At 25°c the decomposition of n2o5 (g) into no2 (g) and o2(g) follows first-order kinetics with k = 3.4×10−5 s−1. how long will it take for a sample originally containing 2.0 atm of n2o5 to reach a partial pressure of 380 torr?

11 hours

Explanation:

Half life: [A] = 1/2 [A]o

ln 1/2[A]o divided by ln [A]o = -kt

ln 1/2 = -kt

kt = - ln 1/2

t = 0.693/k

t = 0.693/ 3.4x10-5 s-¹

t1/2 = 20382,35s

20382,35s x 1h /3600s = 5,66 hours to 1 atm

380 torr equals to = 0.05 atm

5.66 x 2 = 11. 32 hours

c. 75.7

i think that is the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer