Janice needs to have a gastro-intestinal study. So she is given 375 mg of radiocontrast dye containing Barium-142. The half-life
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:
<u>Answer:</u> The molality of the solution is 0.11 m
<u>Explanation:</u>
We are given:
Mole fraction of methanol = 0.135
This means that 0.135 moles of methanol is present in 1 mole of a solution
Moles of ethanol = 1 - 0.135 = 0.865 moles
To calculate the mass for given number of moles, we use the equation:
Moles of ethanol = 0.865 moles
Molar mass of ethanol = 46 g/mol
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute (methanol) = 0.135 g
= Molar mass of solute (methanol) = 32 g/mol
= Mass of solvent (ethanol) = 39.79 g
Putting values in above equation, we get:
Hence, the molality of the solution is 0.11 m