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ankoles [38]
3 years ago
5

One mole of particles of any substance contains how many particles?

Chemistry
1 answer:
melomori [17]3 years ago
4 0

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

According to mole concept:

1 mole of a substance contains 6.02\times 10^{23} number of particles.

This number is known as Avogadro's number.

The mass of 1 mole of a substance or Avogadro number of particles has the mass equal to molar mass of that substance.

Hence, the correct answer is Option d.

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with
grandymaker [24]

\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

8 0
3 years ago
1. the mass of one electron is
kupik [55]

Answer:

Mass of one electron is 9.1 × 10⁻³¹ kg

Mass of one proton is 1.673 × 10⁻²⁷ Kg

Mass of one neutron is 1.675 × 10⁻²⁷ Kg

 

<u>-TheUnknownScientist</u><u> 72</u>

4 0
2 years ago
Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your
k0ka [10]
We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate          ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate          ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate 

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol  + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62


6 0
3 years ago
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When swimmers get out of the pool and sit in the sun, their wet skin dries quickly. What happens to the water?
I am Lyosha [343]

Answer:

It evaporates and moves into the air.

Explanation:

When water is left out for a while, it evaporates into the air! :)

7 0
2 years ago
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vlabodo [156]

Answer:

1. Delete lion, bison, reindeer, and giraffe

2. Delete cactus, oak, evergreen,

Explanation:

1. Lions live in the African savanna, bison live in open fields with lots of grass, reindeer live in Antarctica and other very cold places, and giraffes also live in the areas of the African savanna where there are trees.

2. Cacti live in the desert, oaks and evergreens live in open grassy areas.

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5 0
2 years ago
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