Answer:
There are there are 8.13-10^25
Explanation:
N(CS2)= 1.35* (times) 6.02-10^ 25= 3.13-10^25 molecules
Answer:
57.48%
Explanation:
Calculate the mass of 1 mole of malachite:
MM Cu = 63.55
MM O = 16.00
MM H = 1.01
MM C = 12.01
A mole of malachite has:
2 moles of Cu
5 moles of O
2 moles of H
1 mole of C
MW Malachite = 2*MM(CU) + 5*MM(O) + 2*MM(H) + 1 *MM(C)
MW Malachite = 2*63.55 + 5*16.00 + 2*1.01 + 1*12.01
MW Malachite = 221.13
Mass of Cu in a mole of Malachite = 2*MM(CU) = 127.1
Now divide the mass of Cu by the mass of Malachite
Answer:
2
Explanation:
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The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
To learn more about enthalpy here
brainly.com/question/13981382
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