What is the number of atoms of oxygen in 0.1 mol of CuSO4.5H2O? & explain

1 answer:

4 0

0ne mole of CuSO4.5H2O contains 6.023×10^23 no.of molecules

Again, 1 mole CuSO4.5H2O contains 9 oxygen atoms

Therefore,1 mole CuSO4.5H2O contains 9 × 6.023×10^23 no.of oxygen atoms

Hope this help!

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there is more air near sea level

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The equilibrium constant for the reaction of carbon monoxide with water is 1.845. if 1.00 mol of each reactant is placed in a 2.

SpyIntel [72]

[CO] = 1 mol / 2L = 0.5 M

[
According to the equation:

and by using the ICE table:

CO(g) + H2O(g) ↔ CO2(g) + H2(g)

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change -X -X +X +X

Equ (0.5-X) (0.5-X) X X

when Kc = X^2 * (0.5-X)^2

by substitution:

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3 years ago

Identify h2so4(aq) as an acid or a base. write a chemical equation showing how this is an acid according to the arrhenius defini

yarga [219]

H2SO4 is an acid

the chemical equation showing how H2SO4 is an acid according to the Arrhenius definition is as below

H2SO4 dissociate to give 2H^+ and SO4^2-

that is H2SO4 = 2H^+ + SO4^2-

According to Arrhenius an acid dissociate to give H^+ ions H2SO4 is an acid since it dissociate to give two hydrogen ions

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Read 2 more answers

When NH3(g) reacts with N2O(g) to form N2(g) and H2O(g), 105 kcal of energy are evolved for each mole of NH3(g) that reacts. Wri

Ganezh [65]

Answer : The balanced chemical equation is,

$2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal$

Explanation :

Balanced chemical equation : It is defined as the number of atoms of individual elements present on the reactant side must be equal to the number of atoms of individual elements present on product side.

The given unbalanced chemical reaction is,

$NH_3(g)+N_2O(g)\rightarrow N_2(g)+H_2O(g)+105kcal$

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of individual elements are not balanced.

In order to balanced the chemical reaction, the coefficient 2 is put before the $NH_3$ , the coefficient 3 is put before the $N_2O\text{ and }H_2O$ and the coefficient 4 is put before the $N_2$ .