Answer:
The answer to your question is: ΔH = -283 kJ/mol, first option
Explanation:
Reaction
CO + O₂ ⇒ CO₂
ΔH = ∑H products - ∑H products
ΔH = -393.5 - (-110.5 + 0)
ΔH = -393.5 + 110.5
ΔH = -283 kJ/mol
Answer:
Q1. C
Q2 and Q3 are correct.
Explanation:
Since F=ma, and the force is a constant,
for the greatest acceleration, the mass of the ball must be the least.
Thus ball C has the greatest acceleration.
Let's check:
A) F=ma
a=F/m
a= F/68
B) a=F/72
C) a= F/64 (✓)
The smaller the denominator, the larger the value of a.
(Think: 1/2 >1/3)
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.
Answer:
Potential energy is the stored energy of position possessed by an object.
