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
In order to determine, Order of reaction, we have to add all the exponents written in the Chemical form, on the Reactant species.
Hope this helps!
Answer: Hmmmmm that's crazy....
There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.
Fraction remaining (FR) = 0.5n
n = number of half lives that have elapsed
In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).
0.0156 = 0.5n
log 0.0156 = n log 0.5
-1.81 = -0.301 n
n = 6.0 half lives have elapsed
Explanation:
Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)
