Answer:
There are no acceptable descriptions at all on that list of choices.
((12g/mol)/(28.01g/mol))*100% = 42.84%
Answer: 0.52849 j /g °C
Explanation:
Given the following :
Mass of metal = 36g
Δ Temperature of metal = (28.4 - 99)°C = - 70.6°C
Mass of water = 70g
Δ in temperature of water = (28.4 - 24.0) = 4.4°C
Heat lost by metal = (heat gained by water + heat gained by calorimeter)
Quantity of heat(q) = mcΔT
Where; m = mass of object ; c = specific heat capacity of object
Heat lost by metal:
- (36 × c × - 70.6) = 2541.6c - - - - (1)
Heta gained by water and calorimeter :
(70 × 4.184 × 4.4) + (12.4 × 4.4) = 1288.672 + 54.56 = 1343.232 - - - - (2)
Equating (1) and (2)
2541.6c = 1343.232
c = 1343.232 / 2541.6
c = 0.52849 j /g °C
H2 is known to exist. For dihydrogen, H2, we can identify the frontier molecular orbitals (FMOs). The highest occupied molecular orbital (or HOMO) is the σ (sigma) 1s MO. The lowest unoccupied MO (LUMO) is the σ* (sigma star) 1s MO which is antibonding.
