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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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oksian1 [2.3K]
    The  temperature  at   which  the  process  be   spontaneous  is  calculated  as  follows

delta  G  =  delta H  -T delta S

let  delta G  be =0

therefore  delta H- T  delta s =0

therefore  T=  delta  H/  delta  S
convert  31   Kj  to  J  =  31  x1000=  31000 j/mol

T=31000j/mol /93 j/mol.k =333.33K


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3 years ago
As the concentration of a KOH solution increases, the number of moles of HCl needed to neutralize the KOH solution?
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Explanation:

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Read 2 more answers
If 14.5 g of MnO4- (permanganate) react with manganese (II) hydroxide how many grams of manganese (IV) oxide will be produced? T
Veronika [31]

Answer:

m_{MnO_2}=21.2gMnO_2

Explanation:

Hello,

In this case, given the balanced reaction:

2MnO_4^-+2Mn(OH)_2\rightarrow 4MnO_2+2OH^-+H_2O

We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:

m_{MnO_2}=14.5gMnO_4^-*\frac{1molMnO_4^-}{118.9gMnO_4^-}*\frac{4MnO_2}{2molMnO_4^-}  *\frac{86.9gMnO_2}{1molMnO_2} \\\\m_{MnO_2}=21.2gMnO_2

Best regards.

5 0
4 years ago
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