Answer:
B. It contains 15.0 grams of HNO3 per 100 grams of solution.
Explanation:
The question asks about the mass percent of HNO3 in a solution. Mass percent is the ratio of the mass of the solute (HNO3) to the mass of the solution (HNO3 + water) multiplied by 100. A solution that is 15.0 percent HNO3 by mass means that for every 100 grams of solution, there are 15.0 grams of HNO3 and 85.0 grams of water. This ratio does not change with the volume or the moles of the solution or the solute. The only answer that matches this definition is B. It contains 15.0 grams of HNO3 per 100 grams of solution.
Calculate the change in heat when 16.00 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 25.50 °C.
The change in heat when 16.00 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 25.50°C is -31.25 kJ (exothermic process).
To calculate the change in heat when 16.00 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 25.50 °C, we need to consider two separate processes and add their heat changes together:
The heat change during the condensation of steam to liquid water:
The heat change during this process can be calculated using the heat of vaporization of water, which is 40.7 kJ/mol at 100.0°C.
First, we need to determine the number of moles of water in 16.00 g:
moles of water = mass of water / molar mass of water
moles of water = 16.00 g / 18.015 g/mol
moles of water = 0.8886 mol
The heat change during the condensation of steam to liquid water can be calculated as follows:
q1 = moles of water x heat of vaporization
q1 = 0.8886 mol x 40.7 kJ/mol
q1 = 36.21 kJ
The heat change during the cooling of liquid water from 100.0°C to 25.50°C:
The heat change during this process can be calculated using the specific heat capacity of water, which is 4.184 J/g°C.
The temperature change during this process is:
ΔT = final temperature - initial temperature
ΔT = 25.50°C - 100.0°C
ΔT = -74.50°C
The heat change during the cooling of liquid water can be calculated as follows:
q2 = mass of water x specific heat capacity x ΔT
q2 = 16.00 g x 4.184 J/g°C x (-74.50°C)
q2 = -4,958 J
Therefore, the total heat change for the two processes is:
ΔH = q1 + q2
ΔH = 36.21 kJ + (-4,958 J)
ΔH = 36.21 kJ - 4.958 kJ
ΔH = 31.25 kJ
Therefore, the change in heat when 16.00 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 25.50°C is -31.25 kJ (exothermic process).
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The heat of vaporization for ethanol is 0.826 kJ/g
. Calculate the heat energy in joules required to boil 75.25 g
of ethanol.
Answer:
87469.73J
Explanation:
72.25g/0.826kJ/g=87.4697337kJ
