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eduard
1 year ago
11

When reacted with elemental chlorine at room temperature, liquid phosphorous trichloride (PCl3) is converted to solid phosphorou

s pentachloride (PCl5).
Chemistry
1 answer:
Sergeu [11.5K]1 year ago
6 0

PCl₅ reacts upon contact with water to launch hydrogen chloride and supply phosphorus oxides. the primary hydrolysis product is phosphorus oxychloride     PCl₅ + H₂O ========>  POCl₃ + 2HCl.

Phosphorus trichloride appears as a colorless or slightly yellow fuming liquid with a pungent and irritating odor resembling that of hydrochloric acid. Causes severe burns to skin, eyes and mucous membranes.

Divide the mass of the material through its molar mass. The molar mass of a substance is the mass in grams of 1 mole of that substance. This mass is given via the atomic weight of the chemical unit that makes up that substance in atomic mass units.

One mole is described as the amount of substance containing as many number one entities atoms, molecules, ions, electrons, radicals, and lots of others. As there are atoms in 12 grams of carbon - 12(6. 023×10²³. The mass of one mole of a substance equals to its relative molecular mass expressed in grams.

Learn more about  phosphorous trichloride here:-brainly.com/question/2626850

#SPJ4

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1 Oxygen atom has a mass of 16 grams, thus diatomic oxygen weights twice as much- 32 grams. Hence 32 grams is needed.

6 0
2 years ago
Determine the molar solubility of pbso4 in pure water. ksp (pbso4) = 1.82 x 10-8.
Misha Larkins [42]
Use the ICE table approach as solution:

           PbSO₄   --> Pb²⁺ + SO₄²⁻
I             -                 0          0
C           -                +s         +s
E           -                  s          s

Ksp = [Pb²⁺][SO₄²⁻]
1.82×10⁻⁸ = s²
Solving for s,
s = <em>1.35×10⁻⁴ M</em>
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Explanation:

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2 years ago
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7 0
2 years ago
The first excited vibrational energy level of diatomic chlo- rine (Cl2) is 558 cm^-1 above the ground state. Wave- numbers, the
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Answer:

The answer is "0.0000190 and 2.7 J".

Explanation:

\to P_4=\frac{e^{-\beta}(4+\frac{1}{2}) hev}{2vb}\\\\

         =\frac{e^{-9(1.35)}}{0.278v}\\\\=\frac{e^{-12.5}}{0.278}\\\\=\frac{0.000005285}{0.278}\\\\=0.0000190

Given:

h=6.626 \times 10^{-34}\\\\c=3\times 10^{10}\\\\v=558\ cm^{-1}\\\\K=1.38 \times 10^{-23}\\\\ T=298\\\\

E=\frac{hcv}{KT}\\\\

by putting the value into the above formula so, the value is 2.7 J  

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3 years ago
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