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melomori [17]
2 years ago
7

In an aqueous solution, 42% of a substance dissociates to release hydronium ions. Which of the following statements is true for

the substance?
It is a weak acid.

It is a weak base.

It is a strong acid.

It is a strong base.
Chemistry
1 answer:
const2013 [10]2 years ago
5 0

Answer: The statement it is a weak acid is true for the substance.

Explanation:

An acid that dissociates completely when dissolved in water to give hydrogen (H^{+}) or hydronium (H_{3}O^{+}) ions is called a strong acid.

For example, HCl is a strong acid.

HCl + H_{2}O \rightarrow H_{3}O^{+} + Cl^{-}

An acid that dissociates partially or weakly when dissolved in water to given hydrogen or hydronium ions is called a weak acid.

For example, CH_{3}COOH is a weak acid.

CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H_{3}O^{+}

A strong base is a base which when dissolved in water then it dissociates completely to give hydroxide ions.

For example, NaOH is a strong acid.

A weak base is a base which when dissolved in water then it dissociates partially or weakly to give hydroxide ions.

For example, NH_{3} is a weak base.

Hence, in an aqueous solution where 42% of a substance dissociates to release hydronium ions shows that the dissociation is less than 50%. This means that substance is dissociating weakly so, it is a weak acid.

Thus, we can conclude that the statement it is a weak acid is true for the substance.

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Answer:

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2.)reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water.

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6.)4Al + 3O 2 → 2Al 2O 3

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Explanation:

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2 years ago
A single-effect evaporator is concentrating a feed of 9072 kg/h of a 10 wt % solution of NaOH in water at temperature of 288.8 K
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Answer:

a) steam used = 8440 kg/hr

b)  Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

Explanation:

Saturated steam pressure = 42KPag=1.42 bar

From steam table, Steam temperature = 110 C

Latent heat of this steam = 2230 KJ/kg

Process side pressure = 20 Kpaa = .2 bar

Water latent heat at this Pressue = 2360 KJ/kg

Water boiling Point at this Pressure =60 C

Feed Inlet temperature = 15.6 C

Total Heat required = Heat required to rise feed temperature from 15.6 to 60 degree C + Evaporation of water to concentrate the feed

Total Water evaporation required = 9072*(100-10)/100-9072*.1/.5*.5=7257.6 Kg/hr

Specific heat of feed assumed = 4.2 KJ/kg/K

=> Total heat required = 9072*4.2*(60-15.6)+7257.6*2360=18819682 KJ/hr

LMTD = ((110-60)-(110-15.6))/LN((110-60)/(110-15.6))=69.8 C

U, Overall Heat transfer coefficient = 1988 W/m2/K

Total heat required = U*A*LMTD

=> Area required of evaporator, A=18819682 *1000/3600/1988/59.8=44 m2

Steam used = 18819682/2230=8440 kg/hr

Energy required to condense vaporised feed = 7257.6*2360=17127936 KJ/hr

=> Steam efficiency = (18819682-17127936)/18819682=9%

b) Because as per calculation done in a), Steam is sufficient to evaporate 7257.6 kg/hr water from feed. Since feed is less than this amount, therefore whole water will be evaporated and 100% solid product will be obtained in bottom collection

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3 years ago
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Answer:

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