The molecular weight of Mg(OH)2 : 58 g/mol
<h3>Further explanation</h3>
Given
Mg(OH)2 compound
Required
The molecular weight
Solution
Relative atomic mass (Ar) of element : the average atomic mass of its isotopes
Relative molecular weight (M) : The sum of the relative atomic mass of Ar
M AxBy = (x.Ar A + y. Ar B)
So for Mg(OH)2 :
= Ar Mg + 2 x Ar O + 2 x Ar H
= 24 g/mol + 2 x 16 g/mol + 2 x 1 g/mol
= 24 + 32 + 2
= 58 g/mol
Because the copper roofs, have been outside for a while and oxidization has occurred.
When metals become oxidized they rust, but when copper becomes oxidized, it turns greenish.
So copper roofs oxidize and turn green from being outside and a copper square doesn’t if it’s inside.
To solve this problem,
we can use the Henderson-Hasselbalch Equation which relates the pH to the measure
of acidity pKa. The equation is given as:<span>
<span>pH = pKa + log ([base]/[acid]) ---> 1</span></span>
Where,
[base] = concentration
of C2H3O2
in molarity or moles
<span>[acid] = concentration of HC2H3O2 in molarity or moles</span>
For the sake of easy calculation, let us assume that:
[base] = 1
[acid] = x
<span>
Therefore using equation 1,
4.24 = 4.74 + log (1 / x)
<span>log (1 / x) = - 0.5
1 / x = 0.6065 </span></span>
x =
1.65<span>
The required ratio of C2H3O2 /HC2H3O2 <span>
is 1:1.65 or 3:5. </span></span>
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
