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3 years ago

7

Which olefin should predominate in the product of the dehydration of 2-methyl-2-butanol?

1 answer:

3241004551 [841]3 years ago

5 0

<span>Answer: 2-methyl-2-buthene.. Bcoz it releases H2O (OH- & H+), the OH- need to take one of H+ in the Carbon chain... C-C-C(OH)-C The question is which H will be 'sucked' by OH+?? OH is the border beetwen C-C and C, right?? More C it has, more positive it will be.. H+ is positive, so C-C that more (+) than C, wont bond H better than C (bcoz C is more negative than C-C, so H+ will like C more than C-C).. So, the answer must be C-C=C(CH3)-C and H2O...</span>

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Answer:

The entropy of the final solution decreases, as the reaction disorder is less.

Explanation:

The higher the temperature, the greater the heat of the reaction and the greater the disorder it has, so the entropy will increase ... But this is not the case, since the solution cools, decreasing the entropy proportionally.

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3 years ago

Particle Q has 16protons, 17neutrons and 18 electrons. Particle Q is:

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2 years ago

What is the ratio of carbon, hydrogen, and oxygen in most carbohydrates?

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2 years ago

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2 years ago

Aluminum sulfate, known as cake alum, has a wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous so

NISA [10]

Answer:

a) The chemical reaction is given as:

$6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)$

2.571 grams of aluminum hydroxide is precipitated.

Explanation:

a) The chemical reaction is given as:

$6NaOH(aq)+Al_2(SO_4)_3\rightarrow 2Al(OH)_3(s)+3Na_2SO_4(aq)$

b)

Moles of NaOH = n

Volume of the NaOH = 185.5 mL = 0.1855 L( 1 mL =0.001 L)

Molarity of the solution = 0.533 M

$n=0.533 M\times 0.1855 mL=0.09887 mol$

Moles of aluminum = n

15.8 g of aluminum sulfate per liter.

Volume of solution = 627 mL = 0.627 L (1 mL= 0.001 L)

Mass of aluminium sulfate in solution = 15.8 g/L × 0.627 L =9.9066 g

Moles of aluminum sulfate = $\frac{9.9066 g}{342 g/mol}=0.02897 mol$

Moles of NaOH = 0.09887 mol

According to reaction, 6 moles of NaOH reacts with 1 mole of aluminum sulfate, then 0.09887 moles of NaOH will recat with :

$\frac{1}{6}\times 0.09887 mol=0.01648 mol$

This means that sodium hydroxide moles are in limiting amount.So, amount of aluminum hydroxide will depend upon moles of sodium hydroxide.

According to reaction, 6 moles of sodium hydroxide gives 2 moles of aluminium hydroxide, then 0.09887 moles of sodium hydroxide will give :

$\frac{2}{6}\times 0.09887 mol=0.03296 mol$ of aluminum hydroxide

Mass of 0.03296 moles of aluminum hydroxide:

0.03296 mol × 78 g/mol = 2.571 g

2.571 grams of aluminum hydroxide is precipitated.

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2 years ago