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Veronika [31]
3 years ago
9

Define the following: Bronsted-Lowry acid - Lewis acid- Strong acid - (5 points) Problem 6: Consider the following acid base rea

ction HCI + H20 → H30+ + Cl- a) Is this a strong acid? b) Clearly label the acid, base, conjugate acid and conjugate base. (5 points)
Chemistry
1 answer:
Allushta [10]3 years ago
8 0

Answer: Yes, HCl is a strong acid.

acid = HCl , conjugate base = Cl^- , base = H_2O, conjugate acid = H_3O^+

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

Yes HCl is a strong acid as it completely dissociates in water to give H^+ ions.

HCl\rightarrow H^++Cl^-

For the given chemical equation:

HCl+H_2O\rightarrow H_3O^-+Cl^-

Here, HCl is loosing a proton, thus it is considered as an acid and after losing a proton, it forms Cl^- which is a conjugate base.

And, H_2O is gaining a proton, thus it is considered as a base and after gaining a proton, it forms H_3O^+ which is a conjugate acid.

Thus acid =  HCl

conjugate base = Cl^-

base = H_2O

conjugate acid = H_3O^+.

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damaskus [11]
<h3>Balanced equation : 2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)</h3><h3>Further explanation</h3>

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General formula for alkanes :

\tt \large{\bold{C_nH_{2n+2}}

Hydrocarbon combustion reactions (specifically alkanes)

\large {\box {\bold{C_nH _ (_2_n _ + _ 2_) + \dfrac {3n + 1} {2} O_2 \Rightarrow nCO_2 + (n + 1) H_2O}}}

So that the burning of ethane with air (oxygen):

\tt C_2H_6+\dfrac{7}{2}O_2\rightarrow 2CO_2+3H_2O

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or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction

C₂H₆ (g) + aO₂ (g) ⟶ bCO₂ (g) + cH₂O (ℓ)

C : left 2, right b ⇒ b=2

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6 0
3 years ago
How many moles of CCI are there in 78.2 g of CCI.?
vaieri [72.5K]

Answer:

0.508 mole

Explanation:

NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.

The number of mole present in 78.2 g of CCl₄ can be obtained as follow:

Mass of CCl₄ = 78.2 g

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Mole = mass / molar mass

Mole of CCl₄ = 78.2 / 154

Mole of CCl₄ = 0.508 mole

Therefore, 0.508 mole is present in 78.2 g of CCl₄

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s2 = 2.4 × 10e4
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