Chemistry, 30.11.2019 15:31 minixkevon

The functional group of this molecule is amines ? true ?

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Chemistry, 22.06.2019 04:00, queenkimm26

Tin has ten stable isotopes. the heaviest, 124sn, makes up 5.80% of naturally occuring tin atoms. how many atoms of 124sn are present in 82.0 g of naturally occurring tin? what is the total mass of the 124sn atoms in this sample?

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Chemistry, 22.06.2019 10:10, babyphoraaaaa

For the reaction, 4 a(g) + 3 b(g) => 2 c(g), the following data were obtained at constant temperature. experiment initial[a],mol/l initial [b],mol/l initial rate, m/min 1 0.200 0.150 5.00 2 0.400 0.150 10.0 3 0.200 0.300 10.0 4 0.400 0.300 20.0 which of the following is the correct rate law for the reaction? 1. rate = k[a]2[b]2 2. rate = k[a][b] 3. rate = k[a]2[b] 4. rate = k[a][b]2

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Chemistry, 22.06.2019 21:30, imalexiscv

Plzz a sample of table sugar (sucrose, c12h22o11) has a mass of 7.801 g. ● a) calculate the number of moles of c12h22o11 in the sample b) calculate the number of moles of each element in c12h22o11 (number of moles of c, number of moles of h & number of moles of o) in the sample. (use your answer from part a as your starting point.) show your work and highlight your final answer. calculate the number of atoms of each element in c12h22o11 (number of atoms of c, number of atoms of h & number of atoms of o) in the sample. (use your answers from part b as your starting for each element.) show your work and highlight your final answer.

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Chemistry, 22.06.2019 21:50, BookandScienceNerd

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