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Chemistry, 22.06.2019 07:20
Part b: study of equilibrium on solubility: mg(oh)2(s) ⇌ mg2+(aq) + 2 oh–(aq) cloudy clear (pink) 7. a. b. 8. a. b. 9. 10. 11. 12. when adding concentrated hydrochloric acid, how did the appearance of the equilibrium mixture change? the change in appearance indicated a shift in the point of equilibrium. in which direction did the equilibrium shift? (l) left (r) right explain your answer to question 7a. you should indicate which ion was added to or removed from the equilibrium mixture. when adding edta, how did the appearance of the equilibrium mixture change? the change in appearance indicated a shift in the point of equilibrium. in which direction did the equilibrium shift? (l) left (r) right explain your answer to question 8a. you should indicate which ion was added to or removed from the equilibrium mixture. upon heating in which direction is the equilibrium shifting? upon cooling in which direction is the equilibrium shifting? is the forward reaction a. endothermic explain your answers to questions 9, 10, and 11. (l) left (r) right (l) left (r) right b. exothermic
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Chemistry, 22.06.2019 17:00
What is the approximate vapor pressure when the gas condenses at 70 degrees celsius
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Chemistry, 22.06.2019 17:30
Upon decomposition, one sample of magnesium fluoride produced 1.65 kg of magnesium and 2.56 kg of fluorine. a second sample produced 1.32 kg of magnesium. part a how much fluorine (in grams) did the second sample produce?
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Chemistry, 22.06.2019 21:50
Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes
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