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2 years ago

Which atomic model was proposed as a result of J.J Thomsons work?

Chemistry

1 answer:

Dafna11 [192]2 years ago

6 0

, J.J. Thomson discovered the electron by experimenting with a Crookes, or cathode ray, tube.

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Determine the type of alcohol corresponding to each given description or name. 1-pentanol 3-ethyl-3-pentanol 2-hexanol An alcoho

pochemuha

Answer:

1). 1-pentanol - Primary

2). 3-ethyl-3-pentanol - Tertiary

3). 2-hexanol - Secondary

4). Alcohol with two other carbons attached to the carbon with the hydroxyl group - Secondary

5). Alcohol with one other carbon attached to the carbon with the hydroxyl group - Primary

6). Alcohol with three other carbons attached to the carbon with the hydroxyl group - Tertiary

Explanation:

The distinct types of alcohol have been matched with the categories above as per their descriptions provided. In chemistry, alcohols have been categorized into three different categories namely primary, secondary, and tertiary.

In the primary type, those alcohols are involved in which there is an association of hydroxyl group to a primary atom of carbon along with a minimum of two atoms of hydrogen. Example; ethanol.

In the secondary type, the alcohols have an association of carbon atoms to hydroxyl with a single atom of hydrogen and has a formula of '-CHROH.' Example: 2 - propanol.

In the tertiary alcohols, here the association is between the hydroxyl group with the carbon atom that is saturated and possessing 3 atoms of carbon associated with it. It has a formula of '-CR2OH.' Example: 3-ethyl-3-pentanol, -tert -butyl alcohol, etc.

7 0

2 years ago

Could the cuttlefish tell the difference between the blue and yellow in the lab?

jeka94

Answer:

yes

Explanation:

the cuttle fish tell the difference between blue and yellow

5 0

1 year ago

Which of the following organisms would likely be found on a decaying deer carcass (dead deer)?

TEA [102]

Answer:

c. bacteria and fungus

Explanation:

3 0

3 years ago

Read 2 more answers

Calculate the ph of a 0.021 m nacn solution. [ka(hcn) = 4.9  10–10]

ohaa [14]

Answer is: pH of solution of sodium cyanide is 11.3.
Chemical reaction 1: NaCN(aq) → CN⁻(aq) + Na⁺(aq).
Chemical reaction 2: CN⁻ + H₂O(l) ⇄ HCN(aq) + OH⁻(aq).
c(NaCN) = c(CN⁻) = 0.021 M.
Ka(HCN) = 4.9·10⁻¹⁰.
Kb(CN⁻) = 10⁻¹⁴ ÷ 4.9·10⁻¹⁰ = 2.04·10⁻⁵.
Kb = [HCN] · [OH⁻] / [CN⁻].
[HCN] · [OH⁻] = x.
[CN⁻] = 0.021 M - x..
2.04·10⁻⁵ = x² / (0.021 M - x).
Solve quadratic equation: x = [OH⁻] = 0.00198 M.
pOH = -log(0.00198 M) = 2.70.
pH = 14 - 2.70 = 11.3.

4 0

2 years ago

Use electron configurations to account for the stability of the lanthanide ions Ce⁴⁺ and Eu²⁺.

Citrus2011 [14]

Answer:

Explanation:

The Ce metal has electronic configuration as follows

[Xe] 4f¹5d¹6s²

After losing 4 electrons , it gains noble gas configuration ,. So Ce ⁺⁴ is stable.

Eu has electronic configuration as follows

[ Xe ] 4 f ⁷6s²

[ Xe ] 4 f ⁷

Its outermost orbit contains 2 electrons so Eu²⁺ is stable. Its +3 oxidation state is also stable.

Ce⁺²

7 0

3 years ago