Answer:
10−8 M.
Explanation:
In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × In this problem we are given pH and asked to solve for the hydrogen ion concentration. Using the equation, pH = − log [H+] , we can solve for [H+] as,
− pH = log [H+] ,
[H+] = 10−pH,
by exponentiating both sides with base 10 to "undo" the common logarithm. The hydrogen ion concentration of blood with pH 7.4 is,
[H+] = 10−7.4 ≈ 0.0000040 = 4.0 × 10−8 M.
It is correct, next time re-check your answer and don't second guess yourself. ;3
Answer:
A
Explanation The number of energy levels (n) increases, and so does the distance between the nucleus and the outermost orbital. The increased distance and the increased shielding weaken the nuclear attraction, and so an atom can't attract electrons as strongly.
Answer:
2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O
Explanation:
2 CrO42- + 3N2O + 10 H+ -----> 2Cr3+ + 6NO + 5H2O
Oxidizing agent: -----------------------------> CrO42-
Reducing agent: ----------------------------> N2O
explanation:
in CrO4-2 oxdiation state of Cr = +6
in Cr+3 oxidation state = +3
+6 oxidation state changed from +3 it is reduction .
so CrO4-2 is oxidizing agent
atomatically
N2O should be reducing agent
Answer:
5
Explanation:
Count the digits after the first non zero number, any number after that (even zero) counts.
Hope that helped!!! k
