Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
Answer:
pH = 11.95≈12
Explanation:
Remember the reaction among aqueous acetic acid (
) and aqueous sodium hydroxide (NaOH)

First step. Need to know how much moles of the substances are present
= 0.0025 mol NaOH
0.003 mol NaOH *
/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]
NaOH is in excess. Now, how much?
0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH
Then, that amount in excess would be responsable for the pH.
Third step. Know the pH
Remember that pH= -log[H+]
According to the dissociation of water equilibrium
Kw=[H+]*[OH-]= 10^(-14)
The dissociation of NaOH is
NaOH -> 
Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.
[OH-]= 0.0005 mole / 0.055 L = 0.00909 M
Careful: we have to use the total volumen
Les us to calculate pH
![pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95](https://tex.z-dn.net/?f=pH%3D%20-log%20%5BH%2B%5D%5C%5CpH%3D%20-log%20%5Cfrac%7BK_w%7D%7B%5BOH-%5D%7D%20%5C%5CpH%3D%2011.95)
The more protons you add, the more positively charged the atom becomes
the charge of an atom determines what kind of an atom it is <span />
Answer:
could be reduced by one-quarter.
Explanation:
Answer:C
Explanation:it gets rid of waste