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natima [27]
3 years ago
7

How many orbitals are in Phosphorus

Chemistry
2 answers:
Viefleur [7K]3 years ago
8 0

Answer:

3

three half-filled orbitals each capable of forming a single covalent Bond and an additional lone - pair of electrons

Nastasia [14]3 years ago
3 0
5 orbits are in phosphorus
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fomenos
The answer is b, because all of the other compounds are covelent
6 0
3 years ago
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Complete the reaction by writing the formulas of the products. CH 3 COOH + NH 3 − ⇀ ↽ − CH_{3}COONH_{4}^{+} CH 3 COONH + 4 The K
cricket20 [7]

Answer:

Products are favored.

Explanation:

The acid-base reaction of CH₃COOH (acid) with NH₃ (base) produce:

CH₃COOH + NH₃ ⇄ CH₃COO⁻ +  NH₄⁺ Kr = ?

It is possible to know Kr of the reaction by the sum of acidic dissociations of the half-reactions. That is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ Ka = 1.8x10⁻⁵

NH₃ + H⁺ ⇄ NH₄⁺ 1/Ka = 1/ 5.6x10⁻¹⁰ = 1.8x10⁹

___________________________________

CH₃COOH + NH₃ ⇄ CH₃COO⁻ +  NH₄⁺ Kr =  1.8x10⁻⁵×1.8x10⁹ = <em>3.2x10⁴</em>

<em> </em>

As Kr is defined as:

Kr =  [CH₃COO⁻] [NH₄⁺] / [CH₃COOH] [NH₃]

And Kr is > 1  

[CH₃COO⁻] [NH₄⁺] > [CH₃COOH] [NH₃],

showing <em>products are favored</em>.

4 0
3 years ago
If you have 85.0 grams of CO₂, calculate how many moles of CO₂ would you have?
dedylja [7]
N=m(g)/m.wt
n=85/12(1)+16(2) =1.93 moles
4 0
2 years ago
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Everything in group 2 on the periodic table are called
Ludmilka [50]

Answer:

group two elements are called alkaline earth metals

Explanation:

Because their oxides form in the earth and are water soluble

3 0
3 years ago
How many joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point?
Korolek [52]

Answer: 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

Explanation:

Latent heat of vaporization is the amount of heat required to convert 1 mole of liquid to gas at atmospheric pressure.

Amount of heat required to vaporize 1 mole of lead =  177.7 kJ

Molar mass of lead = 207.2 g

Mass of lead given = 1.31 kg = 1310 g       (1kg=1000g)

Heat required to vaporize 207.2 of lead = 177.7 kJ

Thus Heat required to vaporize 1310 g of lead =\frac{177.7}{207.2}\times 1310=1123kJ=1123000J

Thus 1123000 Joules of energy are required to vaporize 13.1 kg of lead at its normal boiling point

7 0
3 years ago
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