How many grams of kbr are required to make 350. ml of a 0.115 m kbr solution?

Answer is: 4.79 grams og KBr are required.

V(KBr) = 350 mL ÷ 1000 mL/L.

V(KBr) = 0.350 L; volume of potassium bromide solution.

c(KBr) = 0.115 M; molarity of potassium bromide solution.

n(KBr) = c(KBr) · V(KBr).

n(KBr) = 0.115 mol/L · 0.350 L.

n(KBr) = 0.04025 mol, amount of substance.

m(KBr) = n(KBr) · M(KBr).

m(KBr) = 0.04025 mol · 119 g/mol.

m(KBr) = 4.79 g; mass of potassium bromide.

M - molar mass.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

i would say b

explanation:

but i think i would need more information than that