Chemistry, 16.10.2020 23:01 sighgabbie
A student was measuring the time it took for an experiment to take place. They found the time to be 40 seconds. However, they had started timing a few seconds after the experiment had started, and the actual time was 43 seconds. What is the percent error?
Answers: 3
Chemistry, 21.06.2019 23:00
City a and city b had two different temperatures on a particular day. on that day, four times the temperature of city a was 8â° c more than 3 times the temperature of city b. the temperature of city a minus twice the temperature of city b was â’3â° c. what was the temperature of city a and city b on that day? city a was 5â° c, and city b was 4â° c. city a was 3â° c, and city b was â’1â° c. city a was 8â° c, and city b was â’3â° c. city a was 5â° c, and city b was â’5â° c.
Answers: 2
Chemistry, 22.06.2019 06:00
One does not belong why? ice, gold ,wood ,diamond and table salt
Answers: 1
Chemistry, 22.06.2019 20:00
There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. in the first step, calcium carbide and water react to form acetylene and calcium hydroxide: cac2 (s) + 2h2o (g) → c2h2 (g) + caoh2 (s) =δh−414.kj in the second step, acetylene, carbon dioxide and water react to form acrylic acid: 6c2h2 (g) + 3co2 (g) + 4h2o (g) → 5ch2chco2h (g) =δh132.kj calculate the net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide from these reactions. round your answer to the nearest kj .
Answers: 3
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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