The following reaction is found to be endothermic (δh > 0), the entropy change is negative, and therefore the gibbs free energy is positive at all temperatures. does this mean that this reaction could never be spontaneous? for the reaction below at 25 °c, δgºrxn = 326.4 kj. calculate the partial pressure of ozone, o3(g), needed to make this reaction spontaneous if the partial pressure of oxygen is 0.21 atm.

answer: (431,2)

step-by-step explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

did yall get answers for this yet

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2