When 15.3 g of sodium nitrate, NaNO₃, was dissolved in water in a constant-pressure calorimeter, the temperature fell from 25.00 °Celsius to 21.56 °Celsius. If the heat capacity of the solution and the calorimeter is 1071 J/oC, what is the enthalpy change when 1 mol of sodium nitrate dissolves in water?
The solution process is NaNO₃(s) .

ΔH = 20468 J/mol = 20.5 kJ/mol

Explanation:

Step 1 : Data given

Mass of sodium nitrate NaNO3 = 15.3 grams

The temperature fell from 25.00 °Celsius to 21.56 °Celsius

The heat capacity of the solution and the calorimeter is 1071 J/°C

Step 2: Calculate Q

Q = Cp * ΔT

⇒with Q = the heat transfer = TO BE DETERMINED

⇒with Cp = The heat capacity of the solution and the calorimeter is 1071 J/°C

⇒with ΔT = the change of temperature = 25.00 - 21.56 = 3.44 °C

Q= 1071 J/°C * 3.44 °C

Q = 3684.24 J

Step 3: Calculate moles NaNO3

Moles NaNO3 = mass / molar mass NaNO3

Moles NaNO3 = 15.3 grams / 84.99 g/mol

Moles NaNO3 = 0.180 moles

Step 4: Calculate the enthalpy change when 1 mol of sodium nitrate dissolves in water.

ΔH = Q / moles

ΔH = 3684.24 J/ 0.180 moles

ΔH = 20468 J/mol = 20.5 kJ/mol

Since the temperature decreases, this is an endothermic process.

For an endothermic process, the enthalpy change is positive.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.