Answer:
3.07 Cal/g
Explanation:
Step 1: Calculate the heat absorbed by the calorimeter
We will use the following expression.
Q = C × ΔT
where,
- C: heat capacity of the calorimeter (37.60 kJ/K = 37.60 kJ/°C)
- ΔT: temperature change (2.29 °C)
Q = 37.60 kJ/°C × 2.29 °C = 86.1 kJ
According to the law of conservation of energy, the heat released by the candy has the same magnitude as the heat absorbed by the calorimeter.
Step 2: Convert 86.1 kJ to Cal
We will use the conversion factor 1 Cal = 4.186 kJ.
86.1 kJ × 1 Cal/4.186 kJ = 20.6 Cal
Step 3: Calculate the number of Cal per gram of candy
20.6 Cal/6.70 g = 3.07 Cal/g
Answer:
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Explanation:
Answer:
0.158 moles
Explanation:
We are given;
9.50 x 10^22 molecules of CO
We are required to determine the number of moles;
We need to know;
1 mole of a compound = 6.022 × 10^23 molecules
Therefore;
9.50 x 10^22 molecules of CO will be equivalent to;
= 9.50 x 10^22 molecules ÷ 6.022 × 10^23 molecules/mole
= 0.158 moles
Therefore, the number of moles are 0.158 moles
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
