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.

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

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.