First find the oxidation states of the various atoms:
<span>in Cr2O2 2- Cr @ +1; In NH3 N @ +3; in CrO3 Cr @ +3, N2 N @ 0 </span>
<span>Note that N gained electrons, ie, was reduced; Cr was oxidized </span>
<span>Now there is a problem, because B has NH4+ which the problem did not, and is not balanced, showing e- in/out </span>
<span>B.NH4+ → N2 </span>
<span>Which of the following is an oxidation half-reaction? </span>
<span>A.Sn 2+ →Sn 4+ + 2e- </span>
<span>Sn lost electrons so it got oxidized</span>
Answer:

Explanation:
The question will be easier to solve if we interpret it as, " How long will it take until one-fourth of a sample of the element remains,?"
The half-life of the element is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:


Answer: , 4 molecules of ammonia, NH3(g) is produced; 2 molecules of ammonia, NH3(g) is produced respectively
Explanation:
The balanced equation is stated below N2(g) + 3H2(g) → 2NH3(g)
1 mole of N2(g) reacts with 3 moles of H2(g) to yield 2 moles of NH3(g)
1) If 2 molecules of N2 react, then the balanced equation will be
2N2(g) + 6H2(g) → 4NH3(g)
Thus, 4 molecules of ammonia, NH3(g) is produced
2) If 3 molecules of H2 react, then the balanced equation will be
N2(g) + 3H2(g) → 2NH3(g)
Thus, 2 molecules of ammonia, NH3(g) is produced
Answer:
15 ml
Explanation:
Volume = mass / density.
So our answer is 15 / 3 = 15 mL