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3 years ago

7

Treatment of phenol with excess aqueous bromine is actually more complicated than expected. A white precipitate forms rapidly, w

hich on closer examination is not 2,4,6- tribromophenol but is instead 2,4,4,6 tetrabromocyclohexadienone. Explain the formation of this product.

1 answer:

Anettt [7]3 years ago

8 0

Answer:

Please see attachment

Explanation:

Please see attachment

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Sensations of touch are picked up by nerves in the skin's _____.

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Epidermis is the answer to your question

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3 years ago

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The heating curve below shows the temperature change that occurs as a solid is heated. What is occurring at segment BC of the gr

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If Bc is the first horizontal line there, the solid is melting and going into the liquid state

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4. All of the following are examples of natural causes of air pollution

erma4kov [3.2K]

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Explanation:

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3 years ago

If 12.2 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc

PilotLPTM [1.2K]

<h2>Answer: kilograms of cryolite ≈ 50.23 kg</h2>

Explanation:

<u><em>Write the balanced equation</em></u>

Let's start off by writing the reaction for the formation of cryolite:

Al₂O₃ + 6NaOH + 12HF → 2Na₃AlF₆ + 9H₂O

<em><u>Determine the limiting factor</u></em>

moles = mass ÷ molar mass

moles of Al₂O₃ = 12200 g ÷ 101.96 g/mol = 119.65 mol

moles of NaOH = 57400 g ÷ 39.997 g/mol = 1435.11 mol

moles of HF = 57400 g ÷ 20.01 g/mol = 2868.57 mol

Since Al₂O₃ has the smallest number of moles in the reaction, it is the limiting factor. As such we will have to use the moles of Al₂O₃ to determine the moles of cryolite that will be produced.

<em><u>Find the mole ratio between the limiting factor and the cryolite</u></em>

The mole ratio of Al₂O₃ : Na₃AlF₆ based on the balanced equation is 1 : 2

∴ since the moles of Al₂O₃ = 119.65 mol

the moles of Na₃AlF₆ = 119.65 mol × 2 = 239.3 mol

<u><em>Calculate the mass of the Cryolite</em></u>

Mass cryolite = moles × molar mass

= 239.3 mol × 209.9 g/mol

= 50229.07 g

∴ kilograms of cryolite ≈ 50.23 kg

6 0

3 years ago

Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

baherus [9]

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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3 years ago

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