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
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration
.


The given equilibrium reaction is,

Initially c 0
At equilibrium

The expression of
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)

where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:



Therefore, the value of equilibrium constant for this reaction is, 1.1
Almost all properties are common to elements within a single group on the periodic table. They react with water in the same way, they have the same number of valence electrons thereby having the same valency, the number of shells in the atom of the element increases by one as we move down the group.
In general, they have the same chemical properties as chemical properties depend on the number of electrons in the valence shell i.e. the outermost shell in the atom of an element.
Question 4 is C. 13 I believe