A 2.0 g sample of hydrocarbon was burned in the calorimeter. The temperature rose from 29°c to 32°c and heat and combustion is 11. Kj/g. Thr heat capacity of the calorimeter is

THE HEAT CAPACITY OF THE CALORIMETER IS 3666.67 J/C

Explanation:

Mass = 2 g

Temperature difference = 32 C - 29 C = 3 C

Heat of combustion = 11 kJ/g

Heat capacity of the calorimeter = unknown

It is important to note that the heat of combustion of the reaction is the heat absorbed by the calorimeter in raising the mixture by 3 C

So therefore,

Heat = heat capacity * temperature difference

Heat capacity = Heat / temperature difference

Heat capacoty = 11 000 J / 3 C

Heat capacity = 3666.67 J/ C

a) NH₄NO₃ ⇒ N₂O + 2 H₂O

b) 1.69 × 10²³ molecules

Explanation:

Step 1: Write the balanced equation

NH₄NO₃ ⇒ N₂O + 2 H₂O

Step 2: Convert 11.2 g of NH₄NO₃ to moles

The molar mass of NH₄NO₃ is 80.04 g/mol.

11.2 g × 1 mol/80.04 g = 0.140 mol

Step 3: Calculate the moles of H₂O produced

0.140 mol NH₄NO₃ × 2 mol H₂O/1 mol NH₄NO₃ = 0.280 mol H₂O

Step 4: Calculate the number of molecules in 0.280 moles of water

We will use Avogadro's number.

0.280 mol × 6.02 × 10²³ molecules/1 mol = 1.69 × 10²³ molecules