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Fittoniya [83]
3 years ago
10

Identify the reactant that is a Brønsted−Lowry acid in the following reaction: HI(aq)+H2O(l)→I−(aq)+H3O+(aq) Express your answer

as a chemical formula. nothing Request Answer Part B Identify the reactant that is a Brønsted−Lowry base in the following reaction: HI(aq)+H2O(l)→I−(aq)+H3O+(aq) Express your answer as a chemical formula. nothing Request Answer Part C Identify the reactant that is a Brønsted−Lowry acid in the following reaction: F−(aq)+H2O(l)⇌HF(aq)+OH−(aq) Express your answer as a chemical formula. nothing Request Answer Part D Identify the reactant that is a Brønsted−Lowry base in the following reaction: F−(aq)+H2O(l)⇌HF(aq)+OH−(aq) Express your answer as a chemical formula.
Chemistry
1 answer:
Fed [463]3 years ago
6 0

Answer:

(1) HI is Bronsted-Lowry acid and H_{2}O is Bronsted-Lowry base.

(2) H_{2}O is Bronsted-Lowry acid and F^{-} is Bronsted-Lowry base.

Explanation:

A Bronsted-Lowry acid is a species which donates proton (H^{+}).

A Bronsted-Lowry base is a species which accepts proton (H^{+}).

(1) HI(aq)+H_{2}O(l)\rightarrow I^{-}(aq)+H_{3}O^{+}(aq)

Reactants: HI and H_{2}O

Here HI donates proton and H_{2}O accepts proton.

Hence HI is Bronsted-Lowry acid and H_{2}O is Bronsted-Lowry base.

(2) F^{-}(aq)+H_{2}O(l)\rightleftharpoons HF(aq)+OH^{-}(aq)

Reactants: F^{-} and H_{2}O

Here H_{2}O donates proton and F^{-} accepts proton.

Hence H_{2}O is Bronsted-Lowry acid and F^{-} is Bronsted-Lowry base.

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5. Write a net ionic equation that occurs in a Na2HPO4/NaH2PO4 buffer solution when: A) a small amount of HCl is added (2 points
labwork [276]

Answer: (A) H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B) H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

Explanation:

(A) As we know that HCl is a strong acid and when it is added to an aqueous solution then it leads to increase in the concentration of hydrogen ions. And, when an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.

This means that addition of buffer into the given solution will not cause much change in the concentration of H_{3}O^{+} in large amount.

As both the buffer components are salt then they will remain dissociated as follows.

       Na_{2}HPO_{4}(aq) \rightarrow 2Na^{+}(aq) + HPO^{2-}_{4}(aq)

 NaH_{2}PO_{4}(aq) \rightarrow Na^{+}(aq) + H_{2}PO^{-}_{4}(aq)

Hence, net ionic equation will be as follows.

       H_{3}O^{+}(aq) + HPO^{2-}_{4}(aq) \rightarrow H_{2}PO^{2-}_{4}(aq) + H_{2}O(l)

(B)  When we add small amount of sodium hydroxide into the solution then there will occur an increase in concentration of hydroxide ions into the solution. But then due to the presence of buffer there will occur not much change in concentration and the acid will get converted into salt.

     NaOH(aq) \rightarrow Na^{+}(aq) + OH^{-}(aq)

The net ionic equation is as follows.

        H_{2}PO^{-}_{4}(aq) + OH^{-}(aq) \rightarrow HPO^{2-}_{4}(aq) + H_{2}O(l)

7 0
3 years ago
The normal boiling point of a substance occurs when its vapor pressure is 1.00 atm, which means that at this temperature, the li
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Answer:

See explanation below

Explanation:

For start, we need some values here to do this exercise.

In general, you can calculate the normal boiling point of any substance by using the Clausius Clapeyron equation which is the following:

ln(P2/P1) = -ΔHvap/R (1/T2 - 1/T1)

Where:

P1 and P2: pressure of the substance at T1 and T2.

ΔHvap: enthalpy of vaporization of the substance. In the case of bromine is 29.6 kJ/mol

R: constant gas. In this case is 8.3145 J/mol K

T1 and T2: temperature of the substance.

In order to calculate the normal boiling point, we will assign that value to T2, and the pressure would be 1 atm or 1.01x10^5 Pa

T1 and P1 would be temperature and pressure of this substance at any condition. For this example, I will take the fact that Bromine has 22000 Pa at 20 °C (or 293.15 K)

With this data, let's replace in the clausius Clapeyron equation:

ln(1.01x10^5 / 22000) = -29600/8.3145 (1/T2 - 1/293.15)

-1.5241 * 8.3145 / 29600 = (1/T2 - 1/293.15)

-4.281x10^-4 + 1/293.15 = 1/T2

T2 = 1 / 2.9831x10^-3

T2 = 335.22 or 62.07 °C

The real one is 59 °C so, the difference in the result may come with the values of P1 and T1 that may be not accurate.

7 0
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