<h3>Answer:</h3>
#1. Ca²⁺
# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
<h3>Explanation:</h3>
The question above concerns solubility of salts or ions in water.
The solution given contains Ag+, Ca2+, and Co2+ ions.
- In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
- In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.
- The net ionic equation for the reaction is;
Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
- From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.
- In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.
- The net ionic equation for the reaction is;
3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
- According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.
I believe that it is all of the above.
"A Scientific Hypothesis Must Be "Falsifiable". A scientific hypothesis must be testable, but there is a much stronger requirement that a testable hypothesis must meet before it can really be considered scientific."
I hope this helped, have a nice day!
Molar mass is the mass of 1 mol of substance.
Molar masses of compounds can be calculated by the sum of the products of molar masses of individual atoms by number of corresponding individual atoms.
Compound formula is C₉H₈O₄
the molar masses of the atoms making up the compound
C - 12 g/mol x 9 C = 108
H - 1 g/mol x 8 H = 8
O - 16 g/mol x 4 O = 64
therefore molar mass of aspirin = 108 + 8 + 64 = 180 g/mol
answer is 3.180
Answer:
stay the same.
Explanation: Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):
Na: 1s2 2s2 2p6 3s1
Ar: s2 2s2 2p6 3s2 3p6
As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:
Na: 3s1
Ar: 3s2, 3p6
All the new electrons electrons filled the third energy level (3s and 3p). So the energy level does not change, just the orbitals.
