Answer :
Let's work through the problem step-by-step to understand the mass of one mole of M&Ms both in grams and in tons.
### Step 1: Understand the given data
- Mass of a single M&M: 0.915 grams
- Avogadro's number (the number of units in one mole): [tex]\(6.022 \times 10^{23}\)[/tex]
- Conversion factors:
- 1 ton = 2000 pounds (lbs)
- 1 pound (lb) = 454 grams (g)
### Step 2: Calculate the mass of one mole of M&Ms in grams
To find the mass of one mole of M&Ms, we multiply the mass of a single M&M by Avogadro's number:
[tex]\[ \text{Mass of one mole of M&Ms} = 0.915 \,\text{grams} \times 6.022 \times 10^{23} \,\text{M&Ms/mole} \][/tex]
The result is:
[tex]\[ 0.915 \times 6.022 \times 10^{23} = 5.51013 \times 10^{23} \,\text{grams} \][/tex]
So, the mass of one mole of M&Ms is [tex]\(5.51013 \times 10^{23}\)[/tex] grams.
### Step 3: Convert the mass from grams to tons
First, we need to understand how many grams are in one ton.
[tex]\[ 1 \,\text{ton} = 2000 \,\text{lbs} \times 454 \,\text{grams/lb} \][/tex]
Calculating this gives:
[tex]\[ 1 \,\text{ton} = 2000 \times 454 = 908,000 \,\text{grams} \][/tex]
Now, we use this conversion factor to find the mass of one mole of M&Ms in tons. We divide the mass in grams by the number of grams per ton:
[tex]\[ \text{Mass in tons} = \frac{5.51013 \times 10^{23} \,\text{grams}}{908,000 \,\text{grams/ton}} \][/tex]
The result is:
[tex]\[ \frac{5.51013 \times 10^{23}}{908,000} \approx 6.068425110132159 \times 10^{17} \,\text{tons} \][/tex]
### Summary
Here's the summary of our calculations:
| One mole of M&Ms in grams | One mole of M&Ms in tons |
|------------------------------|-------------------------------------|
| [tex]\(5.51013 \times 10^{23}\,g\)[/tex] | [tex]\(6.068425110132159 \times 10^{17}\,tons\)[/tex] |
### Show work:
1. Calculate mass of one mole of M&Ms:
[tex]\[ 0.915 \times 6.022 \times 10^{23} = 5.51013 \times 10^{23} \,\text{grams} \][/tex]
2. Convert grams to tons:
[tex]\[ \frac{5.51013 \times 10^{23}}{908,000} \approx 6.068425110132159 \times 10^{17} \,\text{tons} \][/tex]
This gives us the mass of one mole of M&Ms both in grams and in tons.
### Step 1: Understand the given data
- Mass of a single M&M: 0.915 grams
- Avogadro's number (the number of units in one mole): [tex]\(6.022 \times 10^{23}\)[/tex]
- Conversion factors:
- 1 ton = 2000 pounds (lbs)
- 1 pound (lb) = 454 grams (g)
### Step 2: Calculate the mass of one mole of M&Ms in grams
To find the mass of one mole of M&Ms, we multiply the mass of a single M&M by Avogadro's number:
[tex]\[ \text{Mass of one mole of M&Ms} = 0.915 \,\text{grams} \times 6.022 \times 10^{23} \,\text{M&Ms/mole} \][/tex]
The result is:
[tex]\[ 0.915 \times 6.022 \times 10^{23} = 5.51013 \times 10^{23} \,\text{grams} \][/tex]
So, the mass of one mole of M&Ms is [tex]\(5.51013 \times 10^{23}\)[/tex] grams.
### Step 3: Convert the mass from grams to tons
First, we need to understand how many grams are in one ton.
[tex]\[ 1 \,\text{ton} = 2000 \,\text{lbs} \times 454 \,\text{grams/lb} \][/tex]
Calculating this gives:
[tex]\[ 1 \,\text{ton} = 2000 \times 454 = 908,000 \,\text{grams} \][/tex]
Now, we use this conversion factor to find the mass of one mole of M&Ms in tons. We divide the mass in grams by the number of grams per ton:
[tex]\[ \text{Mass in tons} = \frac{5.51013 \times 10^{23} \,\text{grams}}{908,000 \,\text{grams/ton}} \][/tex]
The result is:
[tex]\[ \frac{5.51013 \times 10^{23}}{908,000} \approx 6.068425110132159 \times 10^{17} \,\text{tons} \][/tex]
### Summary
Here's the summary of our calculations:
| One mole of M&Ms in grams | One mole of M&Ms in tons |
|------------------------------|-------------------------------------|
| [tex]\(5.51013 \times 10^{23}\,g\)[/tex] | [tex]\(6.068425110132159 \times 10^{17}\,tons\)[/tex] |
### Show work:
1. Calculate mass of one mole of M&Ms:
[tex]\[ 0.915 \times 6.022 \times 10^{23} = 5.51013 \times 10^{23} \,\text{grams} \][/tex]
2. Convert grams to tons:
[tex]\[ \frac{5.51013 \times 10^{23}}{908,000} \approx 6.068425110132159 \times 10^{17} \,\text{tons} \][/tex]
This gives us the mass of one mole of M&Ms both in grams and in tons.