Answer:
Sodium hydrogen carbonate
NaBr < H3O+1 = OH-1 < Na^+1 = Br^-1 < H2O
<span>Least is NaBr (100% dissolved so no NaBr remains, only Na^+1 and Br^-1 </span>
<span>H2O yields 10^-7 M H3O^+1 and 10^-7 M OH^-1 (Kw = 1x10^-14 = [H3O+][OH-] </span>
<span>Na^+1 and Br^-1 will bothe be 0.1 M </span>
<span>H2O is slightly less that 1000 g / L in a 0.1 M NaBr solution, so its concentration is about 55.5 M</span>
Answer:
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Answer:
1. Percentage by weight = 0.5023 = 50.23 %
2. molar fraction =0.153
Explanation:
We know that
Molar mass of HClO4 = 100.46 g/mol
So the mass of 5 Moles= 5 x 100.46
Mass (m)= 5 x 100.46 = 502.3 g
Lets assume that aqueous solution of HClO4 and the density of solution is equal to density of water.
Given that concentration HClO4 is 5 M it means that it have 5 moles of HClO4 in 1000 ml.
We know that
Mass = density x volume
Mass of 1000 ml solution = 1 x 1000 =1000 ( density = 1 gm/ml)
m'=1000 g
1.
Percentage by weight = 502.3 /1000
Percentage by weight = 0.5023 = 50.23 %
2.
We know that
molar mass of water = 18 g/mol
mass of water in 1000 ml = 1000 - 502.3 g=497.9 g
So moles of water = 497.7 /18 mole
moles of water = 27.65 moles
So molar fraction = 5/(5+27.65)
molar fraction =0.153
The answer is B. The answer is B. Please pick me as brainlest.
