Answer:- 544.5 mL of water need to be added.
Solution:- It is a dilution problem. The equation used for solving this type of problems is:

where,
is initial molarity and
is the molarity after dilution. Similarly,
is the volume before dilution and
is the volume after dilution.
Let's plug in the values in the equation:



Volume of water added = 907.5mL - 363mL = 544.5 mL
So, 544.5 mL of water are need to be added to the original solution for dilution.
Answer:

Explanation:
Graham’s Law applies to the diffusion of gases:
The rate of diffusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

If you have two gases, the ratio of their rates of diffusion is

The time for diffusion is inversely proportional to the rate.

Data:
t₂ = 222 s
t₁ = 175 s
M₁ = 28.01
Calculation
:

I believe the answer is C
<span>c.
reactivity
this is a physical property depends on how reactive something is</span>
Answer is: <span>volume of 1 M NaOH is 1 ml.
</span>c₁(NaOH) = 1 M.
V₂(NaOH) = 10 ml.
c₂(NaOH) = 0,1 M.
V₁(NaOH) = ?
c₁ - original concentration of the solution, before it gets diluted.
c₂ - final concentration of the solution, after dilution.
V₁ - <span>volume to be diluted.
V</span>₂ - <span>final volume after dilution.
c</span>₁ · V₁ = c₂ · V₂.
V₁(NaOH) = c₂ · V₂ ÷ c₁.
V₁(NaOH) = 0,1 M · 10 ml ÷ 1 M.
V₁(NaOH) = 1 ml.