Chemistry, 28.01.2020 15:50 kdcloyd88191

Elements occur in a number of isotopic forms. in this problem, you will learn about the notation used to distinguish different isotopes. how many protons z and how many neutrons n are there in a nucleus of the most common isotope of silicon, 28/14si?

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Answer from: bobbyxii6033

There are 14 protons (Z) and 14 neutrons (N) in a nucleus of the most common isotope of silicton.

Explanation:

The notation of the most common isotope of silicon is:

$^{28}_{14}Si$

The superscript, 28, to the left of the chemical symbol, Si, is the mass number, usually identified with the symbol A, and it is the sum of the protons and neutrons.

The subscript, 14, to the left of the chemical symbolr, Si, is the atomic number, Z, which is the number of protons.

Then, you have this equation:

mass number = number of protons + neutrons, or, in symbols:A = Z + N

The number of protons is the same for any isotope of the same element. This is, all the isotopic forms of silicon have the same number of prtons: 14

The number of neutrons is determined from the equation A = Z + N, solving for N:

N = A - Z = 28 - 14 = 14.

Hence, the most common isotope of silicon has 14 protons and 14 neutrons.

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

Answer from: Quest

The answer is b metal denting

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Elements occur in a number of isotopic forms. in this problem, you will learn about the notation use...

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