm

Temperature of an ideal gas = 223 K

Number of moles of an ideal gas = 0.540 mol

Ideal gas equation is given as:

PV=nRT

P=\frac{n}{V}\times RT=\frac{0.540 mol}{35.5 L}\times 0.0820 atm l/mol K\times 223 K

P = 0.278 atm

The pressure of an ideal gas 0.278 atm.

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This problem is being solved using Ideal Gas Equation.
                                                   PV  =  nRT
Data Given:
                  Initial Temperature = T₁ = 27 °C = 300 K
                  Initial Pressure      =  P₁ = constant
                  Initial Volume         = V₁ = 8 L
                   Final Temperature = T₂ = 78 °C = 351 K
                  Final Pressure      =  P₂ = constant
                  Final Volume         = V₂ = ?
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Gas constant R and Pressures are constant, so, Ideal gas equation can be written as,
                                          V₁ / T₁  =  V₂ / T₂
Solving for V₂,
                                           V₂  =  (V₁ × T₂) ÷ T₁
Putting Values,
                                           V₂  =  (8 L × 351 K) ÷ 300 K

                                           V₂  =  9.38 L
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snow_lady [41]

Answer:

1. Mass of KCl produced = 774.8 g of KCl

2. Mass of KNO₃ produced = 13.837g

3. Moles of NaOH made = 0.846 moles

4. Moles of LiCl produced = 0.846 moles

5. Moles of CO₂ produced = 207.6 moles

Explanation:

1. From the equation of reaction, 1 mole of ZnCl₂ produces, 2 moles of KCl.

5.02 moles of ZnCl₂ will produce, 2 × 5.02 moles of KCl = 10.4 moles of KCl

Molar mass of KCl = (39 + 35.5) g/mol = 74.5 g/mol

10.4 moles of KCl = 10.4 × 74.5 g

Mass of KCl produced = 774.8 g of KCl

2. Mole ratio of KNO₃ and KOH = 1:1

O.137 moles of KOH will produce 0.137 moles of KNO₃

Molar mass of KNO₃ = 101 g/mol

Mass of KNO₃ produced = 0.137 × 101 g = 13.837g

3. Molar mas of Ca(OH)₂ = 74.0 g

Moles of Ca(OH)₂ in 31.3 g = 31.3/74.0 = 0.423 moles of Ca(OH)₂

Mole ratio of NaOH and Ca(OH)₂ in the reaction = 2 : 1

Moles of NaOH made = 2 × 0.423 = 0.846 moles

4. Molar mass of MgCl₂ = 95.0 g

Moles of MgCl₂ in 40.2 g = 40.2/95.0 = 0.423 moles

From the reaction equation, mole ratio of MgCl₂ and LiCl = 1:2

Moles of LiCl produced = 2 × 0.423 = 0.846 moles

5. From the equation of reaction, 1 mole of C₆H₁₀O₅ produces 6 moles of cO₂

34.6 moles of C₆H₁₀O₅ will produce 34.6 × 6 moles of CO₂

Moles of CO₂ produced = 207.6 moles

4 0
2 years ago
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