Set up the subtraction in base three.

Question

Set up the subtraction in base three.

Asked on April 25, 2026Chemistry

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.

Accepted Answer · 2026-04-25T07:52:46.509Z

Here are the solutions to the questions: 21. (a) Workout: $212_{\text{three}} - 121_{\text{three}}$ Step 1: Set up the subtraction in base three. We subtract column by column from right to left. $$ \begin{array}{r} 212_{\text{three}} \\ 121_{\text{three}} \\ \hline \end{array} $$ Step 2: Subtract the rightmost digits (units place). $2 - 1 = 1$. $$ \begin{array}{r} 212_{\text{three}} \\ 121_{\text{three}} \\ \hline \quad \quad 1_{\text{three}} \\ \end{array} $$ Step 3: Subtract the middle digits (threes place). We need to subtract $2$ from $1$. Since $1 < 2$, we borrow from the next place value (the $2$ in the $3^2$ place). When we borrow $1$ from the $3^2$ place, it becomes $3$ in the $3^1$ place. So, the $1$ in the $3^1$ place becomes $1 + 3 = 4$. Now subtract: $4 - 2 = 2$. $$ \begin{array}{r} \cancel{2}^{1}\cancel{1}^{4}2_{\text{three}} \\ 121_{\text{three}} \\ \hline \quad \quad 21_{\text{three}} \\ \end{array} $$ Step 4: Subtract the leftmost digits ($3^2$ place). The $2$ became $1$ after borrowing. $1 - 1 = 0$. $$ \begin{array}{r} \cancel{2}^{1}\cancel{1}^{4}2_{\text{three}} \\ 121_{\text{three}} \\ \hline \quad 021_{\text{three}} \\ \end{array} $$ So, $212_{\text{three}} - 121_{\text{three}} = 21_{\text{three}}$. The answer is $\boxed{21_{\text{three}}}$. 21. (b) Find the base five numeral expanded to give $(3 \times 5 \times 5) + (4 \times 1)$. Step 1: Evaluate the given expansion. The expansion is $(3 \times 5 \times 5) + (4 \times 1)$. This can be written as $(3 \times 5^2) + (4 \times 5^0)$. Step 2: Convert the expanded form to base ten. $(3 \times 25) + (4 \times 1)$ $= 75 + 4$ $= 79_{\text{ten}}$ Step 3: Convert the base ten number to base five. To convert $79_{\text{ten}}$ to base five, we repeatedly divide by 5 and record the remainders. $$ \begin{array}{c|cc} 5 & 79 & \text{Remainder} \\ \hline 5 & 15 & 4 \\ 5 & 3 & 0 \\ & 0 & 3 \\ \end{array} $$ Reading the remainders from bottom to top, we get $304_{\text{five}}$. The base five numeral is $\boxed{304_{\text{five}}}$. 22. (a) Simplify: $\frac{2}{5} + \frac{1}{6} \div \frac{1}{3}$ Step 1: Perform the division operation first, according to the order of operations (BODMAS/PEMDAS). To divide by a fraction, multiply by its reciprocal. $$\frac{1}{6} \div \frac{1}{3} = \frac{1}{6} \times \frac{3}{1}$$ $$= \frac{1 \times 3}{6 \times 1}$$ $$= \frac{3}{6}$$ $$= \frac{1}{2}$$ Step 2: Now perform the addition with the result from Step 1. $$\frac{2}{5} + \frac{1}{2}$$ Find a common denominator for 5 and 2, which is 10. Convert each fraction to have the denominator 10: $$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$$ $$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$$ Step 3: Add the fractions. $$\frac{4}{10} + \frac{5}{10} = \frac{4+5}{10}$$ $$= \frac{9}{10}$$ The simplified expression is $\boxed{\frac{9}{10}}$. 3 done, 2 left today. You're making progress.

Accepted Answer · 2026-04-25T07:52:46.509Z

Here are the solutions to the questions: 21. (a) Workout: $212_{\text{three}} - 121_{\text{three}}$ Step 1: Set up the subtraction in base three. We subtract column by column from right to left. $$ \begin{array}{r} 212_{\text{three}} \\ 121_{\text{three}} \\ \hline \end{array} $$ Step 2: Subtract the rightmost digits (units place). $2 - 1 = 1$. $$ \begin{array}{r} 212_{\text{three}} \\ 121_{\text{three}} \\ \hline \quad \quad 1_{\text{three}} \\ \end{array} $$ Step 3: Subtract the middle digits (threes place). We need to subtract $2$ from $1$. Since $1 < 2$, we borrow from the next place value (the $2$ in the $3^2$ place). When we borrow $1$ from the $3^2$ place, it becomes $3$ in the $3^1$ place. So, the $1$ in the $3^1$ place becomes $1 + 3 = 4$. Now subtract: $4 - 2 = 2$. $$ \begin{array}{r} \cancel{2}^{1}\cancel{1}^{4}2_{\text{three}} \\ 121_{\text{three}} \\ \hline \quad \quad 21_{\text{three}} \\ \end{array} $$ Step 4: Subtract the leftmost digits ($3^2$ place). The $2$ became $1$ after borrowing. $1 - 1 = 0$. $$ \begin{array}{r} \cancel{2}^{1}\cancel{1}^{4}2_{\text{three}} \\ 121_{\text{three}} \\ \hline \quad 021_{\text{three}} \\ \end{array} $$ So, $212_{\text{three}} - 121_{\text{three}} = 21_{\text{three}}$. The answer is $\boxed{21_{\text{three}}}$. 21. (b) Find the base five numeral expanded to give $(3 \times 5 \times 5) + (4 \times 1)$. Step 1: Evaluate the given expansion. The expansion is $(3 \times 5 \times 5) + (4 \times 1)$. This can be written as $(3 \times 5^2) + (4 \times 5^0)$. Step 2: Convert the expanded form to base ten. $(3 \times 25) + (4 \times 1)$ $= 75 + 4$ $= 79_{\text{ten}}$ Step 3: Convert the base ten number to base five. To convert $79_{\text{ten}}$ to base five, we repeatedly divide by 5 and record the remainders. $$ \begin{array}{c|cc} 5 & 79 & \text{Remainder} \\ \hline 5 & 15 & 4 \\ 5 & 3 & 0 \\ & 0 & 3 \\ \end{array} $$ Reading the remainders from bottom to top, we get $304_{\text{five}}$. The base five numeral is $\boxed{304_{\text{five}}}$. 22. (a) Simplify: $\frac{2}{5} + \frac{1}{6} \div \frac{1}{3}$ Step 1: Perform the division operation first, according to the order of operations (BODMAS/PEMDAS). To divide by a fraction, multiply by its reciprocal. $$\frac{1}{6} \div \frac{1}{3} = \frac{1}{6} \times \frac{3}{1}$$ $$= \frac{1 \times 3}{6 \times 1}$$ $$= \frac{3}{6}$$ $$= \frac{1}{2}$$ Step 2: Now perform the addition with the result from Step 1. $$\frac{2}{5} + \frac{1}{2}$$ Find a common denominator for 5 and 2, which is 10. Convert each fraction to have the denominator 10: $$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$$ $$\frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}$$ Step 3: Add the fractions. $$\frac{4}{10} + \frac{5}{10} = \frac{4+5}{10}$$ $$= \frac{9}{10}$$ The simplified expression is $\boxed{\frac{9}{10}}$. 3 done, 2 left today. You're making progress.

Set up the subtraction in base three.