A: Trial 1, because the average rate of the reaction is lower.
The rate of reaction is the speed with which reactants are converted into products. It is also the rate at which reactants disappear and products appear. The higher the rate of reaction, the greater the amount of product formed in a reaction.
If we look at the graph, we will realize that trial 1 produces a lesser amount of product than trial 2. This implies that the average rate of the reaction in trial 1 is lower than in trial 2.
Lower average rate of reaction implies lower concentration of the reactants since the rate of reaction depends on the concentration of reactants.
Hence trial 1 has a lower concentration of reactants because the average rate of the reaction is lower.
Your answer is B. It gets energy from the Sun.
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Explanation:
The reaction given is;
TiCl4 + H2O --> TiO2 + HCl
The reaction is not balanced, upon balancing it is given as;
TiCl4 + 2H2O → TiO2 + 4HCl
a. How many moles of H2O are needed to react with 6.50 moles of TiCl4?
From the reaction;
1 mol of TiCl4 requires 2 mol of H2O
6.50 mol of TiCl4 would require x mol of H2O
1 = 2
6.5 = x
x = 6.5 * 2 / 1 = 13.0 mol
b. How many moles of HCl are formed when 8.44 moles of TiCl4 react?
From the equation of the reaction;
1 mol of TiCl4 reacts to form 4 mol of HCl
8.44 mol of TiCl4 reacts to form x mol of HCl
1 = 4
8.44 = x
x = 8.44 * 4 / 1 = 33.76 mol
Answer:
b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.
Explanation:
The solubility of NaCH₃CO₂ in water is ~1.23 g/mL. This means that at room temperature, we can dissolve 1.23 g of solute in 1 mL of water (solvent).
<em>What would be the best method for preparing a supersaturated NaCH₃CO₂ solution?</em>
<em>a) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at room temperature while stirring until all the solid dissolves.</em> NO. At room temperature, in 100 mL of H₂O can only be dissolved 123 g of solute. If we add 130 g of solute, 123 g will dissolve and the rest (7 g) will precipitate. The resulting solution will be saturated.
<em>b) add 130 g of NaCH₃CO₂ to 100 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature. </em>YES. The solubility of NaCH₃CO₂ at 80 °C is ~1.50g/mL. If we add 130 g of solute at 80 °C and let it slowly cool (and without any perturbation), the resulting solution at room temperature will be supersaturated.
<em>c) add 1.23 g of NaCH₃CO₂ to 200 mL of H₂O at 80 °C while stirring until all the solid dissolves, then let the solution cool to room temperature.</em> NO. If we add 1.23 g of solute to 200 mL of water, the resulting solution will have a concentration of 1.23 g/200 mL = 0.00615 g/mL, which represents an unsaturated solution.
