Answer:
Standard enthalpy of formation of Carbon disulfide CS2 = 87.3 KJ/mol
Explanation:
forming CS2 means that it should in the product side
C(graphite) + O2 → CO2 ΔH = -393.5
2S(rhombic) + 2O2 → 2SO2 ΔH = -296.4 x 2
CO2 + 2SO2 → CS2 + 3O2 ΔH = -1073.6 x -1
the second reaction is multiplied by 2 so that the SO2 and O2 can cancel out.
the third reaction is reversed (multiplied by -1) so that CS2 will be on the product side.
after adding the reaction and cancelling out similarities, the final reaction is: C(graphite) + 2S(rhombic) → CS2
Add ΔH to find the enthalpy of formation of CS2
ΔHf = (-393.5) + (-296.4 x 2) + (-1073.6 x -1) = 87.3 KJ/mol
be aware of signs
Answer:
by using this formula you will get it
Explanation:
number of mole = number of particles÷ Avogadro's number
n=3.51×10^23÷ 6.02×10^23
n = 0.58 moles
All of these metals from cations, so they lost electrons when forming ions. The answer is D.
Answer:
The answer is C.
Explanation:
This is because you wouldn't be able to function without your lungs and spine. So therefore the answer would have to be "C".
Explanation:
Knowing the pH, you know the concentration of protons:
−log[H+]=pH=3.7
[H+]=10−3.7 M
Now, since the weak (monoprotic) acid dissociates into its conjugate base and a proton, the mols of protons are equimolar with the mols of conjugate base---the protons came FROM the weak acid, so the conjugate base that forms must be equimolar with the protons given out to the solvent.
HA⇌A−+H+
Hence, [A−]=[H+] in the same solution volume. Using the equilibrium constant expression, we get:
Ka=[H+]2eq[HA]eq
Don't forget that the HA form of HA had given away protons, so the mols of protons given away to generate A− is subtracted from the mols of (protons in) HA.
=[H+]2eq[HA]i−[H+]eq
=(10−3.7M)20.02M−10−3.7M
Ka=2.0105×10−6 M
