Answer:
Reverse the 
reaction
Explanation:
Reactions:
Overall:
As can be seen, in the overall reaction we have 
in the reactants like in the second reaction and 
in the products. The is in the first reaction but as a reactant so we need to reverse that reaction:
Combining:
“Metal ions are typically smaller than their corresponding neutral atoms”
Answer) Smaller
Answer:
The intermolecular forces between water molecules are stronger than those between oxygen molecules. In general, the bigger the molecule, the stronger the intermolecular forces, so the higher the melting and boiling points.
The correct answer is industrial smog. This type of smog exists in coal power plants which creates smoke and sulfur dioxide which may mix with fog creating a thick blanket of haze. Sulfur dioxide is one primary component of an industrial smog.
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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