Question

1 question need help please.

Answer 1

The minimum mass of nitrate ion present in a 250 mL cup of water that would be considered unsafe is 13,750 mg or 13.75 g.

To calculate the minimum mass of nitrate ion present in a 250 mL cup of water that would be considered unsafe, you can use the formula:

`\[ \text{Mass} = \text{Concentration} * \text{Volume} \]`

Given:

• Nitrate ion concentration = 55 ppm (parts per million)
• Volume of water = 250 mL

Convert ppm to mg/L (1 ppm = 1 mg/L):

`\[ \text{Concentration} = 55 \, \text{mg/L} \]`

Now, use the formula:

`\[ \text{Mass} = \text{Concentration} * \text{Volume} \]`

`\[ \text{Mass} = 55 \, \text{mg/L} * 250 \, \text{mL} \]`

`\[ \text{Mass} = 13750 \, \text{mg} \]`

Therefore, the minimum mass of nitrate ion present in a 250 mL cup of water that would be considered unsafe is 13,750 mg or 13.75 g.

Answer 2

The minimum mass of nitrate ions in a cup of water (250 mL) that would be considered unsafe for human consumption is 13750 mg or 13.75 grams.

How to find mass?

To calculate the minimum mass of nitrate ions in a cup of water (250 mL) that would be considered unsafe for human consumption, use the formula:

Mass = Concentration × Volume

Given:

Nitrate ion concentration = 55 ppm (parts per million)

Volume of water = 250 mL

First, convert ppm to g/L (grams per liter) since the volume is given in milliliters:

55 ppm = 55 mg/L

Now, calculate the mass:

Mass = Concentration× Volume

Mass = 55 mg/L × 250 mL

Mass = 13750 mg

So, the minimum mass of nitrate ions in a cup of water (250 mL) that would be considered unsafe for human consumption is 13750 mg or 13.75 grams.