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2 years ago

Find gas pressure of 2.5 moles H2 and 4.6 moles Ne at 1400mmHg.

Chemistry

1 answer:

Monica [59]2 years ago

5 0

Answer:

Pressure of neon = 1.17 atm

Pressure of hydrogen = 0.63 atm

Explanation:

Given data:

Number of moles of H₂ = 2.5 mol

Number of moles of Ne = 4.6 mol

Total pressure = 1400 mmHg (1400/760 = 1.8 atm)

Solution:

We will solve this problem through mole fraction method.

Total number of moles = 2.5 + 4.6 = 7.1

Mole fraction of H₂ Mole fraction of Ne

2.5 mol/7.1 mol= 0.35 4.6mol/7.1mol = 0.65

Pressure of hydrogen = 0.35 × 1.8 atm

Pressure of hydrogen = 0.63 atm

Pressure of neon = 0.65 × 1.8 atm

Pressure of neon = 1.17 atm

Total pressure which is already given is sum of partial pressure of hydrogen and neon gas.

0.63 atm + 1.17 atm = 1.8 atm

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Need help with this can someone help me pls the word box are words your supposed to use. Will mark brainiest btw

dybincka [34]

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How about let's just forget about that other stuff and be friends?

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2 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

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Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

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$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

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$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

$n_{(AcO)_2Cu} = 0.0053515 mol$

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

$n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol$

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

$c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M$

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nexus9112 [7]

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