<span>M(NO3)2 ==> [M2+] + 2 [NO3-]
0.202 M ==> 0.202 M
M(OH)2 ==> [M2+] + 2[OH-]
5.05*10^-18 ===> s + [2s]^2
5.05*10^-18 ===> 0.202 + [2s]^2
5.05*10^-18 = 0.202 * 4s^2
4s^2 = 25*10^-18
s^2 = 6.25*10^-18
s = 2.5*10^-9
So, the solubility is 2.5*10^-9</span>
Hey there!:
Molar mass H3PO4 = <span>97.9952 g/mol
Atomic Masses :
H = </span><span>1.00794 a.m.u
</span>P = <span>30.973762 a.m.u
</span>O = 15.9994 a.m.u<span>
H % = [ ( 1.00794 * 3 ) / </span> 97.9952 ] * 100
H% = <span>3.0857 %
P % = [ ( </span>30.973762 * 1 ) / 97.9952 ] * 100
P% = <span>31.6074 %
O % = [ ( </span>15.9994 * 4 ) / 97.9952 ] * 100
O% = <span>65.3069 %
Hope this helps!</span>
Arrhenius base is a substance that , when dissolved in an aqueous solution , increase the concentration of hydroxide (OH) ion in the solution
I hope that's help !
Answer:
a. 113 min
Explanation:
Considering the equilibrium:-
2N₂O₅ ⇔ 4NO₂ + O₂
At t = 0 125 kPa
At t = teq 125 - 2x 4x x
Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x
125 - 3x = 176 kPa
x = 17 kPa
Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
min⁻¹
Initial concentration = 125 kPa
Final concentration = 91 kPa
Time = ?
Applying in the above equation, we get that:-
Answer:
[A]²
Explanation:
Since the formation is independent of D, D is 0 order.
Since a quadruples when it is doubled it can be written as
2A^X= 4
To find the unknown power we can assume A= 1 to make the math simple. So When a = 2 (Because you doubled it) raised to X power it will equal 4
so the unknown power is 2
Making the rate law
[a]²[b]⁰
or simply just
[A]²
