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otez555 [7]
3 years ago
5

What is the answer for those questions plz ,help

Chemistry
1 answer:
marysya [2.9K]3 years ago
6 0

Answer:

1) 1.51 × 10²³  particles of Mg

2) 0.54 × 10⁻³  moles

Explanation:

Given data:

1)

Number of moles of Mg = 0.250 mol

Number of representative particles = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mol = 6.022 × 10²³  particles

0.250 mol × 6.022 × 10²³  particles / 1 mol

1.51 × 10²³  particles of Mg

2)

Given data:

Number of moles of lead = ?

Number of atoms of lead = 3.25×10²⁰ atoms

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mol = 6.022 × 10²³  atoms

1 mol × 3.25×10²⁰ atoms/ 6.022 × 10²³  atoms

0.54 × 10⁻³  moles

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- 622.5kJ

Explanation:

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2)      2NH3 + 3N2O ----> 4N2 + 3H2O        ΔH° 1= - 1010kJ

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2NH3  + 3N2H4  --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

3) -(2NH3 +1/2O2 ---> N2H4 + H2O,     ΔH°3 = -143 kJ)

   -2NH3 - 1/2O2 ---> - N2H4 - H2O,    - ΔH°3 = 143 kJ

 2NH3  + 3N2H4 --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

<u>  -2NH3 - 1/2O2              ---> - N2H4 - H2O,             - ΔH°3 = 143 kJ</u>

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H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ

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<u>9H2+ 9/2 O2    ---> 9H2O,                             9*ΔH° 4 = 9*(- 286) kJ</u>

4N2H4 +4O2 --->4N2+8H2O,         ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

5)

1/4*(4N2H4 +4O2 --->4N2+8H2O,    ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

N2H4 +O2 --->N2+2H2O,        1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)

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1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)=

=1/4(-1010kJ - 3*(-317kJ) - (-143kJ) + 9*(-286kJ))= - 622.5 kJ

   

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Answer :

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