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2 years ago

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

ic acid?

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

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If a wave has a wavelength of 13 meters and a period of 0.005, what's the velocity of the wave? A. 2,600 m/s B. 260 m/s C. 1,300

Rina8888 [55]

Answer:

V = 2600ms⁻¹

Explanation:

Given parameters:

Wavelength(λ) = 13meters

Period (T) = 0.005s

Period(T) is the time it takes for a full cycle of vibration to pass through. It's unit is in seconds (s)

The Velocity of waves is expressed as:

V = fλ

Where f = frequency(s⁻¹)

Frequency of a wave is the number of waves that passes through a point per unit time

f = 1/T

Where T is the period

We can therefore express Velocity of waves as a function of period

V = λ/T

Inputing the parameters, we have:

V = 13m / 0.005s

V = 2600ms⁻¹

7 0

3 years ago

Read 2 more answers

Why do you think large differences in temperature between the wet bulb and dry bulb thermometers are possible only at higher tem

olganol [36]

Answer:

This is the temperature indicated by a moistened thermometer bulb exposed to the air flow. The evaporation is reduced when the air contains more water vapor. The wet bulb temperature is always lower than the dry bulb temperature but will be identical with 100% relative humidity.

Explanation:

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3 years ago

If stomach acid has a ph of 1.3, what is the pOH

Mekhanik [1.2K]

I think the answer is -12.7

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3 years ago

Orbitals which valence electrons are found for calcium

galina1969 [7]

The 4th energy level (row), 's' orbital block, 2nd group (column). The valence electrons are found in the highest energy level of the electron configuration in the 's' and 'p' orbitals. In the case of calcium this is 4s2 . This gives calcium an 's' orbital with a pair of electrons in its valence shell

8 0

3 years ago

Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g)

AURORKA [14]

Explanation:

The net equation will be as follows.

$K(s) + Cl_{2}(g) \rightarrow KCl(s)$

So, we are required to find $\Delta H_{formation}$ for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1: Convert K from solid state to gaseous state

$K(s) \rightarrow K(g)$ , $\Delta H_{1}$ = 89 kJ

Step 2: Ionization of gaseous K

$K(g) \rightarrow K^{+}(g) + e^{-}$ , $H_{2}$ = 418 KJ

Step 3: Dissociation of $Cl_{2}$ gas into chlorine atom .

$\frac{1}{2} Cl_{2}(g) \rightarrow Cl(g)$ , $\Delta H_{3} = \frac{244}{2}$ = 122 KJ

Step 4: Iozination of chlorine atom.

$Cl(g) + e^{-} \rightarro Cl^{-}(g)$ , $H_{4}$ = -349 KJ

Step 5: Add $K^{+}$ ion and $Cl^{-}$ ion formed above to get KCl .

$K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s)$ , $H_{5}$ = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

$\Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}$

= 89 + 418 + 122 - 349 - 717

= - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

5 0

3 years ago