Aquantity of 1.922 g of methanol (ch3oh) was burned in a constant-volume bomb calorimeter. consequently, the temperature rose by 5.14°c. if the heat capacity of the bomb plus water was 8.69 kj / °c, calculate the molar heat of combustion of methanol.

The formula for calculating the amount of energy or heat released is:

ΔH = C ΔT

where ΔH is heat of combustion, C is heat capacity, while ΔT is change in temperature

ΔH = 8.69 kJ / °C * (5.14°C)

ΔH = 44.67 kJ

Then we calculate the moles of CH3OH which has molar mass of 32.04 g/mol:

moles = 1.922 / 32.04 = 0.05999 mol

SO the molar heat of combustion is:

ΔHm = 44.67 kJ / 0.0599875 mol

ΔHm = 744.60 kJ / mol

they are both synthetic elements

explanation:

d

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3