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Wewaii [24]
2 years ago
10

Ethanoic acid, CH3COOH, ionizes to form an ethanoate ion (CH3COO-) in aqueous solution. what else does it form? is it a monoprot

ic acid?
Chemistry
1 answer:
barxatty [35]2 years ago
8 0

Ethanoic acid ionizes in aqueous solutions to form two ions which are CH_3COO^- and H^+

<h3>Ionization of ethanoic acid</h3>

Ethanoic acid goes by the chemical formula CH_3COOH.

In aqueous solutions, it ionizes as a monoprotic acid according to the following equation:

CH_3COOH ---- > CH_3COO^- + H^+

A monoprotic acid is an acid that is able to donate only a proton. Hence, ethanoic acid is said to be monoprotic because it ionizes in aqueous solutions to produce a single H^+

More on ethanoic acid can be found here: brainly.com/question/9991017

#SPJ1

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8 0
2 years ago
zn (s) 2 hcl (aq) -----&gt; zncl2 (aq) h2(g) if 520 ml of h2 is collected over water at 28oc and the atmospheric pressure is 1.0
nadya68 [22]

The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.

Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.

We recognise that

h * 2 = PT - P * h * 20 = 1atm - 0.037atm

= 0.963 atm

1 * h * 2 = Ph * 2V / R * T

= 0.963 atm x 0.520 L / 0.0821 L atm/

molK * 301

= 0.02 mol h2

= 0.02molZn

So 0.02 mol Zn x 65.39 g/mol

= 1.33 g Zn

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brainly.com/question/28880469

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3 0
1 year ago
Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el
Likurg_2 [28]

Answer: 281 hours

Explanation:-

1 electron carry charge=1.6\times 10^{-19}C

1 mole of electrons contain=6.023\times 10^{23} electrons

Thus  1 mole of electrons carry charge=\frac{1.6\times 10^{-19}}{1}\times 6.023\times 10^{23}=96500C

Cu^{2+}+2e^-\rightarrow Cu

96500\times 2=193000Coloumb of electricity deposits 1 mole or 63.5 g of copper

0.0635 kg of copper is deposited by 193000 Coloumb

11.5 kg of copper is deposited by=\frac{193000}{0.0635}\times 11.5=34952756 Coloumb

Q=I\times t

where Q= quantity of electricity in coloumbs  = 34952756 C

I = current in amperes = 34.5 A

t= time in seconds = ?

34952756 C=34.5A\times t

t=1013123sec=281hours

Thus it will take 281 hours to plate 11.5 kg of copper onto the cathode if the current passed through the cell is held constant at 34.5 A.

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Calcium Flouride. It's an ionic bond. Cation + anion with the suffix -ide
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