Chemistry, 06.05.2021 04:50 zmerriweather167

Proton is to as Neutron is to as abbreviations are to

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Chemistry, 22.06.2019 17:10

Some liquids can be distilled, but only at temperatures that are so high that it is impractical, or so high the compound decomposes. explain why distillation such compounds at significantly less than atmospheric pressure (some degree of vacuum) would solve this problem.

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Chemistry, 22.06.2019 18:00

Answer asap need it by wednesday morning carry out the following calculations on ph and ka of from data. i. calculate the ph of 0.02m hcl ii. calculate the ph of 0.036m naoh iii. calculate the ph of 0.36m ca(oh)2 iv. calculate the ph of 0.16m ch3cooh which has ka = 1.74 x 10-5 mol dm-3 v. calculate ka for weak acid ha which has a ph of 3.65 at 0.30m concentration vi. calculate the ka of a solution made by mixing 15.0 cm3 0.2m ha and 60.0 cm3 0.31m a-. [ph= 3.80] vii. calculate the ph of a solution made by mixing 15.0 cm3 0.1m naoh and 35.0 cm3 0.2m hcooh. [ka = 1.82 x 10-4 m]

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Chemistry, 22.06.2019 20:30

Consider the following unbalanced equation for the combustion of hexane: αc6h14(g)+βo2(g)→γco2(g)+δh2o(g) part a balance the equation. give your answer as an ordered set of numbers α, β, γ, use the least possible integers for the coefficients. α α , β, γ, δ = nothing request answer part b determine how many moles of o2 are required to react completely with 5.6 moles c6h14. express your answer using two significant figures. n n = nothing mol request answer provide feedback

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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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