A1.00 g sample of n-hexane (c6h14) undergoes complete combustion with excess o2 in a bomb calorimeter. the temperature of the 1502 g of water surrounding the bomb rises from 22.64°c to 29.30°c. the heat capacity of the hardware component of the calorimeter (everything that is not water) is 4042 j/°c. what is δu for the combustion of n-c6h14? one mole of n-c6h14 is 86.1 g. the specific heat of water is 4.184 j/g·°c.

-5,921x10⁶J/mol

Explanation:

Internal energy change (ΔU) for the reaction of combustion in the bomb calorimeter is:

ΔU = q calorimeter + q solution

Where:

q calorimeter is Ccal×ΔT (Ccal=4042J/°C) and (ΔT is 29,30°C-22,64°C=6,66°C)

q solution is c×m×ΔT (c= 4.184 J/g°C), (m=1502g H₂O), (ΔT is 29,30°C-22,64°C=6,66°C)

Replacing:

ΔU = 26920J + 41854J = 68774 J

This energy is per g of n-hexane, now, per mole of n-hexane:

= -5,921x10⁶J/mol

-negative because the energy is produced-

I hope it helps!

c. is the answer

by love gaming

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2