Answer:
d d d d d d d d d dd d d d d .
f q q q q q
Answer:
balanced in ACID not BASE
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Answer
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Explanation:
Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)
add H^1+ (acid) to capture the O and make 7 water molecules
Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O
Cr goes from +6 to +3 by gaining 3 e
Hg goes from 0 to +2 by losing 2 e
we need 3 Hg for every 2 Cr
so
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
2 Cr on the right and left
Net 12 positive charges on the right and the left
3 Hg on the right and left
14 H on the right and left
the equation is balanced
we cannot balance the equation in a basic solution with OH^1-
we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid
Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)
v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol
n(Na₂S)=n{Cd(NO₃)₂}=cv
m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv
m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
<span>Both provide approaches to confirming the result of experimentation. Repetition can be developed by one scientist or team continually achieving the expected result but replication requires an independent person or team shows thay can arrive at the same answer independently</span>
