# Question #65722

##### 1 Answer

#### Explanation:

You are on the right track with this one.

Use the known ratio for the first mixture of

If you take

#3 xx m_1 -># the mass of chemical#"B"#

#1 xx m_1 -># the mass of chemical#"C"#

This means that you have

#m_1 + 3 * m_1 + m_1 = "100 g"#

#5 * m_1 = "100 g" implies m_1 = "100 g"/5 = "20 g"#

Do the same for the second

#2 xx m_2 -># the mass of chemical#"B"#

#7 xx m_2 -># the mass of chemical#"C"#

This means that you have

#m_2 + 2 * m_2 + 7 * m_2 = "100 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_B = (3 xx "20 g") + (2 xx "10 g") = "80 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

Let's say that **third mixture**, the one that contains

This means that the mass of **total mass** of the third mixture will be

#x + (9 xx x) = 10x#

Now, the mass of chemical **not change** when you add this third mixture, since it only contains **final mass** of

#m_A = "30 g"#

Now look at the **final mixture** will contain

#3 xx "30 g" = "90 g" -># the final mass of chemical#"B"#

#6 xx "30 g" = "180 g" -># the final mass of chemical#"C"#

This means that you can say, using the mass of

#x + "80 g" = "90 g" implies x = "10 g"#

The same can be said for

#9x + "90 g" = "180 g" implies x = "10 g"#

Therefore, the third mixture must contain

#m_"third mixture" = "10 g" + "90 g" = color(green)("100 g")#