8. a) Isotopes are atoms of the same element that have the same number of protons (atomic number) but different numbers of neutrons (mass number). From the given data, the atomic number for element
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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8. a) Isotopes are atoms of the same element that have the same number of protons (atomic number) but different numbers of neutrons (mass number). From the given data, the atomic number for element K is 6.
The isotopes of element K are:
$$ \boxed{^{12}_{6}\text{K}, ^{13}_{6}\text{K}, ^{14}_{6}\text{K}} $$
b) To calculate the relative atomic mass of element A, we use the formula:
$$ \text{Relative Atomic Mass} = \sum (\text{Isotope mass} \times \frac{\text{% abundance}}{100}) $$
Step 1: Substitute the values for isotope A1.
$$ \text{Contribution from A1} = 62.93 \times \frac{69.09}{100} = 62.93 \times 0.6909 = 43.471837 $$
Step 2: Substitute the values for isotope A2.
$$ \text{Contribution from A2} = 64.93 \times \frac{30.91}{100} = 64.93 \times 0.3091 = 20.069123 $$
Step 3: Add the contributions to find the relative atomic mass.
$$ \text{Relative Atomic Mass of A} = 43.471837 + 20.069123 = 63.54096 $$
Rounding to two decimal places:
$$ \text{Relative Atomic Mass of A} = \boxed{63.54} $$
9. Step 1: Calculate the mass of carbon in 4.2 g of carbon (IV) oxide ($\text{CO}_2$).
Molar mass of C = 12.0 g/mol
Molar mass of $\text{CO}_2 = 12.0 + (2 \times 16.0) = 44.0 \text{ g/mol}$
$$ \text{Mass of C} = \frac{\text{Mass of C in CO}_2}{\text{Molar mass of CO}_2} \times \text{Mass of CO}_2 $$
$$ \text{Mass of C} = \frac{12.0}{44.0} \times 4.2 \text{ g} = 0.2727 \times 4.2 \text{ g} \approx 1.1454 \text{ g} $$
Step 2: Calculate the mass of hydrogen in 1.71 g of water ($\text{H}_2\text{O}$).
Molar mass of H = 1.0 g/mol
Molar mass of $\text{H}_2\text{O} = (2 \times 1.0) + 16.0 = 18.0 \text{ g/mol}$
$$ \text{Mass of H} = \frac{\text{Mass of 2H in H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} \times \text{Mass of H}_2\text{O} $$
$$ \text{Mass of H} = \frac{2 \times 1.0}{18.0} \times 1.71 \text{ g} = \frac{2.0}{18.0} \times 1.71 \text{ g} = 0.1111 \times 1.71 \text{ g} \approx 0.1900 \text{ g} $$
Step 3: Calculate the moles of carbon and hydrogen.
$$ \text{Moles of C} = \frac{\text{Mass of C}}{\text{Molar mass of C}} = \frac{1.1454 \text{ g}}{12.0 \text{ g/mol}} \approx 0.09545 \text{ mol} $$
$$ \text{Moles of H} = \frac{\text{Mass of H}}{\text{Molar mass of H}} = \frac{0.1900 \text{ g}}{1.0 \text{ g/mol}} = 0.1900 \text{ mol} $$
Step 4: Determine the simplest whole-number ratio of moles.
Divide by the smallest number of moles (0.09545 mol).
For C: $\frac{0.09545}{0.09545} = 1$
For H: $\frac{0.1900}{0.09545} \approx 1.99 \approx 2$
The ratio of C:H is approximately 1:2.
The empirical formula of the hydrocarbon is $\boxed{\text{CH}_2}$.
10. i) To obtain pure water from a mixture of water and common salt, the best method is distillation.
