Answer : The amount of heat changes is, 56.463 KJ
Solution :
The conversions involved in this process are :
Now we have to calculate the enthalpy change.
where,
= enthalpy change or heat changes = ?
n = number of moles of water = 1 mole
= specific heat of solid water =
= specific heat of liquid water =
= specific heat of liquid water =
m = mass of water
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
= enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole
Now put all the given values in the above expression, we get
(1 KJ = 1000 J)
Therefore, the amount of heat changes is, 56.463 KJ
Answer:
A There are more particles in the air for water vapor to cluster around.
Explanation:
- As the condensation process takes place more particles are added to the atmosphere for water vapour and condensation further leads to the build-up of clouds.
- If the sky is polluted with pollutants the condensation can raise the level of pollution by blocking the rays of the sun from reaching the ground. This affects the visibility of most of the cities.
Because there is a balloon to stop everything from coming out
Answer:
0.305 mol
Explanation:
Ca(NO) is not a molecule. I think you meant to type Ca(NO3)2, which is calcium nitrate.
The moles of a compound is equal to is mass divided by its molar mass.
The molar mass of Ca(NO3)2 is 164.09 g/mol.
50.0 / 164.09 = 0.305
<h3>
Answer:</h3>
259,000 Joules
<h3>
Explanation:</h3>
To calculate the heat energy needed to change 100 grams of liquid water at 75°C to vapor at 225°C, we will do this in steps;
<h3>
Step 1: Heat energy required to raise the temp of water from 75°C to 100°C</h3>
Mass of water = 100 g
Heat capacity of water = 4.184J/g °C
Temperature change (75°C to 100°C) = 25°C
Using the formula to get Quantity of heat;
Q = mcΔT
= 100 g × 4.184 J/g °C × 25°C
= 10,460 joules
<h3>
Step 2: Heat required to change water at 100°C to vapor at 100°C</h3>
Mass of water = 100 g
Heat of vaporization = 2,256 J/g
Heat required to change water to vapor without a change in temperature is given by;
Q = m × Lv
= 100 × 2,256 J/g
= 225,600 Joules
<h3>
Step 3: Heat energy required to raise the temperature of ice from 100°C to 225°C.</h3>
Mass of vapor = 100 g
Specific heat of gas = 1.84 J/g °C
Change in temperature (100°C to 225°C) = 125°C
Therefore;
Q = mcΔT
= 100 g × 1.84 J/g °C × 125°C
= 23,000 joules
<h3>Step 4: Total heat energy required</h3>
Total energy = 10,460 J + 225,600 J + 23,000 J
= 259,060 J
= 259,000 J(approx)
Therefore, heat energy required is 259,000 Joules