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neonofarm [45]
2 years ago
9

A mothball, composed of naphthalene (c10h8), has a mass of 1.64 g . part a how many naphthalene molecules does it contain?

Chemistry
1 answer:
OLga [1]2 years ago
3 0
The molar mass of Naphthalene is 128g/mol
Therefore; a mass of 1.64 g of Naphthalene contains'
   = 1.64g/128 g
    = 0.0128 moles
But, from the Avogadro's law 1 mole of a substance contains 6.022 × 10^23 particles
Therefore 1 mole of Naphthalene contains 6.022×10^23 molecules
Hence; 0.0128 moles × 6.022 ×10^23 molecules
          = 7.716 × 10^21 molecules
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how many liters of nitrogen gas are needed to react with 90.0 g of potassium at STP in order to produce potassium nitride accord
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The atomic mass of potassium is 19 g/mol
90.0 g of potassium contains ;
   = 90.0/19 
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The reaction between potassium and nitrogen is given by the equation;
 6K + N2 = 2K3N
The mole ratio of potassium and nitrogen is 6:1
The number of moles of nitrogen;
    = 4.737/6
    = 0.7895 moles
But, 1 mole of nitrogen occupies 22.4 liters at STP
Therefore; 0.7895 moles × 22.4 liters
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Why is water able to dissolve salts such as sodium chloride
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2 years ago
A gas mixture has a total pressure of 0.51 atm and consists of He and Ne. If the partial pressure of He in the mixture is 0.32 a
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Explanation:

Relation between total pressure and mole fraction is as follows.

                   P_{1} = P^{o}x_{1}

where,     P_{1} = partial pressure of gas 1

                x_{1} = mole fraction of gas 1

                P^{o} = total pressure of the gases

Also,  it is known that x_{1} + x_{2} = 1

or,               x_{2} = 1 - x_{1}

It is given that total pressure of gas mixture is 0.51 atm and partial pressure of helium is 0.32 atm.

Hence, calculate the mole fraction of helium gas present in the given mixture as follows.

                P_{He} = P^{o}x_{He}

                0.32 atm = 0.51 atm \times x_{He}

                    x_{He} = 0.63

As, x_{He} + x_{Ne} = 1

so,             x_{Ne} = 1 - x_{He}

                               = 1 - 0.63

                               = 0.37

Hence, calculate the partial pressure of Ne as follows.

                    P_{total} = P_{He}x_{He} + P_{Ne}x_{Ne}

            0.51 atm = 0.32 atm \times 0.63 + P_{Ne}\times 0.37    

                   0.51 atm = 0.2016 atm + 0.37 \times P_{Ne}

                       0.3084 atm = 0.37 \times P_{Ne}

                            P_{Ne} = \frac{0.3084 atm}{0.37}

                                        = 0.834 atm

Thus, we can conclude that the partial pressure of the Ne in the mixture is 0.834 atm.

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I’m pretty sure it’s B
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Explanation:

Fe2+ lose electrons to become Fe 3+ hence oxidized

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