From the calculation, the molar mass of the solution is 141 g/mol.
<h3>What is the molar mass?</h3>
We know that;
ΔT = K m i
K = the freezing constant
m = molality of the solution
i = the Van't Hoft factor
The molality of the solution is obtained from;
m = ΔT/K i
m = 3.89/5.12 * 1
m = 0.76 m
Now;
0.76 = 26.7 /MM/0.250
0.76 = 26.7 /0.250MM
0.76 * 0.250MM = 26.7
MM= 26.7/0.76 * 0.250
MM = 141 g/mol
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The heat of solution is -51.8 kJ/mol
<h3>What is the heat of solution?</h3>
We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.
Number of moles of KOH = 11.9-g/56 g/mol = 0.21 moles
Temperature rise = 26.0 ∘c
Mass of the water = 100.0 grams
Heat capacity = 4.184 j/g⋅°c
Then;
ΔH = mcθ
ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ
Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol
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Answer:
9.6 moles O2
Explanation:
I'll assume it is 345 grams, not gratis, of water. Hydrogen's molar mass is 1.01, not 101.
The molar mass of water is 18.0 grams/mole.
Therefore: (345g)/(18.0 g/mole) = 19.17 or 19.2 moles water (3 sig figs).
The balanced equation states that: 2H20 ⇒ 2H2 +02
It promises that we'll get 1 mole of oxygen for every 2 moles of H2O, a molar ratio of 1/2.
get (1 mole O2/2 moles H2O)*(19.2 moles H2O) or 9.6 moles O2
B) 6
one above one below and 2 on the left and right sides
