When sodium amide i.e.
reacts with water i.e. 
results in the formation of sodium hydroxide i.e. 
and ammonia 
.
The chemical reaction is given by:
Now, when ammonia i.e. reacts with water results in the formation of ammonium hydroxide i.e. 
The chemical reaction is given by:
Thus, the products of the above reactions are ammonia and ammonium hydroxide (without sodium ion).
The structures of the products are shown in figure (1): ammonium hydroxide and figure (2) ammonia.
Balanced equation : C. CH₄ + 4Cl₂⇒ CCl₄+ 4HCl
<h3>Further explanation </h3>
Equalization of chemical reactions can be done using variables. Steps in equalizing the reaction equation:
1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.
2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index (subscript) between reactant and product
3. Select the coefficient of the substance with the most complex chemical formula equal to 1
Reaction
CH₄ + Cl₂⇒ CCl₄+ HCl
aCH₄ + bCl₂⇒ CCl₄+ cHCl
C, left=a, right=1⇒a=1
H, left=4a, right=c⇒4a=c⇒4.1=c⇒c=4
Cl, left=2b, right=4+c⇒2b=4+c⇒2b=4+4⇒2b=8⇒b=4
The equation becomes :
CH₄ + 4Cl₂⇒ CCl₄+ 4HCl
Answer:
1x10^–9 M
Explanation:
From the question given,
Concentration of hydronium ion, [H3O+] = 1x10^-5 M.
Concentration of Hydroxide ion, [OH-] =..?
The concentration of the hydroxide ion, [OH-] can be obtained as follow:
[H3O+] x [OH-] = 1x10^–14
1x10^-5 M x [OH-] = 1x10^–14
Divide both side by 1x10^-5
[OH-] = 1x10^–14 / 1x10^-5
[OH-] = 1x10^–9 M
Unmmmm i’m not quite understanding brhajsnsbfbfhrndnsnsjsmsbgbtieoou that but yes
