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Lynna [10]
3 years ago
7

Please help with that problem Important İ have quiz tomorrow

Chemistry
1 answer:
iogann1982 [59]3 years ago
6 0

Answer:

Final pressure of the gas remaining in the first container is 3.5 atm

Explanation:

Since it is given that temperature is constant , we can apply -

PV=constant , where

P = Pressure of the gas

V = Volume of the container in which the gas is contained

Initially,

For container 1 -

P_{i} = 4 atm

V_{1} = 6 L

Finally,

For container 2 -

P_{f,2} = 3 atm

V_{2} = 1 L

For container 1 -

P_{f,1} = ?

V_{1} = 6 L

∴ P_{i}V_{1} = P_{f,1}V_{1} + P_{f,2}V_{2}

∴ 4×6 =  (P_{f,1}×6) +(3×1)

∴ P_{f,1} = 3.5 atm

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Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide
Rama09 [41]
<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

          = 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

                            = 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>
  • We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

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