Chemistry, 15.10.2020 23:01, pooperjooper

What is the independent variable in this experiment?

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Chemistry, 22.06.2019 11:40, nick5442

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Chemistry, 22.06.2019 18:30, bibiansolis

The table lists the lattice energies of some compounds. compoundlattice energy (kj/mol)lif –1,036licl –853naf –923kf –821nacl –786which statement about crystal lattice energy is best supported by the information in the table? the lattice energy increases as cations get smaller, as shown by lif and kf. the lattice energy increases as the cations get larger, as shown by lif and licl. the lattice energy decreases as cations get smaller, as shown by nacl and naf. the lattice energy decreases as the cations get smaller, as shown by naf and kf.

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Chemistry, 22.06.2019 20:00, emilyswinge4421

Listenbase your answer to the question on the information below. nuclear radiation is harmful to living cells, particularly to fast-growing cells, such as cancer cells and blood cells. an external beam of the radiation emitted from a radioisotope can be directed on a small area of a person to destroy cancer cells within the body. cobalt-60 is an artificially produced radioisotope that emits gamma rays and beta particles. one hospital keeps a 100.0-gram sample of cobalt-60 in an appropriate, secure storage container for future cancer treatment. which choice represents the correct product for the beta decay of the co-60? fe-60ni-60fe-61ni-61

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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