Answer:
0.2 is conpound Co2 STP.
Explanation:
on combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical formula of compound.?
B- 8.2980
C- 11.2603
F- 17.4228
Li- 5.3917
Na- 5.1391
I would say your answer is Na.
Answer:
t = 7.58 * 10¹⁹ seconds
Explanation:
First order rate constant is given as,
k = (2.303
/t) log [A₀]
/[Aₙ]
where [A₀] is the initial concentraion of the reactant; [Aₙ] is the concentration of the reactant at time, <em>t</em>
[A₀] = 615 calories;
[Aₙ] = 615 - 480 = 135 calories
k = 2.00 * 10⁻²⁰ sec⁻¹
substituting the values in the equation of the rate constant;
2.00 * 10⁻²⁰ sec⁻¹ = (2.303/t) log (615/135)
(2.00 * 10⁻²⁰ sec⁻¹) / log (615/135) = (2.303/t)
t = 2.303 / 3.037 * 10⁻²⁰
t = 7.58 * 10¹⁹ seconds
