To calculate the average atomic weight, each exact atomic weight is multiplied by its percent abundance, then add the results together and round off to an appropriate number of significant figures. So according to what’s given, the average atomic mass of elements M should be (23.9850 x 0.7899) + (24.9858 x 0.1000) + (25.9826 x 0.1101) = 24.30
When atoms of beryllium-9 are bombarded with alpha particles, neutrons are produced, the new isotope will be C-6.
In research, the physicist James Chadwick attacked Beryllium using alpha particles produced by the nuclear reactions of polonium naturally. High radiation penetration through some kind of lead shield was seen, which the then-current particle theories were unable to account for.
There was indeed the production of a carbon-12 nucleus as well as the production of a neutron whenever one blasted beryllium with alpha particles.
The reaction of alpha decay can be written as:
→
Therefore, when atoms of beryllium-9 are bombarded with alpha particles, neutrons are produced, the new isotope will be C-6.
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There are 9.8 moles of sodium (Na) in a sample of 5.87 x 10²⁴ atoms of sodium.
<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>
The number of moles of a substance can be calculated by dividing the number of molecules in the substance by Avogadro's number. That is;
no. of moles = no. of atoms ÷ 6.02 × 10²³
According to this question, there are 5.87 x 10²⁴ atoms of sodium. The number of moles is as follows:
no. of moles = 5.87 x 10²⁴ atoms ÷ 6.02 × 10²³
no. of moles = 0.98 × 10¹
no. of moles = 9.8moles
Therefore, there are 9.8 moles of sodium (Na) in a sample of 5.87 x 10²⁴ atoms of sodium.
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Well, I say the answer is Thermosphere! Hope I helped :)
Answer: N (the Nitrogen)
Explanation:
Reduction refers to a decrease in oxidation number/state due to the gaining of electrons. As such the species that is being reduced will show a decrease in oxidation state.
Based on the redox rules,
Zn(s) has oxidation number of 0 [<em>rule 1: the oxidation number of an element in its free (uncombined) state is zero</em>]
Zn²⁺ has oxidation number of +2 [<em>rule 2: The oxidation number of a monatomic (one-atom) ion is the same as the charge on the ion</em>]
Now, since Nitrogen is enbedded in a polyatomic ion in both cases, you have to do a bit a calculation to obtain the oxidation state.
For NO₃⁻ : N + (-2 × 3) = -1
N - 6 = -1
N = 5
<em>[Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 6: The oxidation state of hydrogen in a compound is usually +1]</em>
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For NH₄⁺ :
N + (4 x 1) = 1
N + 4 = 1
N = -3
[<em>Rule 3: The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion; Rule 5: The oxidation number of oxygen in a compound is usually –2]</em>
Therefore, Zn moves from oxidation state of 0 to +2 (oxidation), while N moves from +5 to -3 (reduction).