The rate constant for a certain reaction is kkk = 8.70×10−3 s−1s−1 . If the initial reactant concentration was 0.150 MM, what will the concentration be after 19.0 minutes?

7.40 x 10⁻⁶ M

Step-by-step explanation:

Our strategy  here is to use the definition for the rate which is

rate = - ΔA/Δt  = k[A]^n

where  k is the rate rate constant raised to the power of n, the reaction order respect to A and [A] is the concentration, M, of A.

Now if the rate is first order we can see that k will have units of s⁻¹ which is what we are told since the rate has units of M/s ( assuming the time is in seconds) . Therefore we know we have a first order reaction:

rate = k[A]

and its integrated rate law wil be

ln [A]/[A]₀ = - kt

where t is the time, [A] and [A]₀ are the concentrations after time t and inital concentration respectively. Thus, we have all the information required to solve this question:

ln( [A]/0.150 ) = - 8.70 x 10⁻³ s⁻¹ x (19.0 min x 60 s/min)

ln ( [A]/0.150 ) = - 9.92

Taking inverse log function to solve for [A]

[A] /0.150 = 4.9 x 10⁻⁵

[A] = 7.40 x 10⁻⁶ M

Note: One can always deduce the order of the reaction if we are given the units of the rate constant k.

For example for a second order reaction it is M⁻1s⁻1 if we are working in  units of second. For third order it is M⁻²s⁻¹, and hopefully you can see the pattern.

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.

me! what did you put for this question?

step-by-step explanation: