Question #65722

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Question #65722

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Answer 1 · 2016-02-20T20:59:14

$100 g$ $"100 g"$

Explanation:

You are on the right track with this one.

Use the known ratio for the first mixture of $A$ $"A"$ , $B$ $"B"$ , and $C$ $"C"$ to determine the mass of each chemical in the $100-g$ $"100-g"$ sample.

If you take $m_{1}$ $m_1$ to be the mass of chemical $A$ $"A"$ , you can say that

$3 \times m_{1} \to$ $3 xx m_1 ->$ the mass of chemical $B$ $"B"$

$1 \times m_{1} \to$ $1 xx m_1 ->$ the mass of chemical $C$ $"C"$

This means that you have

$m_{1} + 3 \cdot m_{1} + m_{1} = 100 g$ $m_1 + 3 * m_1 + m_1 = "100 g"$

$5 \cdot m_{1} = 100 g \Rightarrow m_{1} = \frac{100 g}{5} = 20 g$ $5 * m_1 = "100 g" implies m_1 = "100 g"/5 = "20 g"$

Do the same for the second $100-g$ $"100-g"$ mixture of $A$ $"A"$ , $B$ $"B"$ , and $C$ $"C"$ . Let's say that $m_{2}$ $m_2$ is the mass of chemical $A$ $"A"$ in this mixture. You will have

$2 \times m_{2} \to$ $2 xx m_2 ->$ the mass of chemical $B$ $"B"$

$7 \times m_{2} \to$ $7 xx m_2 ->$ the mass of chemical $C$ $"C"$

This means that you have

$m_{2} + 2 \cdot m_{2} + 7 \cdot m_{2} = 100 g$ $m_2 + 2 * m_2 + 7 * m_2 = "100 g"$

$10 \cdot m_{2} = 100 g \Rightarrow m_{2} = 10 g$ $10 * m_2 = "100 g" implies m_2 = "10 g"$

So, you're adding these two mixture together. The masses of the three chemicals will be

$m_{A} = 20 g + 10 g = 30 g$ $m_A = "20 g" + "10 g" = "30 g"$

$m_{B} = (3 \times 20 g) + (2 \times 10 g) = 80 g$ $m_B = (3 xx "20 g") + (2 xx "10 g") = "80 g"$

$m_{C} = 20 g + (7 \times 10 g) = 90 g$ $m_C = "20 g" + (7 xx "10 g") = "90 g"$

Now, you know that the ratio for the three chemicals in the final mixture must be $1 : 3 : 6$ $1:3:6$ . Keep this in mind.

Let's say that $x$ $x$ represents the mass of chemical $B$ $"B"$ in the third mixture, the one that contains $B$ $"B"$ and $C$ $"C"$ in a $1 : 9$ $1:9$ ratio.

This means that the mass of $C$ $"C"$ will be $(9 \times x)$ $(9 xx x)$ , and the total mass of the third mixture will be

$x + (9 \times x) = 10 x$ $x + (9 xx x) = 10x$

Now, the mass of chemical $A$ $"A"$ will not change when you add this third mixture, since it only contains $B$ $"B"$ and $C$ $"C"$ . This means that the final mass of $A$ $"A"$ will be

$m_{A} = 30 g$ $m_A = "30 g"$

Now look at the $1 : 3 : 6$ $1:3:6$ ratio that must exist between $A$ $"A"$ , $B$ $"B"$ , and $C$ $"C"$ . According to this ratio, the final mixture will contain

$3 \times 30 g = 90 g \to$ $3 xx "30 g" = "90 g" ->$ the final mass of chemical $B$ $"B"$

$6 \times 30 g = 180 g \to$ $6 xx "30 g" = "180 g" ->$ the final mass of chemical $C$ $"C"$

This means that you can say, using the mass of $B$ $"B"$ in the third mixture, $x$ $x$ , and the mass of $B$ $"B"$ in the first two mixtures, that

$x + 80 g = 90 g \Rightarrow x = 10 g$ $x + "80 g" = "90 g" implies x = "10 g"$

The same can be said for $C$ $"C"$

$9 x + 90 g = 180 g \Rightarrow x = 10 g$ $9x + "90 g" = "180 g" implies x = "10 g"$

Therefore, the third mixture must contain $10 g$ $"10 g"$ of $B$ $"B"$ and $90 g$ $"90 g"$ of $C$ $"C"$ . The total mass of this third mixture must thus be

$m_{third mixture} = 10 g + 90 g = 100 g$ $m_"third mixture" = "10 g" + "90 g" = color(green)("100 g")$

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Question #65722

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