Chemistry, 01.07.2021 17:40 vaeh41

IUPAC name of compound (ii)

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In the 1800s, one of the statements in john dalton's atomic theory was that atoms are indivisible. later experimental evidence led to the discovery of subatomic particles such as neutrons, electrons, and protons. what happened to the indivisible atom part of dalton's atomic theory, and why?

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Chemistry, 22.06.2019 12:30

Nebulae are enormous clouds in outer space. they are made mostly of hydrogen gas, helium gas, and dust. some nebulae glow brightly, while others do not. the stars that people see are huge, bright balls of glowing gas. they are made mostly of hydrogen and helium. which statement correctly describes other ways in which nebulae and stars are different? a. stars can form inside a nebula but a nebula can never be produced by any star. b. a star always has a higher density than a nebula. c. stars can never form inside a nebula but a nebula can be produced by any star. d. a nebula always has a higher density than a star.

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Chemistry, 22.06.2019 17:50

Cryolite, na3alf6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. start this question by first balance the chemical equation.1.) balance the equation: - alo3(s)+naoh(l)+hf(> na3alf6+h2o(g). 2.) if 17.5 kilograms of al2o3(s), 51.4 kilograms of naoh(l), and 51.4 kilograms of hf(g) react completely, how many kilograms of cryolite will be produced? 3.)which reactants will be in excess, (al2o3, naoh, or hf) 4.)what is the total mass of the excess reactants left over after the reaction is complete in kg?

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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

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IUPAC name of compound (ii)
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