Determine the total net force and the direction of movement for each situation
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Chemistry, 15.09.2021 23:10 brionna246
Determine the total net force and the direction of movement for each situation
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Chemistry, 21.06.2019 19:00
Asmall amount of a solid is added to water. the observation made after fifteen minutes is shown in the figure. which of these solids has been probably added to water? a) oil b) sand c) sugar d) wood chips
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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 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
Answers: 1
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