Answer:
a) 119 g/mol
Explanation:
-We apply the formula for freezing point depression to obtain the molality of the solution:
#We use the molality above to calculate the molar mass:
Hence, the molar mass of the compound is 119 g/mol
If the alum reaction was done starting with 12.000 g of al, how many moles of hydrogen gas would be produced according to the ba
lanced equation?
1 answer:
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Answer:
Hydrogen: -141 kJ/g
Methane: -55kJ/g
The energy released per gram of hydrogen in its combustion is higher than the energy released per gram of methane in its combustion.
Explanation:
According to the law of conservation of the energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qc + Qb = 0
Qc = -Qb [1]
We can calculate the heat absorbed by the bomb calorimeter using the following expression.
Q = C . ΔT
where,
C is the heat capacity
ΔT is the change in the temperature
<h3>Hydrogen</h3>Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (14.3°C) = -162 kJ
The heat released per gram of hydrogen is:
Qc = -Qb = -C . ΔT = -(11.3 kJ/°C) . (7.3°C) = -82 kJ
The heat released per gram of methane is:
<em>Chemists use the mole unit to represent 6.022 × 10 23 things, whether the things are atoms of elements or molecules of compounds. This number, called Avogadro's number, is important because this number of atoms or molecules has the same mass in grams as one atom or molecule has in atomic mass units. </em>
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