Answer:
Final concentration = 0.019 M
Explanation:
Initial Concentration [A]o = 0.27M
Rate constant, k = 0.75 s^-1
Final concentration [A] = ?
Time, t = 1.5s
The relationship between the variables is given by the equation;
ln[A] = ln[A]o - kt
ln[A] = ln(0.27) - (0.75)(1.5)
ln[A] = - 1.309 - 1.125
ln[A] = - 2.434
[A] = 0.019 M
Volume Ba(OH)2 = 23.4 mL in liters :
23.4 / 1000 => 0.0234 L
Molarity Ba(OH)2 = 0.65 M
Volume HNO3 = 42.5 mL in liters:
42.5 / 1000 => 0.0425 L
number of moles Ba(OH)2 :
n = M x V
n = 0.65 x 0.0234
n = 0.01521 moles of Ba(OH)2
Mole ratio :
<span>Ba(OH)2 + 2 HNO3 = Ba(NO3)2 + 2 H2O
</span>
1 mole Ba(OH)2 ---------------- 2 moles HNO3
0.01521 moles ----------------- moles HNO3
moles HNO3 = 0.01521 x 2 / 1
moles HNO3 = 0.03042 / 1
= 0.03042 moles HNO3
Therefore:
M ( HNO3 ) = n / volume ( HNO3 )
M ( HNO3 ) = 0.03042 / 0.0425
M ( HNO3 ) = 0.715 M
Positively charged and the sodium ion would be a cation
Answer:
The concentration of methyl isonitrile will become 15% of the initial value after 10.31 hrs.
Explanation:
As the data the rate constant is not given in this description, However from observing the complete question the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .
Now the ratio of two concentrations is given as

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.
k is the rate constant which is given as 
So time t is given as

So the concentration will become 15% of the initial value after 10.31 hrs.