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mixer [17]
3 years ago
10

Which element will most easily lose an electron? A. Calcium (Ca) B. Potassium (K) C. Boron (B) D. Krypton (Kr)

Chemistry
2 answers:
Artemon [7]3 years ago
8 0
<h2>Answer:</h2>

Option B. Potassium(K).

<h2>Explanation:</h2>

Electronic configuration of the given elements are:

  1. Ca - [Ar] 4s²
  2. K - [Ar] 4s¹
  3. B - [He] 2s2 2p1
  4. Kr -  [Ar] 3d¹⁰4s²4p⁶
  1. Krypton(K) have 36 electrons and it is a noble gas and hence all of its shells are completely filled with electrons and hence it will never loose electrons in normal conditions.
  2. Boron(B) have 5 electrons and 3 electrons in its outer shell. In order to attain a stable configuration it will loose 3 electrons and it is difficult to loose 3 electrons at a time for an atom.
  3. Calcium(Ca) have 20 electrons and 2 electrons in its outermost shell, in order to attain a stable configuration it will loose 2 electrons. it is quite difficult but easier than Boron.
  4. Potassium(K) have 21 electrons and 1 electron in its outermost orbit and in order to attain a stable configuration it will loose 1 electron. It is much easier to donate 1 electron than 2 or 3 electrons.

Result: Potassium will loose an electron most easily from the given elements.

geniusboy [140]3 years ago
6 0

Answer:

Potassium

Explanation:

Answer via Educere/ Founder's Education

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At elevated temperatures, molecular hydrogen and molecular bromine react to partially form hydrogen bromide:
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<u>Answer:</u> The moles of bromine gas at equilibrium is 0.324 moles.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}        .......(1)

Calculating the initial moles of hydrogen and bromine gas:

  • <u>For hydrogen gas:</u>

Moles of hydrogen gas = 0.682 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

\text{Molarity of solution}=\frac{0.682mol}{2.00L}=0.341M

  • <u>For bromine gas:</u>

Moles of bromine gas = 0.440 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

\text{Molarity of solution}=\frac{0.440mol}{2.00L}=0.220M

Now, calculating the molarity of hydrogen gas at equilibrium by using equation 1:

Equilibrium moles of hydrogen gas = 0.566 mol

Volume of solution = 2.00 L

Putting values in equation 1, we get:

\text{Molarity of solution}=\frac{0.566mol}{2.00L}=0.283M

Change in concentration of hydrogen gas = 0.341 - 0.283 = 0.058 M

This change will be same for bromine gas.

Equilibrium concentration of bromine gas = (\text{Initial concentration}-\text{Change in concentration})=0.220-0.058=0.162M

Now, calculating the moles of bromine gas at equilibrium by using equation 1:

Molarity of bromine gas = 0.162 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

0.162M=\frac{\text{Moles of bromine gas}}{2.00L}\\\\\text{Moles of bromine gas}=(0.162mol/L\times 2.00L)=0.324mol

Hence, the moles of bromine gas at equilibrium is 0.324 moles.

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3 years ago
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