The decomposition of nitramide in aqueous solution at 25 °C NH2NO2(aq)N2O(g) + H2O(l) is first order in NH2NO2 with a rate constant of 4.70×10-5 s-1. If an experiment is performed in which the initial concentration of NH2NO2 is 0.384 M, what is the concentration of NH2NO2 after 31642.0 s have passed? M

[A] = 0.0868 M

Explanation:

Rate constant = 4.70×10-5 s-1

First order reaction

Initial concentration, [A]o = 0.384 M

Final concentration, [A] = ?

Time, t = 31642.0 s

All these variables are related by the following equation;

[A] = [A]o e^(-kt)

[A] = 0.384  e^(-4.70×10-5 x  31642.0)

[A] = 0.384 e^(-1.4872)

[A] = 0.384 * 0.2260

[A] = 0.0868 M

sry but any answer choices

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g