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Tcecarenko [31]
3 years ago
7

How many moles are in 8.30x10^23 molecules of H2o

Chemistry
2 answers:
sasho [114]3 years ago
8 0

Answer: 1.38 moles

Explanation: According to Avogadro's law, 1 mole of every substance contains Avogadro number of particles (atoms, molecules or ions) i.e. 6.023\times 10^{23} particles.

Given : 1 mole molecule of water contains 6.023\times 10^{23} molecules of water.

6.023\times 10^{23} molecules are present in = 1 mole

8.30\times 10^{23}  molecules will be present in =\frac{1}{6.023\times 10^{23}}\times 8.30\times 10^{23}=1.38 moles

Thus 8.30\times 10^{23} molecules is equal to 1.38 moles.

LUCKY_DIMON [66]3 years ago
4 0
You multiply avogadro's number to what you were given.
8.30x10^23 * 6. 0221409x10^23
=1.357*10^25

That should be the right answer but I'm not sure. It has been awhile since I have done this.

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The order of entropy will be,

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As we are moving from gaseous state to liquid state to solid state, the entropy will be decreases due to the decrease in the disorderedness.

(a) Condensation of steam.

Condensation : It is a type of process in which the phase changes from gaseous state to liquid state at constant temperature.

In this process, phase changes from gaseous state to liquid state that means the degree of disorderedness decreases. So, the entropy will also decreases.

(b) Reaction of magnesium with oxygen to form magnesium oxide.

Mg(s)+O_2(g)\rightarrow MgO(s)

In this reaction, the randomness of reactant molecules are more (solid+gas) and as we move towards the formation of product the randomness become less (solid) that means the degree of disorderedness decreases. So, the entropy will also decreases.

(d) Reaction of nitrogen and hydrogen to form ammonia.

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In this reaction, the randomness of reactant molecules are more (2 moles of gas) and as we move towards the formation of product the randomness become less (1 mole of gas) that means the degree of disorderedness decreases. So, the entropy will also decreases.

(d) Sublimation of dry ice.

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6 0
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Explanation:

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<em>Hope this helps... </em>

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3 years ago
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