First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
<span>The rate of infusion is 2.1L/19h or 2100mL/19h (as 1L = 100 mL).
To convert 19 hours to minutes we multiply as follows:
19 hours = (19 hours) x (60 minutes/1 hour) = 1140 minutes
So the rate of infusion becomes:
2100mL /1140 min
In order to converted mL to drops (gtt) we multiply the rate of infusion with the drop factor to get the drip rate:
(2100mL/1140min) x (20 gtt/mL) = 36.8 gtt/min</span>
Answer:
Group VIIIA in which the noble/inert gases are found
I’m pretty sure the answer is A.
Answer:
6 mol/min
Explanation:
2A+2B→4C
The relationship between the reactants and products of this equation is given by;
1/2 -d[A]/dt = 1/2 -d[B]/dt = 1/4 d[C]/dt
Our focus is on A and C
From the question;
d[A]/dt = 3mol/min
We have;
1/2 (3) = 1/4 d[C]/dt
d[C]/dt = 4/2 * 3 = 6 mol/min
