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
<span>1.86 moles of hydrogen gas.
Since what the HCl is reacting with hasn't been mentioned, I'll assume zine. In that case, the balanced reaction is
Zn + 2HCl ==> ZnCl2 + H2
So for every 2 moles of HCl used, 1 mole of hydrogen gas will be generated. So let's figure out how many moles of HCl we have and then divide by 2.
Molarity is defined as moles/liter. So a 2.75 M HCl solution has 2.75 moles of HCl per liter. So the total number of moles we have is:
2.75 mole/L * 1.35 L = 3.7125 mol
And since we get 1 mole H2 per mole of HCl, we get:
3.7125 mol / 2 = 1.85625 mol
Rounding to 3 significant figures gives us 1.86 moles of hydrogen gas.</span>
I think the answer is B. the sum of the enthalpy changes of the intermidiate reactions
I think the correct answer would be HCl + NaHCO3 -> NaCl + H2O + CO2, 2HCl + CaCO3 -> CaCl2 + CO2 + H2O, 2HCl + Mg(OH)2 -> MgCl2 + 2H2O. Hydrochloic acid would react with the basic substances in the stomach which are magnesium hydroxide, sodium bicarbonate and <span>calcium carbonate.</span>
