Answer:
the answer for this question is the option D
I suppose it false, since the oxidation involves the loss or removal of the electrons such forth it does not gain electrons.
Answer:
The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M
Explanation:
We use the formulas:
pH= - log(H30+) and Kwater=(H30+)x(OH-)
pH= - log(H30+) ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M
Kwater=(H30+)x(OH-)
(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7
Answer:
Demo Mole Quantities
58.5g NaCl(mol/58.5g)(6.02 x 1023/mol) = 6.02 x 1023 Na
+
Cl21 pre-1982 pennies (after 1982 pennies are mostly zinc with copper coating)
63.5g Cu( mol/ 63.5g)(6.02 x 1023/mol) = 6.02 x 1023 Cu
19.0g Al (mol/27.0g)(6.02 x 1023/mol) = 4.24 x 1023 Al
Explanation:
