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Stels [109]
3 years ago
13

Which of the following correctly lists the name of the element, the symbol for the ion, and the name of the ion?

Chemistry
2 answers:
Yanka [14]3 years ago
8 0

The correct answer is B.

Hope this helps you.

Mrac [35]3 years ago
6 0

Answer:

zinc, Zn²⁺, zincate ion

Explanation:

Let us check with all the given examples:

a) Fluorine (F). It is the most electronegative element and thus cannot form positive ion. It will gain electron and form a negative ion (F⁻) known as Fluoride ion.

b) Zinc: it is a metal and give two electrons easily to attain noble gas configuration. Hence it will form Zincate ion (Zn⁺²).

c) Copper,: it is a metal and can form positive ion easily.

It can give either two or one electron to form

Cu⁺²: Cupric ion [More common]

Cu⁺: Cuprous ion

d) Barium : it is alkaline earth metal and will give only two electron to form dipositive ion barium ion (Ba⁺²)

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An ionized helium atom has a mass of 6.6 × 10-27 kg and a speed of 5.3 × 105 m/s. It moves perpendicular to a 0.78-T magnetic fi
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Explanation:

In a magnetic field, the radius of the charged particle is as follows.

             r = \frac{mv}{qB}

where,   m = mass,      v = velocity

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Therefore, q will be calculated as follows.

         q = \frac{mv}{rB}

            = \frac{6.6 \times 10^{-27} \times 5.3 \times 10^{5}}{0.014 m \times 0.78 T}

            = \frac{34.98 \times 10^{-22}}{0.01092}

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Thus, we can conclude that the charge of the ionized atom is +2e.

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3 years ago
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Explanation:

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4 years ago
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Answer:

See explanation

Explanation:

The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.

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In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.

Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.

Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.

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