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:
The number of moles of the gas is 9.295 moles or 9.30 moles
Explanation:
We use PV = nRT
Where P = 4.87 atm;
V = 67.54 L
R= 0.0821Latm/molK
T = 158 C = 158 +273 K = 431 K
the number of moles can be obtained by substituting the values in the respective columns and solve for n
n = PV / RT
n = 4.87 * 67.54 / 0.0821 * 431
n = 328.9198 / 35.3851
n = 9.295moles
The number of moles is approximately 9.30moles.
Answer:
All of them!
Explanation:
Since Mg, Li, Ca, and Cs are all in groups 1 and 2 of the periodic table, they are alkali/alkaline earth metals and will all lose electrons during ionic bonding.
Answer: The results agree with the law of conservation of mass
Explanation:
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. On the reactant side, the total mass of reactants is 14.3g and the total product masses is also 14.3g. That implies that no mass was !most in the reaction. The sum of masses on the left hand side corresponds with sum of masses on the right hand side of the reaction equation.
16.4 grams is the mass of solute in a 500 mL solution of 0.200 M
.
sodium phosphate
Explanation:
Given data about sodium phosphate
atomic mass of Na3PO4 = 164 grams/mole
volume of the solution = 500 ml or 0.5 litres
molarity of sodium phosphate solution = 0.200 M
The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:
The formula is
molarity =
putting the values in the equation, we get
molarity x volume = number of moles
0.200 X 0.5= number of moles
number of moles = 0.1 moles
Atomic mass x number of moles = mass
putting the values in the above equation
164 x 0.1 = 16.4 grams
16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.