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:
Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]
Explanation:
The equation for the reaction of trimethylamine when it is dissolved in water is:
C₃H₉N + H₂O ⇄ C₃H₉NH⁺ + OH⁻ Kb
1 mol of trimethylamine catches a proton from the water in order to produce trimethylamonium.
It is a base, because it give OH⁻ to the medium
Expression for Kb (Molar concentration)
Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]
What happens is that the sodium solution when put in water reacts and creates thermal energy .
Answer:
1g or 10^-3kg
Explanation:
as you know , the density =mass \volume
so you have the mass from the number it self
<h2>
so from this equation, you will get 1 g and you can to SI to be
kg</h2>
Density = Mass/Volume => Mass = Density*Volume => Mass = 3g/ml*10ml =30g
