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Nostrana [21]
3 years ago
6

Which of the following shows a Bronsted-Lowry acid reacting?

Chemistry
2 answers:
7nadin3 [17]3 years ago
8 0
<h3>Answer:</h3>

         Option-C:  HCl + H₂O  →   H₃O⁺ + Cl⁻

Explanation:

       Bronsted-Lowery concept of Acid and Base defines Acid as that specie which tends to donate H⁺ (Hydrogen Ion) and bases are those species which accepts H⁺ from Acids.

In selected option, HCl is reacting as Acid as it donates H⁺ to water (lowery bronsted base).

Also, the correspong acid is converted into conjugate base (i.e. Cl⁻) and base is converted into conjugate acid (i.e. H₃O⁺)

Akimi4 [234]3 years ago
7 0

Answer:

HCl + H2O Right arrow. H3O+ + Cl–

Explanation:D is the answer

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In a titration, 0.01M KOH was used to neutralize 18 milliliters of 2 M HBr. What was the volume of the base used?
NikAS [45]

Answer:

3.60 ml

Explanation:

First of all we must put down the equation of the reaction. This will serve as a guide to our solution;

KOH(aq) + HBr(aq) -----> KBr(aq) + H2O(l)

The following were given in the question;

Concentration of acid CA= 2M

Volume of acid VA= 18ml

Concentration of base CB= 0.01 M

Volume of base VB= ????

Number of moles of acid NA= 1

Number of moles of base NB= 1

From;

CAVA/CBVB = NA/NB

CAVANB= CBVBNA

Therefore;

VB= CAVANB/CBNA

Substituting values;

VB= 2 × 18 ×1 / 0.01×1

VB= 3.60 ml

Therefore; 3.60 ml of base was used.

4 0
3 years ago
7. How much energy in KJ is released when 15.0 g of steam at 100.0 degrees Celsius is condensed and then frozen to ice at 0.0 de
Aliun [14]

Answer:

-6.27 kj

Explanation:

Given data:

Energy released = ?

Mass of steam = 15 g

Initial temperature = 100°C

Final temperature = 0°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT =  0°C - 100°C = -100°C

Q = 15g × 4.18 j/g.°C  × -100°C

Q = -6270 j

J to KJ:

-6270 j/1000 = -6.27 kj

4 0
4 years ago
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