The reaction ch3-n ≡ c → ch3-c ≡ n is a first-order reaction. at 230.3°c, k = 6.29 ⋅ 10-4s-1. if [ch3 = n≡c] is 1.00 ⋅ 10-3 initially, [ch3 = n≡c] is after 1.000 ⋅ 103 s.

The answer is 5.33 x 10^-4

to find the answer follow these steps:

the reaction is first order reaction

here k = 6.29 x 10^-4 s^-1

initial concentration = 1.00 x 10^-3

After time t = 1.000 x 10^3 s concentration is = ?

So, - 6.29 x 10^-4 x 1.00 x 10^3 + ln (1.00 x 10^-3) = - 7.54 

C = e ^ - 7.54= 5.33 x 10^-4 M

your answer would be a. 12.5 kg

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm