Chemista
four Element
and I have
alomo
2, 6, 16, and 20 respectivel
which of th...
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Chemistry, 05.03.2021 14:00 Violinartist7267
Chemista
four Element
and I have
alomo
2, 6, 16, and 20 respectivel
which of those Element is a metal
A, B,C
numbers
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Answers: 3
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Other questions on the subject: Chemistry
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Chemistry, 21.06.2019 18:00, Falconpride4079
Construct the hypothetical phase diagram for metals a and b between room temperature (20c) and 700c, given the following information: * the melting temperature of metal a is 480c. • the maximum solubility of b in a is 4 wt% b, which occurs at 420c. • the solubility of b in a at room temperature is 0 wt% b. • one eutectic occurs at 420c and 18 wt% b–82 wt% a. • a second eutectic occurs at 475c and 42 wt% b–58 wt% a. • the intermetallic compound ab exists at a composition of 30 wt% b–70 wt% a, and melts congruently at 525c.• the melting temperature of metal b is 600c. • the maximum solubility of a in b is 13 wt% a, which occurs at 475c. • the solubility of a in b at room temperature is 3 wt% a.
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Chemistry, 22.06.2019 07:00, maryjane8872
Which set of characteristics best describes igneous rock? a) largest type of rock, made of organic matter, hardest type of rock b) least abundant type of rock, made of other rocks, made mostly of minerals c) found on all continents, contains wavy bands of stripes, contains fossils d) most abundant type in earth's crust, made of magma/lava, contains no fossils
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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
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