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Sagot :
Let's break down the solution step-by-step for each of the questions.
### 1. Mass Percent of Chromium in the River
First, we need to convert the mass of chromium from milligrams (mg) to grams (g):
[tex]\[ 52 \, \text{mg} = 52 \div 1000 = 0.052 \, \text{g} \][/tex]
Next, we find the mass percent of chromium in the river water:
[tex]\[ \text{Mass percent} = \left(\frac{\text{mass of chromium}}{\text{mass of sample}}\right) \times 100 \][/tex]
[tex]\[ \text{Mass percent} = \left(\frac{0.052 \, \text{g}}{780.0 \, \text{g}}\right) \times 100 \][/tex]
[tex]\[ \text{Mass percent} = 0.006666666666666667 \, \% \][/tex]
Expressed in scientific notation:
[tex]\[ \text{Mass percent} = 6.67 \times 10^{-3} \, \% \][/tex]
### 2. Parts Per Million (ppm) of Chromium
To find the parts per million (ppm) of chromium:
[tex]\[ \text{ppm} = \left(\frac{\text{mass of chromium}}{\text{mass of sample}}\right) \times 1,000,000 \][/tex]
[tex]\[ \text{ppm} = \left(\frac{0.052 \, \text{g}}{780.0 \, \text{g}}\right) \times 1,000,000 \][/tex]
[tex]\[ \text{ppm} = 66.66666666666667 \][/tex]
Expressed in scientific notation:
[tex]\[ \text{ppm} = 6.67 \times 10^{1} \, \text{ppm} \][/tex]
### 3. Parts Per Billion (ppb) of Chromium
To find the parts per billion (ppb) of chromium:
[tex]\[ \text{ppb} = \left(\frac{\text{mass of chromium}}{\text{mass of sample}}\right) \times 1,000,000,000 \][/tex]
[tex]\[ \text{ppb} = \left(\frac{0.052 \, \text{g}}{780.0 \, \text{g}}\right) \times 1,000,000,000 \][/tex]
[tex]\[ \text{ppb} = 66666.66666666667 \][/tex]
Expressed in scientific notation:
[tex]\[ \text{ppb} = 6.67 \times 10^{4} \, \text{ppb} \][/tex]
### Summary of Answers:
1. Mass Percent: [tex]\(\boxed{6.67} \times 10^{-3} \, \% \)[/tex]
2. Parts Per Million: [tex]\(\boxed{6.67} \times 10^{1} \, \text{ppm} \)[/tex]
3. Parts Per Billion: [tex]\(\boxed{6.67} \times 10^{4} \, \text{ppb} \)[/tex]
### 1. Mass Percent of Chromium in the River
First, we need to convert the mass of chromium from milligrams (mg) to grams (g):
[tex]\[ 52 \, \text{mg} = 52 \div 1000 = 0.052 \, \text{g} \][/tex]
Next, we find the mass percent of chromium in the river water:
[tex]\[ \text{Mass percent} = \left(\frac{\text{mass of chromium}}{\text{mass of sample}}\right) \times 100 \][/tex]
[tex]\[ \text{Mass percent} = \left(\frac{0.052 \, \text{g}}{780.0 \, \text{g}}\right) \times 100 \][/tex]
[tex]\[ \text{Mass percent} = 0.006666666666666667 \, \% \][/tex]
Expressed in scientific notation:
[tex]\[ \text{Mass percent} = 6.67 \times 10^{-3} \, \% \][/tex]
### 2. Parts Per Million (ppm) of Chromium
To find the parts per million (ppm) of chromium:
[tex]\[ \text{ppm} = \left(\frac{\text{mass of chromium}}{\text{mass of sample}}\right) \times 1,000,000 \][/tex]
[tex]\[ \text{ppm} = \left(\frac{0.052 \, \text{g}}{780.0 \, \text{g}}\right) \times 1,000,000 \][/tex]
[tex]\[ \text{ppm} = 66.66666666666667 \][/tex]
Expressed in scientific notation:
[tex]\[ \text{ppm} = 6.67 \times 10^{1} \, \text{ppm} \][/tex]
### 3. Parts Per Billion (ppb) of Chromium
To find the parts per billion (ppb) of chromium:
[tex]\[ \text{ppb} = \left(\frac{\text{mass of chromium}}{\text{mass of sample}}\right) \times 1,000,000,000 \][/tex]
[tex]\[ \text{ppb} = \left(\frac{0.052 \, \text{g}}{780.0 \, \text{g}}\right) \times 1,000,000,000 \][/tex]
[tex]\[ \text{ppb} = 66666.66666666667 \][/tex]
Expressed in scientific notation:
[tex]\[ \text{ppb} = 6.67 \times 10^{4} \, \text{ppb} \][/tex]
### Summary of Answers:
1. Mass Percent: [tex]\(\boxed{6.67} \times 10^{-3} \, \% \)[/tex]
2. Parts Per Million: [tex]\(\boxed{6.67} \times 10^{1} \, \text{ppm} \)[/tex]
3. Parts Per Billion: [tex]\(\boxed{6.67} \times 10^{4} \, \text{ppb} \)[/tex]
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