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valina [46]
3 years ago
8

State the number of electrons that must be gained by atoms of each of the following elements to obtain a stable electron configu

ration. a) CI b) Se C) N D) I E) S
Chemistry
1 answer:
Molodets [167]3 years ago
8 0

Answer:

The answer to your question is below

Explanation:

a) CI Chlorine is located in group VIIA, so its must gain one electron to be stable.

b) Se Selenium is located in group VIA, it must gain 2 electrons to be stable.

c) N Nitrogen is located in group VA, it must gain 3 electrons to be stable.

d) I Iodine is located in group VIIA, so it must gain 1 electron to be stable.

e) S sulfur is located in group VIA, so it must gain 2 electrons to be stable.

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What are the challenges for a "sustainable" world of lithium ion batteries?
Neporo4naja [7]

The Lithium-ion Battery Problem

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7 0
2 years ago
I NEED HELP!!
solniwko [45]

Carboxylic acids and alcohols have higher boiling point than other hydrocarbons due to their polarity and from the fact that they form very strong intermolecular hydrogen bonding. This is due to the large difference in their electronegativity that forms between the oxygen and the hydrogen atom. 

3 0
3 years ago
4. A 0.51 kg solution contains 87 mg of potassium iodide. Calculate the W/W concentration
anzhelika [568]

Taking into account the definition of percentage composition, the percent composition of potassium iodide in this sample is 0.017%.

<h3>Definition of percent composition </h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

To calculate the percentage of composition, it is necessary to know the mass of the element in a known mass of the compound.

<h3>Percentage Composition in this case</h3>

In this case, you know that a 0.51 kg (or 510000 mg, being 1 kg= 1000000 mg) solution contains 87 mg of potassium iodide.

Dividing the mass amount of potassium iodide present in the compound by the mass of the sample and multiplying it by 100 to obtain a percentage value, the percentage composition of potassium iodide is obtained:

percentage composition of potassium iodide=\frac{87 mg}{510000 mg} x100

<u><em>percentage composition of potassium iodide= 0.017%</em></u>

Finally, the percent composition of potassium iodide in this sample is 0.017%.

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3 0
2 years ago
2C2H2(g) 5O2(g) → 4CO2(g) 2H2O(g) This is a balanced equation for the combustion of acetylene(C2H2). How many moles of oxygen(O2
omeli [17]

The moles of oxygen required to completely react with 1-mole acetylene is 2.5 mol.

The moles of reactant and product in a chemical reaction to the whole number ratio is given by the stoichiometric coefficient of the balanced chemical equation.

<h3>Computation for the moles of oxygen</h3>

The balanced chemical equation for the reaction is :

\rm 2\;C_2H_2\;+\;5\;O_2\;\to\;4\;CO_2\;+\;2\;H_2O

From the balanced chemical equation, the 2 moles of acetylene react with 5 moles of oxygen.

The moles of oxygen react with 1 mole of acetylene are:

\rm 2\;mol\;C_H_2=5\;mol\;O_2\\\\1\;mol\;C_2H_2=\dfrac{5}{2}\;\times\;1\;mol\;O_2\\\\ 1\;mol\;C_2H_2=2.5\;mol\;O_2

The moles of oxygen required to completely react with 1-mole acetylene is 2.5 mol.

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8 0
3 years ago
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