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trasher [3.6K]
3 years ago
14

A lithium atom contains 3 protons, 4 neutrons and 3 electrons. What would be formed it one proton if one proton is added to this

atom
Chemistry
2 answers:
yarga [219]3 years ago
4 0

Answer: Beryllium ion (Be+)

Explanation:

Addition of one proton to Li would lead to formation of Beryllium ion since Beryllium contains 4 protons and 4 electrons to counter the charges and make the element neutral. But in this case, Be+ is formed to show a deficit of one electron (e-).

marusya05 [52]3 years ago
3 0
<h2>Answer:</h2>

Beryllium ion.

<h2>Explanation:</h2>
  • Adding one proton to an atom of lithium which have 3 protons, 4 neutrons and 3 electrons will form a beryllium ion.
  • The new atom formed after adding a proton will have 4 protons and 4 neutrons since to have a mass number of 9 then it has to form an ion.
  • Beryllium is of gray colour, strong, light-weight, and used as hardening agent in different alloys. It has one of the highest m.p(melting point) in the light metals.

Result: On adding a proton in lithium will form a beryllium ion.

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M=10,V=2 What is the density
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Answer:

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Explanation:

Steps:

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m

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While the reaction in the lab text gives the product 4-methylcyclohexene, it is possible that several products are formed in thi
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Answer:

Explanation:

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6 0
3 years ago
Antimony has two naturally occuring isotopes, 121 Sb and 123 Sb . 121 Sb has an atomic mass of 120.9038 u , and 123 Sb has an at
valina [46]

Answer:

Percentage abundance of 121 Sb is = 57.2 %

Percentage abundance of 123 Sb is = 42.8 %

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

Since the element has only 2 isotopes, so the let the percentage of first be x and the second is 100 -x.

For first isotope, 121 Sb :

% = x %

Mass = 120.9038 u

For second isotope, 123 Sb:

% = 100  - x  

Mass = 122.9042 u

Given, Average Mass = 121.7601 u

Thus,  

121.7601=\frac{x}{100}\times 120.9038+\frac{100-x}{100}\times 122.9042

120.9038x+122.9042\left(100-x\right)=12176.01

Solving for x, we get that:

x = 57.2 %

<u>Thus, percentage abundance of 121 Sb is = 57.2 % </u>

<u>percentage abundance of 123 Sb is = 100 - 57.2 %  = 42.8 %</u>

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