Chemistry, 07.03.2020 01:27 arturo698839
A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution.
Answers: 2
Chemistry, 22.06.2019 12:10, kaitlynbernatz2778
If a molecule with a molecular formula of c13h18 is treated with an excess of h2 in the presence of finally divided pt metal under conditions required for maximum hydrogenation of the molecule to give a molecule with a formula c13h24, how many rings are in the molecule?
Answers: 3
Chemistry, 22.06.2019 14:50, rebeccamckellpidge
The table compares the number of electrons in two unknown neutral atoms. comparison of electrons atom number of electrons a 9 d 11 use this information to determine the number of valence electrons in the atoms. which of the following correctly compares the stability of tthe table compares the number of electrons in two unknown neutral atoms. comparison of electrons atom number of electrons a 9 d 11 use this information to determine the number of valence electrons in the atoms. which of the following correctly compares the stability of the two atoms? both are unreactive. both are highly reactive. a is unreactive and d is reactive. a is reactive and d is unreactive.
Answers: 3
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
Answers: 1
A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Ca...
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