Answer:its A ( the transport of nutrients,water,oxygen to body cells
Explanation:
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<u>Answer:</u> The amount of water required to prepare given amount of salt is 398.4 mL
<u>Explanation:</u>
To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

We are given:
Molarity of solution = 0.16 M
Given mass of manganese (II) nitrate tetrahydrate = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Putting values in above equation, we get:

Volume of water = Volume of solution = 398.4 mL
Hence, the amount of water required to prepare given amount of salt is 398.4 mL
Step 1: write the equation:
P₄(s) + 6F₂(g) → 4PF₃(g)
Step 2: Molar mass of P₄ = 30.97 g/mol × 4 = 123.88 g/mol
Step 3: Number of moles of phosphorus
n = m/M
n = 8.5 g/123.88g/mol
n = 0.07 moles
Step 4: 0.07 × 12 = 0.84 moles of fluorine.
Fluorine is diatomic gas so we multiplied the number of moles by 12.
Step 5: To find the mass of fluorine we multiply the number of moles with the molar mass.
Mass of fluorine = 0.84 × 228
= 191.52 grams.