Answer:
1) Salts X and Y
2) The solubility of the salts
3) a) The solvent
b) The solvent temperature
Explanation:
1) The independent variable is the variable that is suspected to be the cause of the subject of the investigation
The given investigation is meant to investigate the solubility of different salts
Therefore, the solubility is expected to be dependent on the type of salt, and the independent variable is the type of salt, X or Y
2) The dependent variable is the effect meant to be observed in the investigation, which is the solubility of the salt in water at room temperature
3) The control variables are the variables which are held constant during the investigation, including;
a) The solvent used if the investigation; water
b) The temperature of the solvent; Room temperature
Answer:
3 × 10⁴ kJ
Explanation:
Step 1: Write the balanced thermochemical equation
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ
Step 2: Calculate the moles corresponding to 865.9 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
865.9 g × 1 mol/18.02 g = 48.05 mol
Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced
According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.
48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ
The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C.
Answer:
Electrolysis
Explanation:
The electrolysis of water is one such experiment that shows that water is made up of hydrogen and oxygen atoms only in the ratio of 2 to 1.
In the electrolysis of water, electricity is passed through acidified water to cause it to decompose.
The electrolysis of water is also known as the electrolysis of dilute tetraoxosulphate (VI) acid.
At the cathode, H⁺ ions are discharged and hydrogen gas is liberated:
2H⁺ + 2e⁻ → H₂
At the anode, both the sulfate ion and hydroxyl ions migrate to this electrode. Only the OH⁻ is selected for preferential discharge due to its lower position in that activity series.
4OH⁻ → 2H₂O + O₂ + 4e⁻
Oxygen gas is produced at the anode.
This electrolysis demonstrates the volumetric composition of water that is, 2 volumes of hydrogen at the cathode and 1 volume of oxygen at the anode.
