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Zina [86]
3 years ago
10

What amount (moles) is represented by each of these samples?

Chemistry
1 answer:
Zinaida [17]3 years ago
5 0
<h3>Answer:</h3>

                  a)  Moles of Caffeine  =  1.0 × 10⁻⁴ mol

                  b) Moles of Ethanol   =  4.5 × 10⁻³ mol

<h3>Solution:</h3>

Data Given:

                  Mass of Caffeine  =  20 mg  =  0.02 g

                  M.Mass of Caffeine  =  194.19 g.mol⁻¹

                  Molecules of Ethanol  =  2.72 × 10²¹

Calculate Moles of Caffeine as,

                               Moles  =  Mass ÷ M.Mass

Putting values,

                               Moles  =  0.02 g ÷ 194.19 g.mol⁻¹

                                Moles  =  1.0 × 10⁻⁴ mol

Calculate Moles of Ethanol as,

                                                         As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Particles and Avogadro's Number is given as,

                          Number of Moles  =  Number of Molecules ÷ 6.022 × 10²³

Putting values,

                          Number of Moles  =  2.72 × 10²¹ Molecules ÷ 6.022 × 10²³

                          Number of Moles  =  4.5 × 10⁻³ Moles

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A sample compound contains 9.11 g Ni and 5.89g F. What is the empirical formula of this compound?
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The empirical formula of the compound is C. NiF₂.

<em>Step 1</em>. Calculate the <em>moles of each element</em>

The empirical formula is the simplest whole-number ratio of atoms in a compound.

The ratio of atoms is the same as the ratio of moles.

So, our job is to calculate the molar ratio of Ni to F.

Moles of Ni = 9.11 g Ni × (1 mol Ni /(58.69 g Ni) = 0.1552 mol Ni

Moles of F = 5.89 g F × (1 mol F/19.00 g F) = 0.3100 mol F

<em>Step 2</em>. Calculate the <em>molar ratio</em> of the elements

Divide each number by the smallest number of moles

Ni:F = 0.1552:0.3100 = 1:1.997 ≈ 1:2

<em>Step 3</em>: Write the <em>empirical formula</em>

EF = NiF₂

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The Society of Automotive Engineers has established an accepted numerical scale to measure the viscosity of motor oil. For examp
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Answer:

Most viscous to least viscous: (c)> (b)> (a)

Explanation:

For hydrocarbons, viscosity increases with increasing molar mass. Because increasing molar mass signifies increase in number of electrons in molecules.

We know that in non-polar hydrocarbons, only van der waal intermolecular force exists. Van der waal force is proportional to number of electrons in a molecule.

Therefore with increasing molar mass, van der waal force increases. hence molecules gets more tightly bind with each other resulting increase in viscosity.

Here molar mass order : (c)> (b)> (a)

Therefore viscosity order : (c)> (b)> (a)

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30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr
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Answer: Thus concentration of Ca(OH)_2 in mol/dm^3  is 0.011 and in g/dm^3 is 0.814

Explanation:

To calculate the concentration of Ca(OH)_2, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3         1cm^3=0.001dm^3

Putting values in above equation, we get:

1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3

The concentration in g/dm^3 is 0.011mol/dm^3\times 74g/mol=0.814g/dm^3

Thus concentration of Ca(OH)_2 is 0.011mol/dm^3 and 0.814g/dm^3

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.2. causing the demand for money to fall.

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