Calculate the molarity of bromide ions in 250. mL of a solution containing 25.9 g NaBr and 0.155 moles of HBr.

Explanation:

From the question we were given 0.155 moles of HBr, but Br and H are in ratio 1:1, then there are 0.155 moles of Br- ions.

We were also told that the solution contain NaBr, of 25.9 g. Then it must be converted to moles.

molar mass of NaBr =(22.99g + 79.90 )

= 102.89 g per mol.

the moles of NaBr can be calculated as 25.9 / 102.89

=0.252 moles

But Na and Br are in a ratio 1:1 , then there are 0.252 moles of Br-.

Then to get two Br- mol , we will add the first and second mol of Br- together

= 0.155 + 0.252

=0.407 moles.

The given solution has volume of 250 mL, but we know that there are 1000 ml in a liter, then if we convert to L for unit consistency we have

= 250/1000

= 0.25 L

molarity=0.407 moles/0.25 L

= 1.628 M.

Therefore, Br ion molarity is 1.628 M.

i know insulin is primary but i dont know the others im sorry

explanation: