Balance each one by adding electrons to make the charges on both sides the same:
Sn--> Sn2+ + 2 e-
Ag+ + 1 e- --> Ag
Now, you have to have the same number of electrons in the two half-reactions, so multiply the second one by 2 to get:
2 Ag+ + 2 e- --> 2 Ag
Now, just add the two half reactions together, cancelling anything that's the same on both sides:
2 Ag+ + Sn --> Sn2+ + 2 Ag
And you're done.
When edible oils are idle and stored for a long amount of time, they undergo oxidation due to the exposure to oxygen. This oxidation causes rancidity in oils.
Answer:
(i) specific heat
(ii) latent heat of vaporization
(iii) latent heat of fusion
Explanation:
i. Q = mcΔT; identify c.
Here, Q is heat, m is the mass, c is the specific heat and ΔT is the change in temperature.
The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C is known as the specific heat.
ii. Q = mLvapor; identify Lvapor
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg liquid into 1 kg vapor at constant temperature.
iii. Q = mLfusion; identify Lfusion
Here, Q is the heat, m is the mass and L is the latent heat of fusion.
Here, Q is the heat, m is the mass and L is the latent heat of vaporization.
The amount of heat required to convert the 1 kg solid into 1 kg liquid at constant temperature.
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
