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1 year ago

how to identify amphiprotic compounds, if i give you the chemical structure or name of the compound/molecule.

Chemistry

1 answer:

Sonja [21]1 year ago

7 0

Amphiprotic compounds are able to both donate and accept a proton.

Amphiprotic compounds contain a hydrogen atom and lone pair of valence electron.

For example, HSO₃⁻ (hydrogen sulfate ion) is an amphiprotic compound.

Balanced chemical equation for reaction when HSO₃⁻ donate protons to water:

HSO₃⁻(aq) + H₂O(l) ⇄ SO₄²⁻(aq) + H₃O⁺(aq).

Ka = [SO₄²⁻] · [H₃O⁺] / [HSO₃⁻]

Balanced chemical equation for reaction when HSO₃⁻ accepts protons from water:

HSO₃⁻(aq) + H₂O(l) ⇄ H₂SO₄(aq) + OH⁻(aq).

Kb = [H₂SO₄] · [OH⁻] / [HSO₃⁻]

Water (H₂O), amino acids, hydrogen carbonate ions (HCO₃⁻) are examples of amphiprotic species.

Another example, water is an amphiprotic substance:

H₂O + HCl → H₃O⁺ + Cl⁻

H₂O + NH₃ → NH₄⁺ + OH⁻

More about amphiprotic compounds: brainly.com/question/3421406

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Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

$(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)$

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

$Q_1=m\times \Delta H_{fusion}$

where,

$Q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

$\Delta H_{fusion}$ = enthalpy change for fusion = $3.35\times 10^5J/Kg$

Now put all the given values in $Q_1$ , we get:

$Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J$

<u>For process 2 :</u>

$Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})$

where,

$Q_2$ = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

$c_{p,l}$ = specific heat of liquid water = $4186J/Kg^oC$

$T_1$ = initial temperature = $0^oC$

$T_2$ = final temperature = $100^oC$

Now put all the given values in $Q_2$ , we get:

$Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC$

$Q_2=4.186\times 10^5J$

From this we conclude that, $Q_1$ that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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3 years ago

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