1. Answer;
Copper (ii) carbonate.
The name of the compound CuCO3 is copper (ii) carbonate.
Explanation;
Cu is the chemical symbol for the copper and CO3 is the chemical symbol for the carbonate group and each one of them has valency of two. Therefore, a compound CuCO3 is formed.
2. Answer;
Yes
Ca2+ reacted with Na2S to form CaS and Na+
Explanation:
Calcium ions reacts with sodium sulfide to form calcium sulfide and sodium ions.
For example; a salt of calcium, calcium carbonate reacts with sodium sulfide to form sodium carbonate and calcium sulfide.
3. Answer;
NaCl and Ag+ do not form a product
Explanation;
The reaction between sodium chloride and silver metal will not take place. This is because silver (Ag) is less reactive than sodium metal and therefore cannot displace sodium from its salt. In other words, silver metal is lower in the reactivity series as compared to sodium metal which indicates sodium metal is more reactive than silver.
4. Answer;
Formation of a white precipitate ; this indicates that silver sulfide is insoluble in water.
Explanation;
When an aqueous solution containing Ag+ ions is added to aqueous solution of sodium sulfide (Na2S), there will be formation of white precipitate. Formation of white precipitate indicates that a reaction has taken place to form a water insoluble compound. The water insoluble compound occurs as a precipitate. The white precipitate is silver sulfide (Ag2S)
2 Ag+ (aq) + Na2S(aq) ----- Ag2S (s) + 2 Na+ (aq)
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
The concentration of sodium and sulphate ions are [
] = 0.4 M, [
] = 0.2 M
Explanation:
The molar concentration is defined as the number of moles of a molecule or an ion in 1 liter of a solution.
In the given solution, the concentration of the salt sodium sulphate is 0.2M. So, 0.2 moles of sodium sulphate is present in 1 liter of solution.
Assuming 100% dissociation,
1 molecule of sodium sulphate gives 2 ions of sodium and 1 ion of sulphate.
So 0.2 moles of sodium sulphate will give 0.4 moles of sodium ions and 0.2 moles of sulphate ions.
Answer:
0.213 J/g°C
Explanation:
To calculate specific heat of the metal, the formula is used:
Q = m × c × ∆T
Where Q = amount of heat
m = mass
c = specific heat
∆T = change in temperature
According to this question, Q = 37.7 J, m= 12.5 g, initial temperature= 19.5 °C, final temperature = 33.6°C, c=?
Q = m × c × ∆T
37.7 = 12.5 × c × (33.6-19.5)
37.7 = 12.5c × 14.1
37.7 = 176.25c
c = 37.7/176.25
c = 0.2139
Hence, the specific heat of the metal is 0.213 J/g°C
Answer:
<u>So, the right answer is</u>
No. of moles of FeS₂ = 0.25 mole
Explanation:
From the balanced
4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2
it is clear that 4 mol FeS₂ react with O₂ to give Fe₂O₃ and 8 mol of SO₂
First, we have to convert mass of SO₂ into No. of moles as following:
SO₂ has molar mass = 64 g/mol
No. of moles of SO₂ = (mass / molar mass) = (32 g / 64 g/mol) = 0.5 mol
we know that
4 mol FeS₂ gives→ 8 mol of SO₂
1 mol FeS₂ gives→ 2 mol of SO₂
??? mol FeS₂ gives→ 0.5 mol of SO₂
No. of moles of FeS₂ = (0.5 mol * 1 mol ) / 2 mol = 0.25 mol
<u>So, the right answer is</u>
No. of moles of FeS₂ = 0.25 mol
