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:
D is the best choice. Those percentages, are giving you the information about how concentrated are the solutions. As 0.015 is so concentrated, this solution will damage the structures more quickly
Explanation:
Molarity of the resulting solution will be 1.33 M.
<u>Explanation:</u>
First we have to find the number of moles for each of the solution using the formula, moles = molarity × volume
For cup 1 = 1 M ×0.05 L = 0.05 moles
For cup 2 = 2.5 M × 0.05 L= 0.125 moles
For cup 3 = 0.5 M × 0.05 L = 0.025 moles
Total moles = 0.05 + 0.125 + 0.025 = 0.2 moles
We have to find the total volume as, 0.05 + 0.05 + 0.05 = 0.15 L
Now we have to find the molarity as, moles / volume = 0.2 moles/ 0.15 L = 1.33 M
When sugar is heated it melts and then caramelize giving of water. After this, it turns blank (carbon) and then coverts to co2 (carbon dioxide) .
So, heating of sugar is an endothermic , decomposition and oxidation reaction.
When Ammonium Chloride is heated , it directly goes from solid state to vapour state without changing into liquid state.
So, heating of Ammonium Chloride is an example of sublimation reaction.
I think it is decomposition because it processes in which chemical species break up into simpler parts.
