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PIT_PIT [208]
3 years ago
9

The ionic equation for MgCl2 and KOH.

Chemistry
2 answers:
saw5 [17]3 years ago
5 0
The reaction of Mg Cl2 and KOH can be described as a double substitution type of reaction. This means the cations of the reactants are exchanged in places when the products are formed. In this case, the balanced reaction is expressed
MgCl2 (s) + 2KOH (aq) = Mhg (OH)2 (aq) + 2KCl (s)
natta225 [31]3 years ago
4 0

solution:

the ionic equation of MgCl2 and KOH are as follows:

MgCl2+KOH

for the we have to multiply by 2 in KOH.

Now,

MgCl2+2KOH --> Mg(OH)2 + 2KCL

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In an endothermic reaction the ____ have more energy than the ____?
Ksju [112]

Answer: products; reactants

Explanation: as the endothermic reactions are tye one which absorbs energy

3 0
3 years ago
Li + HNO3 > LiNO3 + H2 how do I balance it? Show work pls
azamat

Answer: The balanced equation is 2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}.

Explanation:

The given reaction equation is as follows.

Li + HNO_{3} \rightarrow LiNO_{3} + H_{2}

Number of atoms present on reactant side are as follows.

  • Li = 1
  • H = 1
  • NO_{3} = 1

Number of atoms present on product side are as follows.

  • Li = 1
  • H = 2
  • NO_{3} = 1

To balance this equation, multiply Li by 2 and HNO_{3} by 2 on reactant side. Also, multiply LiNO_{3} by 2 on product side.

Hence, the equation can be rewritten as follows.

2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}

Now, number of atoms present on reactant side are as follows.

  • Li = 2
  • H = 2
  • NO_{3} = 2

Number of atoms present on product side are as follows.

  • Li = 2
  • H = 2
  • NO_{3} = 2

As there are same number of atoms on both reactant and product side. Hence, the equation is now balanced.

Thus, we can conclude that the balanced equation is 2Li + 2HNO_{3} \rightarrow 2LiNO_{3} + H_{2}.

3 0
2 years ago
Phosphorous acid, h3po3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. calculate the ph f
Varvara68 [4.7K]

Answer:

Explanation:

(a)

Before the addition of KOH :-

Given pKa1 of H3PO3 = 1.30

we know , pKa1 = - log10Ka1

Ka1 = 10-pKa1

Ka1 = 10-1.30

Ka1 = 0.0501

similarly pKa2 = 6.70 ,therefore Ka2 = 1.99 x 10-7

because Ka1 >> Ka2 , therefore pH of diprotic acid i.e H3PO3 can be calculated from first dissociation only .

ICE table is :-

H3PO3 (aq) <-------------> H+ (aq) + H2PO3-(aq)

I 2.4 M 0 M 0 M

C - x + x + x

E (2.4 - x )M x M x M

x = degree of dissociation

Now expression of Ka1 is :

Ka1 = [ H+ ] [ H2PO3-] / [ H3PO3]

0.0501 = x2 / 2.4 - x

on solving for x by using quadratic formula , we have

x = 0.32

Now [ H+ ] = [ H2PO3-] = 0.32 M

pH = - log [H+]

pH = - log 0.32

pH = - ( - 0.495)

pH = 0.495

Hence pH before the addition of KOH = 0.495

(b)

After the addition of 25.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.025 L = 0.06 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

Now 0.06 moles of KOH is equal to the half of the moles required for the first equivalent point . therefore pH at this point is equal to pKa1 .

Hence pH = 1.30 M

(c)

After the addition of 50.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.050 L = 0.12 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

because Number of moles of H3PO4 = Number of moles of KOH

therefore , this point is the first equivalence point

and pH = pKa1 + pKa2 / 2

pH = 1.30 + 6.70 / 2

pH = 4.00

Hence pH = 4.00

(d)

After the addition of 75.0 mL of 2.4 M KOH :-

Number of moles of KOH = 2.4 M x 0.075 L = 0.18 mol

Number of moles of H3PO3 = 2.4 M x 0.050 L = 0.12 mol

This is the half way of the second equivalence point , therefore pH is equal to pKa2 .

Hence pH = 6.70

5 0
3 years ago
Sublimation and boiling both happen at the surface of the substance. <br> a. True <br> b. False
Burka [1]
This statement is: False

Boiling is vaporization that occurs both below and at the surface of the liquid.
However, evaporation happens at the surface of the substance.

So, if you need to correct this statement, the correction would be:
Sublimation and evaporation happen at the surface of the substance.
7 0
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Describe the formation of covalent bond in methane (5 marks) ​
Lady bird [3.3K]

Answer:

Explanation:

The structure of the methane, CH4, molecule exhibits single covalent bonds. Covalent bonding involves the sharing of electrons. In the methane molecule, the four hydrogen atom share one electron each with the carbon atom

8 0
3 years ago
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