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Klio2033 [76]
3 years ago
7

Elaborate on how the isotopes and their relative abundances affect the average atomic mass of an element.

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

Answer:

c. isotope mass both varies in neutron numbers and is weighted by its relative abundance to give the average atomic mass.

Explanation:

Isotopes differ in the number of neutrons only so they have the same atomic number but different mass number. Mass number is the sum of the number of protons and neutrons.  

Relative weighted mass of an element  

= [(relative abundance of isotope 1 x molecular mass of isotope 1) + (relative abundance of isotope 1 x molecular mass of isotope 2)] / 100

balandron [24]3 years ago
3 0
I would say c . isotope mass both varies in neutron numbers and its weighted by its relative abundance to give the average atomic mass. <span />
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The formula equation of Sodium carbonate + hydrochloric acid
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Answer :

The oxidation state of oxygen (O) in OF_2  is, (+2)

The oxidation state of carbon (C) in CO  is, (+2)

The oxidation state of nitrogen (N) in K_3N  is, (-3)

Explanation :

Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

Rules for Oxidation Numbers :

The oxidation number of a free element is always zero.

The oxidation number of a monatomic ion equals the charge of the ion.

The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.

The oxidation number of  oxygen (O)  in compounds is usually -2, but it is -1 in peroxides.

The oxidation number of a Group 1 element in a compound is +1.

The oxidation number of a Group 2 element in a compound is +2.

The oxidation number of a Group 17 element in a binary compound is -1.

The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

(a) The given compound is, OF_2

Let the oxidation state of 'O' be, 'x'

x+2(-1)=0\\\\x-2=0\\\\x=+2

The oxidation state of oxygen (O) in OF_2  is, (+2)

(b) The given compound is, CO

Let the oxidation state of 'C' be, 'x'

x+(-2)=0\\\\x-2=0\\\\x=+2

The oxidation state of carbon (C) in CO  is, (+2)

(c) The given compound is, K_3N

Let the oxidation state of 'N' be, 'x'

3(+1)+x=0\\\\3+x=0\\\\x=-3

The oxidation state of nitrogen (N) in K_3N  is, (-3)

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3 years ago
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