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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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Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume
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Answer : 7.87 X 10^{-6}.

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, \frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )

on solving with the given values ,Q_{s}= 1.86 eV and T as 645 + 273 K and rest are the constants.

\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )

we get the answer as 7.87 X 10^{-6}.
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Which type of orbit does the earth have with the sun ?
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Answer:

elliptical orbit

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Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the
Sonbull [250]

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

  • If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If you know pH and pka:

10^(pH-pka) = \frac{[A^-]}{[HA]}

The ratio will be: 10^(pH-pka)

  • At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

0 = log₁₀ \frac{[A^-]}{[HA]}

10^0 = \frac{[A^-]}{[HA]}

1 = \frac{[A^-]}{[HA]}

As ratio is 1,  [conjugate base] = [acid] in solution.

  • At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

  • At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

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Determine the maximum number of electrons that can be found in each of the following subshells: a. 3s b. 3d c. 4p d. 4f.
monitta

Answer:

a. 2

b. 10

c. 6

d. 14

Explanation:

The maximum number of the electrons which can be filled in the s orbital are:- 2

The maximum number of the electrons which can be filled in the p orbital are:- 6

The maximum number of the electrons which can be filled in the d orbital are:- 10

The maximum number of the electrons which can be filled in the f orbital are:- 14

Thus,

a. 3s can have maximum of 2 electrons.

b. 3d can have maximum of 10 electrons.

c. 4p can have maximum of 6 electrons.

d. 4f can have maximum of 14 electrons.

3 0
3 years ago
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