Answer:
Explanation:
Given:
Ka of HClO2 = 1.1 × 10-2
Ka of HCHO2 = 1.8 × 10-4
Ka of HCN = 4.9 × 10-10
Ka of HNO2 = 4.6 × 10-4
Ka of HF = 3.5 × 10-4
All at a concentration of 0.1 M
A.
HA --> H+ + A-
Ka = [H+] × [A-]/[HA]
[H+] = [A-] = x
[HClO2] = 0.1 M
HClO2 --> ClO2^- + H+
1.1 × 10-2 = x^2/0.1
x = sqrt(1.1 × 10^-3
= 0.033 M
pH = -log[H+]
= 1.48
B.
HA --> H+ + A-
Ka = [H+] × [A-]/[HA]
[H+] = [A-] = x
[HCHO2] = 0.1 M
HCHO2 --> CHO2^- + H+
1.8 × 10-4 = x^2/0.1
x = sqrt(1.8 × 10^-5)
= 0.0042 M
pH = -log[H+]
= 2.37
C.
HA --> H+ + A-
Ka = [H+] × [A-]/[HA]
[H+] = [A-] = x
[HCN] = 0.1 M
HCN --> CN- + H+
4.9 × 10-10 = x^2/0.1
x = sqrt(4.9 × 10-11)
= 0.000007 M
pH = -log[H+]
= 5.16
D.
HA --> H+ + A-
Ka = [H+] × [A-]/[HA]
[H+] = [A-] = x
[HNO2] = 0.1 M
HNO2 --> NO2^- + H+
4.6 × 10-4 = x^2/0.1
x = sqrt(4.6 × 10-5)
= 0.0068 M
pH = -log[H+]
= 2.17
E.
HA --> H+ + A-
Ka = [H+] × [A-]/[HA]
[H+] = [A-] = x
[HF] = 0.1 M
HF --> F- + H+
3.5 × 10-4 = x^2/0.1
x = sqrt(3.5 × 10-5)
= 0.0059 M
pH = -log[H+]
= 2.23
a) pH = 1.48
b) pH = 2.37
c) pH = 5.16
d) pH = 2.17
e) pH = 2.33
HCN is most basic because;
smallest Ka
Highest pH value = 5.16
Answer:
B. It contains 15.0 grams of HNO₃ per 100 grams of solution.
Explanation:
- The percent of a solute in a solution is the mass of the solute (g) per 100 g of solution.
<em>So, an aqueous solution that is 15.0 percent HNO₃ by mass means that It contains 15.0 grams of HNO₃ per 100 grams of solution.</em>
Answer: 30
Explanation:
Sulfur S 1
Oxygen O 4
Nitrogen N 2
Hydrogen H 8
15 in total. For 2 moles, 15*2 = 30
Answer:
0.96 g/cm3, and it will float!
Explanation:
I've explained how to do this before (remember me? lol), but ig I'll do it again..
By looking at the graph you can see that Object C has a mass of ~24 grams and a volume of ~25 cm3
Density = Mass/Volume -> 24 grams/25 cm3 = 0.96 g/cm3
Density of water is 1 g/cm3
Object C is less dense than water and therefore will float (just barely, though)
:)
C it is c the answer is c it’s c
