Answer:
<em>The correct option is A) Arrhenius</em>
Explanation:
According to the Arrhenius concept of acids and bases, an acid must produce H+ ions when it is present in a solution and the base must produce OH- ions when placed in a solution.
Ammonia does not contain OH- ions of its own when dissolved in water.
The reaction of ammonia dissolving is water can be written as:
NH3 + H2O ⇌ NH4+ + OH−
As we can see from the equation, ammonia does form OH- ions but it does not have OH- ions on its own.
Hence, according to the Arrhenius concept, NH3 is not a base.
The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.
<h3>How do we calculate pH of weak base?</h3>
pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:
pH = pKb + log([HB⁺]/[B])
pKb = -log(1.8×10⁻⁶) = 5.7
Chemical reaction for C₆H₅NH₂ is:
C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻
Initial: 0.25 0 0
Change: -x x x
Equilibrium: 0.25-x x x
Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]
Kb = x² / 0.25 - x
x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:
1.8×10⁻⁶ = x² / 0.25
x² = (1.8×10⁻⁶)(0.25)
x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]
On putting all these values on the above equation of pH, we get
pH = 5.7 + log(0.67×10⁻³/0.25)
pH = 3.13
Hence pH of the solution is 3.13.
To know more about Henderson Hasselbalch equation, visit the below link:
brainly.com/question/13651361
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A
nswer: -
C. Energy is released by the reaction
Explanation:-
An exothermic reaction is one in which during the progress of the reaction heat is evolved.
So energy is released by the reaction.
It cannot be created as energy is neither created nor destroyed as per the Law of conservation of energy. Energy is not transferred either.
The energy released during the progress of the reaction originates from the chemical bonds of the reactants as they break during their conversion into products.
Answer:
207.03°C
Explanation:
The following data were obtained from the question:
V1 (initial volume) = 6.80 L
T1 (initial temperature) = 52.0°C = 52 + 273 = 325K
P1 (initial pressure) = 1.05 atm
V2 (final volume) = 7.87 L
P2 (final pressure) = 1.34 atm
T2(final temperature) =?
Using the general gas equation P1V1/T1 = P2V2/T2, the final temperature of the gas sample can be obtained as follow:
P1V1/T1 = P2V2/T2
1.05 x 6.8/325 = 1.34 x 7.87/T2
Cross multiply to express in linear form as shown below:
1.05 x 6.8 x T2 = 325 x 1.34 x 7.87
Divide both side by 1.05 x 6.8
T2 = (325 x 1.34 x 7.87) /(1.05 x 6.8)
T2 = 480.03K
Now, let us convert 480.03K to a number in celsius scale. This is illustrated below:
°C = K - 273
°C = 480.03 - 273
°C = 207.03°C
Therefore, the final temperature of the gas will be 207.03°C
