Chapter 6.3, Problem 42E

a.

To determine

To find the quantity of ferric oxide in a piece of red glass.

Answer to Problem 42E

The red glass containing $60\text{ grams}$ of sand contains $3.6\text{ grams}$ of ferric oxide.

Explanation of Solution

Given information:

The following table is given:

  

A piece of red glass contains $60\text{ grams}$ of sand.

Calculation :

The ratio of sand to ferric oxide is $50:3\text{ or }\frac{50}{3}$

Let the quantity of ferric oxide used be $\text{x grams}$ .

The quantity of ferric oxide used in red glass can be calculated by:

  $\begin{array}{l}\frac{50}{3}=\frac{\text{60}}{\text{x}}\\ 50\text{x}=180\left[\text{Cross products property}\right]\\ \text{x}=\frac{180}{50}\left[\text{Divide both sides by 50}\right]\\ \text{x}=3.6\text{ grams}\end{array}$

Hence, the red glass containing $60\text{ grams}$ of sand contains $3.6\text{ grams}$ of ferric oxide.

b.

To determine

To find the quantity of chalk in a piece of yellow glass.

Answer to Problem 42E

The yellow glass containing $\text{31 kilograms}$ of soda ash contains $3.72\text{ kilograms}$ of chalk.

Explanation of Solution

Given information:

The following table is given:

  

A piece of yellow glass contains $\text{31 kilograms}$ of soda ash.

Calculation :

The ratio of soda ash to chalk is $25:3\text{ or }\frac{25}{3}$

Let the quantity of chalk used be $\text{x grams}$ .

The quantity of chalk used in yellow glass can be calculated by:

  $\begin{array}{l}\frac{25}{3}=\frac{31}{\text{x}}\\ 25\text{x}=93\left[\text{Cross products property}\right]\\ \text{x}=\frac{93}{25}\left[\text{Divide both sides by 25}\right]\\ \text{x}=3.72\text{ kilograms}\end{array}$

Hence, the yellow glass containing $\text{31 kilograms}$ of soda ash contains $3.72\text{ kilograms}$ of chalk.

c.

To determine

To compare the quantity of sand in red glass and yellow glass.

Answer to Problem 42E

Yellow glass with $\text{4 grams}$ of wood charcoal contains more sand.

Explanation of Solution

Given information:

The following tables are given:

  

A piece of red glass contains $\text{200 grams}$ of red lead and a piece of yellow glass contains $\text{4 grams}$ of wood charcoal.

Calculation :

The ratio of red lead to sand is $100:50\text{ or }\frac{100}{50}$

Let the quantity of sand used in red glass be $\text{x grams}$

The quantity of sand in red glass can be calculated by:

  $\begin{array}{l}\frac{100}{50}=\frac{\text{200}}{\text{x}}\\ 100\text{x}=10000\left[\text{Cross products property}\right]\\ \text{x}=\frac{10000}{100}\left[\text{Divide both sides by 100}\right]\\ \text{x}=100\text{ grams}\end{array}$

Red glass has $100\text{ grams}$ of sand.

The ratio of wood charcoal to sand is $1:65\text{ or }\frac{1}{65}$

Let the quantity of sand used in yellow glass be $\text{x grams}$

The quantity of sand in yellow glass can be calculated by:

  $\begin{array}{l}\frac{1}{65}=\frac{4}{\text{x}}\\ \text{x}=4\times 65\left[\text{Cross products property}\right]\\ \text{x}=260\text{ grams }\end{array}$

Yellow glass has $260\text{ grams}$ of sand.

Hence, yellow glass with $\text{4 grams}$ of wood charcoal contains more sand.