<span>When M(OH)2 dissolves we have
M(OH)2 which produces M2+ and 2OHâ’
pH + pOH=14
At ph =7; we have
7+pOH=14
pOH=14â’7 = 7
Then [OHâ’]=10^(â’pOH)
[OH-] = 10^(-7) = 1* 10^(-7)
At ph = 10. We have,
pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4)
Finally ph = 14. We have, pOH = 0
And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)...
So [OH-] = 1</span>
Answer: a) 
b) 1 mole of 
is produced.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The skeletal equation is:
The balanced equation will be:
Thus the coefficients are 2, 3 , 10 , 4 , 3 , 2 and 5.
b) Oxidation: 
Reduction: 
Net reaction: 
When 1 mole of 
is produced, 1 mole of is produced.
Answer:
HBr + H2O = H3O+ + Br-
So our conjugate acid is the H3O+ to H2O
Explanation:
A conjugate acid of a base results when the base accepts a proton.
Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.
Answer:
Some of the physical changes used by the industrial chemist in order to identify it is by scratching it with other metals in order to find the hardness of it. Trying to deform it in order to find the malleability, and to heat it and measure the temperature in order to find the melting point.
Some of the chemical changes used by the industrial chemist in order to identify it is by inserting it in water to observe that whether it reacts with it or not, if the reaction is violent, then the metal belongs to either group I or group II. The other method is to insert it in acids of distinct strength and to observe its reaction. The metals belonging to the second group react briskly with acids. The other metals react gradually with acids and others are almost inert.
