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3 years ago

Lithium chloride forms three hydrates. They are LiCl.H2O, LiCl.2H2O and LiCl.3H2O.

Chemistry

1 answer:

Stels [109]3 years ago

7 0

Answer:

The answer is LiCl.2H2O

Explanation:

Li=7

Cl=35.5

O=16

LiCl.H2O

7+35.5+16+2

60.5

%comp=60.5/78.5×100

22.9

LiCl.2H20

7+35.5+2(2+16)

42.5+36

78.5

%comp=36/78.5×100

45.9

LiCl.3H20

7+35.5+3(2+16)

42.5+54

96.5

54/96.5×100

56.0

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc

Anni [7]

Answer:

Ka = 4.76108

Explanation:

CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

[ ]initial change [ ]eq

CO(g) 0.27 M 0.27 - x 0.27 - x

H2(g) 0.49 M 0.49 - x 0.49 - x

CH3OH(g) 0 0 + x x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

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3 years ago

Consider the following equilibrium: H2CO3+H2O = H3O+HCO3^-1. What is the correct equilibrium expression?

mylen [45]

Equilibrium expression is $Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

Explanation:

Equilibrium expression is denoted by Keq.

Keq is the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

Example -

aA + bB = cC + dD

So, Keq = conc of product/ conc of reactant

$Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹

$Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}$

The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.

Thus,

$Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

Therefore, Equilibrium expression is $Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

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3 years ago