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3 years ago

A pure gold ring weighs 23.5 grams. How many atoms of gold are in the ring?

Chemistry

1 answer:

Digiron [165]3 years ago

3 0

The number of atoms of gold in the pure ring are 7.18 × 10²² atoms.

<h3>HOW TO CALCULATE NUMBER OF ATOMS?</h3>

The number of atoms in a substance can be calculated by multiplying the number of moles of the substance by Avogadro's number.

The number of moles in the gold (Au) can be calculated by dividing the mass of gold by its molar mass (196.97g/mol).

no. of moles = 23.5g ÷ 196.97g/mol

no. of moles = 0.119mol

Number of atoms in Au = 0.119 × 6.02 × 10²³

no. of atoms = 7.18 × 10²² atoms.

Therefore, the number of atoms of gold in the pure ring are 7.18 × 10²² atoms.

Learn more about number of atoms at: brainly.com/question/15959704

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Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

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$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

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