The ions in the unknown salt are:
and 
The chemical formula of the salt is determined as shown in the picture.
The chemical formula of the salt is 
.
The correct answer is 1.778 x 10^-11M. Hope this helps.
Answer:
Explanation:
Amount HCl = 0.455mole
Molar mass = 1 + 35.5 = 36.5g/mol
Amount = mass/molar mass
:- mass = Amount × Molar mass
Mass = 16.97g
Answer:
The amount of base needed is the amount that would give one mole of the hydroxide ion needed to neutralise one mole of the hydroxonium ion from the acid.
Explanation:
The chemical reaction between an acid and a base to form salt and water only is called a Neutralization reaction. Chemically
H⁺ + OH⁻ = H₂0
Hence, one mole of hydroxonium ion (H⁺) will combine with one mole of hydroxide ion (OH⁻) to give salt and water only.
In a completely neutralized reaction, the resulting salt is formed when there is complete dissociation of the acid and base to give salt and water with a pH of 7.
In the given question, the stated pH of between 8-9 tells us that the salt produced in this particular neutralization reaction is basic or alkaline. This usually occurs when a strong base reacts with a weak acid, producing a higher concentration of the hydroxide ion at equilibrium.
Hence the amount of base needed is the amount that would give one mole of the hydroxide ion needed to neutralise one mole of the hydroxonium ion from the acid.
If the concentration or molarity of the acid is known, then the exact amount of base required to neutralize it can be calculated. This is usually done via titrating the acid against drop wise solution of the base. Neutralization usually occurs when there is a change in colour of the resulting solution. The pH of the resulting solution can be determined using a litmus paper.
A blue litmus paper is indicative of a basic solution while a red litmus paper is indicative of an acidic solution.
1) (Hvap)(moles of water)=236.9783574kJ
(40.67)(105/18.02)
2) (change in temperature)(mass)(Cliquid)=43.9345172kJ
(100)(105/18.02)(75.4)/1000
3) (Hfus)(moles of water)=35.01942286kJ
(6.01)(105/18.02)
4) (change in temperature)(mass)(Csolid)=3.181465039kJ
(15)(105/18.02)(36.4)/1000
Total released=319.1137625kJ
