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lana [24]
2 years ago
5

Convert from moles to particles 6.02x1023 particles 1 mole Question 2. 2.7 moles of lithium Question 3. 1.8 moles of sodium chlo

ride​
Chemistry
1 answer:
spayn [35]2 years ago
4 0

Considering the definition of Avogadro's number:

  • the number of molecules of lithium is 1.62621×10²⁴ molecules.
  • the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

<h3>Definition of Avogadro's number</h3>

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Amount of molecules in this case</h3>

You can apply the following rule of three, considering the Avogadro's number:  If 1 mole of lithium contains 6.023×10²³ molecules, 2.7 moles of lithium contains how many molecules?

amount of molecules of lithium= (6.023×10²³ molecules × 2.7 moles)÷1 mole

<u><em>amount of molecules of lithium=1.62621×10²⁴ molecules</em></u>

Finally, the number of molecules of lithium is 1.62621×10²⁴ molecules.

On the other hand you can apply the following rule of three, considering the Avogadro's number:  If 1 mole of sodium chloride​ contains 6.023×10²³ molecules, 1.8 moles of  sodium chloride​ contains how many molecules?

amount of molecules of sodium chloride= (6.023×10²³ molecules × 1.8 moles)÷1 mole

<u><em>amount of molecules of sodium chloride=1.08414 ×10²⁴ molecules</em></u>

Finally, the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

Learn more about Avogadro's Number:

brainly.com/question/11907018

brainly.com/question/1445383

brainly.com/question/1528951

#SPJ1

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The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add
JulijaS [17]

Answer:

a)The Ksp was found to be equal to 13.69

Explanation:

Terminology

Qsp of a dissolving ionic solid — is the solubility product of the concentration of ions in solution.

Ksp however, is the solubility product of the concentration of ions in solution at EQUILIBRIUM with the dissolving ionic solid.

Note that if Qsp > Ksp , the solid at a certain temperature, will precipitate and form solid. That means the equilibrium will shift to the left in order to attain or reach equilibrium (Ksp).

Step-by-step solution:

To solve this: 

#./ Substitute the molar solubility of KCl as given into the ion-product equation to find the Ksp of KCl.

#./ Find the total concentration of ionic chloride in each beaker after the addition of HCl. We pay attention to the amount moles present at the beginning and the moles added.

#./ Find the Qsp value to to know if Ksp is exceeded. If Qsp < Ksp, nothing will precipitate.

a) The equation of solubility equilibrium for KCL is thus;

KCL_(s) ---> K+(aq) + Cl- (aq)

The solubility of KCl given is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

The Ksp was found to be equal to 14.

In pure water KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x= molar solubility [K+],/[Cl-] :. × , x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl requires to make 100mL saturated solutio

37M moles/L

The Ksp was found to be equal to 14.

4.0 M HCl = KCl =[K+][Cl-]

Let y= molar solubility :. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b= molar solubility :. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - rule of thumb

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Therefore in a solution with a common ion, the solubility of the compound reduces dramatically.

8 0
3 years ago
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If a house viewed through infrared goggles ,where would the most seen
Serjik [45]

Answer:

The answer is D - near the roof.

Explanation:

Attics don't have a ton of insulation or hard cinder-block to block the goggles form seeing through it. This is why the attic can be a place where the temperature varies with the outside temperature.

5 0
3 years ago
4.
telo118 [61]

The series which is in order of increasing boiling point is CH3CH2CH3 CH3COCH3 CH2CH2CH3OH

However, the boiling point of an organic substance is the temperature at which the vapor pressure of the liquid organic substance equals the pressure surrounding the liquid and the liquid changes into a vapor.

<h3>What are organic compounds?</h3>

Organic compounds are substance containing carbon and hydrogen. Some few organic compounds include:

  • Alkanes
  • Alkenes
  • Alkynes
  • Alkanols
  • Alkanals
  • Alkanones
  • Esters
  • Amines

So therefore, the series which is in order of increasing boiling point is CH3CH2CH3 CH3COCH3 CH2CH2CH3OH

Learn more about organic compounds:

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8 0
2 years ago
The function of this organelle is to create energy (ATP) through the process of cellular respiration.
dedylja [7]

Answer:

C

Explanation:

I hope this is correct and have a great day

3 0
3 years ago
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1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to
Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant,  NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

8 0
3 years ago
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