Answer:
53.5g of NH4Cl
Explanation:
First, we need to obtain the number of mole of NH4Cl. This is illustrated below:
Volume = 0.5L
Molarity = 2M
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 2 x 0.5
Mole = 1mole
Now, let us convert 1mole of NH4Cl to gram. This is illustrated below:
Molar Mass of NH4Cl = 53.5g/mol
Number of mole = 1
Mass =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass = 1 x 53.5
Mass = 53.5g
Therefore, 53.5g of NH4Cl is contained in the solution.
Explanation:
Reaction equation is as follows.

Here, 1 mole of
produces 2 moles of cations.
![[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58](https://tex.z-dn.net/?f=%5BNa%5E%7B%2B%7D%5D%20%3D%202%5BNa_%7B2%7DSO_%7B3%7D%5D%20%3D%202%20%5Ctimes%200.58)
= 1.16 M
= 0.58 M
The sulphite anion will act as a base and react with
to form
and
.
As, 
= 
=
According to the ICE table for the given reaction,

Initial: 0.58 0 0
Change: -x +x +x
Equilibrium: 0.58 - x x x
So,
![K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B%5BHSO%5E%7B-%7D_%7B3%7D%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BSO%5E%7B2-%7D_%7B3%7D%5D%7D)


x = 0.0003 M
So, x =
= 0.0003 M
= 0.58 - 0.0003
= 0.579 M
Now, we will use
= 0.0003 M
The reaction will be as follows.

Initial: 0.0003
Equilibrium: 0.0003 - x x x


= 
= 
Therefore, 
As, x <<<< 0.0003. So, we can neglect x.
Therefore, 
= 
x = 
x =
= 
![[H^{+}] = \frac{10^{-14}}{[OH^{-}]}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B%5BOH%5E%7B-%7D%5D%7D)
= 
=
M
Thus, we can conclude that the concentration of spectator ion is
M.
i think that the answer is d. millions of small rocky objects
The mass number is the sum of the protons and neutrons in an atom. 235-92 = 143 neutrons.
Element 92 is Uranium.
Uranium is also the highest atomic number element which exists in nature. All of the rest above element 92 only exist nuclear high-energy reactions (the transuranium elements), and most experience nuclear decay very quickly and form other elements with fewer protons.