The answer would be
C) To reduce destruction by providing early warning of severe weather
This can help us because then we can plan our day-to-day activities better and can definitely help keep us out of danger.
Compete Question:
What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1?
Passage: "16 mmol of CDP in 1 L of buffer"
Answer:
6.4 × 10-2 g
Explanation:

we are given from the question that 16 mmol of CDP is in 1 L of buffer
this mean that we have
moles of CDP in 1 liter of buffer.
so the mass of CDP in one liter of buffer will be calculate as,
mass of CDP =
× 403g mol−1
=
= 6.4 g/L
But because the question
asks us about the mass of CDP in 10 mL of solution, we will go further to calculate it like this:
6.4 g/L × 10 mL
6.4 g/L × 0.01 L =
Copper (II) Sulfate, if you mean nomenclature
Answer:
It's coefficient to the front of each element that requires it.
Explanation:
It is coefficient to the front of each element or compound that requires it. Essentially you are multiplying the amount of atoms or compounds on one side to match the amount on the other side.