I would go with B. Not everyone can learn this skill.
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)
This description applies and is suitable for what a chemical precipitate is. A precipitate is a product that is formed from a certain chemicals reaction that yields a solid that is insoluble in the reaction vessel. It is usually white and opaque.
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
Answer:
116 g
Explanation:
From the question given above, the following data were obtained:
Number of mole of calcium = 2.9 moles
Mass of calcium =.?
The mole and mass of a substance are related according to the following formula:
Mole = mass / molar mass
With the above formula, we can obtain the mass of calcium. This can be obtained as follow:
Number of mole of calcium = 2.9 moles
Molar mass of calcium = 40 g/mol
Mass of calcium =.?
Mole = mass / molar mass
2.9 = mass of calcium / 40
Cross multiply
Mass of calcium = 2.9 × 40
Mass of calcium = 116 g
Therefore, the mass of 2.9 moles of calcium is 116 g.
