<h3>Answer:</h3>
162.43 g of FeCl₂
<h3>
Explanation:</h3>
Step 1: Calculate mass of Fe;
As,
Density = Mass ÷ Volume
Or,
Mass = Density × Volume
Where Volume is the volume of water displaced = 10.4 mL
Putting values,
Mass = 7.86 g.mL⁻¹ × 10.4 mL
Mass = 81.744 g of Fe
Step 2: Calculate amount of FeCl₂;
The balance chemical equation is as follow,
Fe + 2 HCl → FeCl₂ + H₂ ↑
According to this equation,
55.85 g (1 mol) Fe produced = 110.98 g (1 mol) of FeCl₂
So,
81.744 g Fe will produce = X g of FeCl₂
Solving for X,
X = (81.744 g × 110.98 g) ÷ 55.85 g
X = 162.43 g of FeCl₂
Density of liquid= 
.
so, density of liquid= 
= 1.2 gm/cm³.
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
Answer:
NO3-
Explanation:
Given the reaction equation;
Au(s) + 3NO3-(aq) + 6H+(aq)→Au3+(aq) + 3NO2(g) + 3H2O (l).
We can consider the oxidation states of species on the left and right hand sides of the reaction equation;
Au is in zero oxidation state on the left hand side and an oxidation state of +3 on the righthand side.
NO3- is in oxidation state of +5 on the righthand side and NO2 is in + 4 oxidation state.
H+ is in + 1 oxidation state on both the left and right hand sides of the reaction equation.
Since reduction has to do with a decrease in oxidation number, it follows that NO3- was reduced in the reaction.
