To what volume should you dilute 55 mL of 12 M stock HNO3 solution to obtain a 0.145 HNO3 solution?

4552 mL

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V₁) = 55 mL

Molarity of stock solution (M₁) = 12 M

Molarity of diluted solution (M₂) = 0.145 M

Volume of diluted solution (V₂) =?

The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

12 × 55 = 0.145 × V₂

660 = 0.145 × V₂

Divide both side by 0.145

V₂ = 660 / 0.145

V₂ ≈ 4552 mL

Thus, the volume of the diluted solution is 4552 mL

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

water is used to store them because it would keep the radiation from getting out in the open and causing harm to the workers