In the reaction as follows: NH2- + CH3OH → NH3 + CH3O−, NH2- is the Brønsted-Lowry base.
BRØNSTED-LOWRY BASE:
- According to Bronsted-Lowry definition of a base and acid, a base is substance that accepts an hydrogen ion or proton (H+) while an acid is a substance that donates a proton.
- According to this reaction given as follows: NH2 + CH3OH → NH3+ CH3O-
- NH2- is a reactant that accepts a hydrogen ion (H+) to become NH3+
- NH3+CH3OH is a reactant that donates hydrogen ion (H+)
- Since NH2- accepts a proton, this means that in the reaction as follows: NH2 + CH3OH → NH3 + CH3O−, NH2- is the Brønsted-Lowry base.
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False, it must be either empty or filled with electrons order to be most stable.
Answer:
The answer is C. The high solvation energy for LI+
Explanation:
LiF has lower solubility because of the high solvation energy of Li+ ion. This is due to the smaller size and very big charge compared to Cs+ ion which has a bigger size and solvent molecules easily surround it.
Solvation energy is simply the amount energy that is required to make a solute dissolve in a solvent.
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>
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<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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