Answer:
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Explanation:
ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.
In our case we have:
mass of MgBr₂ = 12.41 g
volume of water (which is equal to the final solution volume) = 2.55 L
Now we devise the following reasoning:
if 12.41 g of MgBr₂ are dissolved in 2.55 L of water
then X g of MgBr₂ are dissolved in 1 L of water
X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂
if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻
then in 4.867 g of MgBr₂ we have Y g of Br⁻
Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)
4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
I got 0 ?? if that’s any help
Answer:
This cannot be determined without knowing the actual mass of the objects.
Explanation:
its like trying to compare the letter A and letter B
Hello!
The molarity of the HBr solution is 0,172 M.
Why?
The neutralization reaction between LiOH and HBr is the following:
HBr(aq) + LiOH(aq) → LiBr(aq) + H₂O(l)
To solve this exercise, we are going to apply the common titration equation:
Have a nice day!