Answer:
Since with LiBr no precipitation takes place. So, Ag+ is absent
When we add Li2SO4 to it, precipitation takes place.
Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate
Thus, Ca2+ is present.
When Li3PO4 is added, again precipitation takes place.Reaction is:
Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate
A. Ca2+ and Co2+ are present in solution
B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)
C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)
Answer: This can be quickly solved with "traintracks"
Explanation:
You start w/ grams of water and want to find moles of oxygen gas produced.
So you want to Convert:
Grams of water -> moles of water -> moles of oxygen gas.
The two things you need to know to set up the tracks are:
1)Molar mass of water- H2O
Hydrogen - 1.008(x2)
Oxygen - 16.00
Water - 18.016
Answer: Option (c) is the correct answer.
Explanation:
When an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.
This is because a buffer has the ability to not get affected by the addition of small amounts of an acid or a base. As a result, it helps in maintaining the pH of the solution.
In the give case, when we add the HCl then more number of protons will dissociate. This causes the acetate to react with the protons and leads to the formation of acetic acid.
We know that acetic acid is a weak acid and it dissociates partially or feebly. Therefore, no change in pH will take place.
Thus, we can conclude that equation 
represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH.
Answer:
P + Q2+ -> no reaction
R + P2+ -> R2+ + P
Q + S2+ -> Q2+ + S
S + P2+ -> S2+ + P
S + R2+ -> S2+ + R
S + Q2+ -> no reaction
What is the correct order of reactivity of the metals?
Most -----------------------> least
A. P R S Q
B. Q R S P
C. Q S R P
D. S Q P Rhhgu
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.
A(t) = A(o) (1/2)^(t/d)
where t is the certain period of time. Substituting the known values,
A(t) = (20 mg)(1/2)^(85.80/14.30)
Solving,
A(t) = 0.3125 mg
Hence, the answer is 0.3125 mg.
