In terms of the disappearance of the reactants, the rate equations are R= -kd[I-]/dt and -kd[OCl-]/dt.
<h3>What is an ionic reaction?</h3>
The term ionic reaction refers to the reaction that takes place between two ions. In this case, the ionic reaction is; I-(aq) + OCl-(aq) -------> Cl-(aq) + OI-(aq).
The rate equations in terms of the disappearance of the reactants is;
R= -kd[I-]/dt
And
R = -kd[OCl-]/dt
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A.water.
An aqueous solution is a solution in water.
Hope this helped :)
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
Sodium borohydride is a relatively selective reducing agent Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.
The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
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To get the theoretical yield of ammonia NH3:
first, we should have the balanced equation of the reaction:
3H2(g) + N2(g) → 2NH3(g)
Second, we start to convert mass to moles
moles of N2 = N2 mass / N2 molar mass
= 200 / 28 = 7.14 moles
third, we start to compare the molar ratio from the balanced equation between N2 & NH3 we will find that N2: NH3 = 1:2 so when we use every mole of N2 we will get 2 times of that mole of NH3 so,
moles of NH3 = 7.14 * 2 = 14.28 moles
finally, we convert the moles of NH3 to mass again to get the mass of ammonia:
mass of NH3 = no.moles * molar mass of ammonia
= 14.28 * 17 = 242.76 g
<span>Conductor, and there you go, i hope this helped but if its wrong, i am extremly sorry</span>
