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Leokris [45]
3 years ago
10

This is an easy since question

Chemistry
2 answers:
Fiesta28 [93]3 years ago
6 0

Answer:

A

Explanation:

Elden [556K]3 years ago
3 0
A are volcanoes active in the area
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What is the mole fraction of KCI in a
Margaret [11]

Answer:

Mole fraction of KCl = 0.4056

Explanation:

We'll begin by calculating the number of mole of each compound. This can be obtained as follow:

For NaCl:

Mass NaCl = 0.564 g

Molar mass of NaCl = 58.44 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 0.564 / 58.44

Mole of NaCl = 0.0097 mole

For KCl:

Mass KCl = 1.52 g

Molar mass of KCl = 74.55 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1.52 / 74.55

Mole of KCl = 0.0204 mole

For LiCl:

Mass LiCl = 0.857 g

Molar mass of LiCl = 42.39 g/mol

Mole of LiCl =?

Mole = mass /Molar mass

Mole of LiCl = 0.857 / 42.39

Mole of LiCl = 0.0202 mole

Next, we shall determine the total mole in the mixture. This can be obtained as follow:

Mole of NaCl = 0.0097 mole

Mole of KCl = 0.0204 mole

Mole of LiCl = 0.0202 mole

Total mole =?

Total mole = Mole of NaCl + Mole of KCl + Mole of LiCl

Total mole = 0.0097 + 0.0204 + 0.0202

Total mole = 0.0503 mole

Finally, we shall determine the mole fraction of KCl in the mixture. This can be obtained as follow:

Mole of KCl = 0.0204 mole

Total mole = 0.0503 mole

Mole fraction of KCl =?

Mole fraction of KCl = Mole of KCl /Total mole

Mole fraction of KCl = 0.0204 / 0.0503

Mole fraction of KCl = 0.4056

4 0
3 years ago
Read 2 more answers
For the equilibrium system described by this
ziro4ka [17]
Th equilibrium shifts to the left
6 0
3 years ago
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Which electron in sulfur is most shielded from nuclear charge?
natima [27]

Answer: 3p Orbitals

Explanation:

Electrons present in the 3p orbitals are farthest from the nucleus. Therefore, the electrons present in the 3p orbital will be shielded by the electrons present in the inner orbitals. Hence, 3p orbital in sulfur is most shielded from the nuclear charge".

5 0
3 years ago
What general rule can be followed when choosing a type of solvent to dissolve a particular solid?
Fittoniya [83]
 <span>the polarity of the solute or the solvent. 

for example: 
oil will not mix with water because oil molecules are nonpolar however water moleculses are polar. so, they will not mix with each other. 

when we put sodium chloride in water, sodium chloride will be easily dissoved. because both sodium chloride and water are polar. 

in other case, if we put sodium chloride and hexane together. sodium chloride will not dissove in hexane, because hexane is a nonpolar solvent. 

finally, if we try to mix hexane and bromine together, they will mix uniformly. because both hexane and bromine are nonpolar. (note: most diatomic molecules are nonpolar, such as hydrogen gas, oxygen gas, chlorine gas, etc. ) 

so just remember, nonpolar and nonpolar will dissovle each other. and polar and polar will dissolve each other.</span>
4 0
4 years ago
Calculate the concentration (M) of sodium ions in a solution made by diluting 10.0 mL of a 0.774 M solution of sodium sulfide to
Luda [366]

The concentration of sodium ions in a solution made by diluting 10.0 mL of a 0.774 M solution of sodium sulfide to a volume of 100 mL would be 0.1548 M.

<h3>Dilution</h3>

According to the dilution principle, the number of moles of solutes in a solution before and after dilution must be the same. This is expressed as the following equation:

m_1v_1 = m_2v_2

Where m_1 is the molarity before dilution, v_1 is the volume before dilution, m_2 is the molarity after dilution, and v_2 is the volume after dilution.

In this case, m_1 = 0.774 M, v_1 = 10.0 mL, v_2 = 100 mL.

m_2 = m_1v_1/v_2

     = 0.774 x 10/100

      = 0.0774 M

Thus, the new concentration of the sodium sulfide solution would be 0.0774 M.

Mole of the final solution: 0.0774 x 0.1 = 0.00774 mol

Sodium sulfide formula = Na_2S ---> 2Na^+ + S^{2-

Equivalent mole of sodium ion = 0.00774 x 2

                                                    = 0.01548 mol

The concentration of sodium ions = mol/volume

                                                  = 0.01548/0.1

                                                  = 0.1548 M

In other words, the concentration of the sodium ions in the diluted solution would be 0.1548 M.

More on dilution can be found here: brainly.com/question/21323871

#SPJ1

6 0
1 year ago
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