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3 years ago

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How many moles of Cs are contained in 3 moles of Cs3PO4?

1 answer:

Alenkinab [10]3 years ago

6 0

Answer:

$\boxed {\boxed {\sf 9 \ moles \ Cs}}$

Explanation:

We are given the compound: Cs₃PO₄

According to the formula, 1 mole of cesium phosphate contains 3 moles of cesium, 1 mole of phosphate, and 4 moles of oxygen.

Therefore, there are 3 moles of cesium for 1 mole of cesium phosphate.

$\frac {3 \ mol \ Cs}{1 \ mol \ Cs_3PO_4}$

We want to calculate the moles of cesium in 3 moles of cesium phosphate, so we multiply the ratio by 3.

$3 \ mol \ Cs_3PO_4 *\frac {3 \ mol \ Cs}{1 \ mol \ Cs_3PO_4}$

$3 *\frac {3 \ mol \ Cs}{1 }= 9 \ mol \ Cs$

3 moles of cesium phosphate contains <u>9 moles of cesium.</u>

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3 years ago

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ivanzaharov [21]

Answer:

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Explanation:

From the question we are told that

The pressure of (NO) is $P_{NO} = 9.20 \ atm$

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The reaction is

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The Standard state free energy for $Cl_2$ is constant with a value

$\Delta G^o _{Cl_2} = 0kJ/mol$

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$\Delta G^o _{NOCl} =66.1kJ/mol$

Now substituting this into the equation

$\Delta G^o = 2 * 66.1 - 0 - 2 * 87.6$

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The pressure constant is evaluated as

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Substituting values

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$= 0.0765$

The free energy for this reaction is evaluated as

$\Delta G = \Delta G^o + RT ln Q$

Where R is gas constant with a value of $R = 8.314 J / K \cdot mol$

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Substituting value

$\Delta G = -43 *10^{3} + 8.314 *298 * ln [0.0765]$

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Answer:

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