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VLD [36.1K]
1 year ago
9

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

#SPJ4

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Answer:

D. Visible Light

Explanation:

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Answer:

Size of the nucleus of an atom is very small as compared to the size of the atom.

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3 years ago
Calculate the hight of a column of liquid glycerol in meters required to exert the same pressure as 3.02 m if CCl4?
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Answer:

Approximately 3.81\; \rm m.

Explanation:

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Let g denote the gravitational field strength. (Typically g \approx 9.81\; \rm N \cdot kg^{-1} near the surface of the earth.) For a column of liquid with a height of h, if the density of the liquid is \rho, the pressure at the bottom of the column would be:

P = \rho\cdot g \cdot h.

The pressure at the bottom of this carbon tetrachloride column would be:

\begin{aligned} P &= \rho \cdot g \cdot h \\ & \approx 1.59\times 10^{3} \; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1} \times 3.02 \; \rm m \approx 4.71 \times 10^{4} \; \rm N \cdot m^{-2} \end{aligned}.

Rearrange the equation P = \rho\cdot g \cdot h for h:

\displaystyle h = \frac{P}{\rho \cdot g}.

Apply this equation to calculate the height of the liquid glycerol column:

\begin{aligned}h &= \frac{P}{\rho \cdot g} \\ &\approx \frac{4.71 \times 10^{4}\; \rm N \cdot m^{-2}}{1.26 \times 10^{3}\; \rm kg \cdot m^{-3} \times 9.81\; \rm N \cdot kg^{-1}} \approx 3.81\; \rm m\end{aligned}.

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2 years ago
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