1. D
2. DR
3. SR
4. DR
5. C
6. SR
7. S
8. D
10. C
Answer:
Mercury is useful in lighting because it contributes to the bulbs' efficient operation and life expectancy. Fluorescent and other mercury-added bulbs are generally more energy efficient and last longer than incandescent and other equivalent forms of lighting.
Explanation:
Answer:
V of Sulfur tetrafluoride is 17.2 L
Explanation:
Given data;
T = -6°C = 267K [1° C = 273 K]
n = 786 mmol of SF4 which is 0.786 mol
P = 1 atm
from ideal gas law we have
PV = nRT
where n is mole, R is gas constant, V is volume
V of Sulfur tetrafluoride is 17.2 L
when the top of the pencil moves in one direction, the bottom end on the ground will move in another, but due to the frictional force it will move by a lesser amount, and the center of mass of the pencil will be accelerated in the horizontal direction due to the force of friction
Answer:
C. at low temperature and low pressure.
Explanation:
- <em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
<em />
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.</em>
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<em><u>Effect of pressure:</u></em>
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.
<em>So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), </em><em>so the equilibrium partial pressure of CO (g) can be maximized at low pressure.</em>
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<u><em>Effect of temperature:</em></u>
- The reaction is exothermic because the sign of ΔH is (negative).
- So, we can write the reaction as:
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.</em>
- Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, <em>so the equilibrium partial pressure of CO (g) can be maximized at low temperature.</em>
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<em>C. at low temperature and low pressure.</em>
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