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UNO [17]
3 years ago
6

If 7.45g of N2 and 36.5mL of 0.150M O2 react together, how many grams of N2O gas will be produced?

Chemistry
1 answer:
ira [324]3 years ago
4 0

Answer:

0.482 g of N₂O are produced.

Explanation:

The reaction is: 2N₂ + O₂ →  2N₂O

In order to produce 2 moles of N₂O, we need to make react 1 mol of oxygen and 2 moles of nitrogen.

We determine the moles of each reactant:

7.45 g / 28g/mol = 0.266 moles of nitrogen

36.5 mL . 0.15 M = 5.475 mmoles →  . 1mol /1000 mmol = 0.00547 moles

Certainly the oxygen is the limiting reactant.

In order to determine the grams produced, we need to know the limiting reactant.

2 moles of N₂ need 1 mol of O₂

Then 0.266 moles of N₂ may react to (0.266 . 1) /2 = 0.133 moles

We only have 0.00547 moles, so there is not enough O₂.

1 mol of O₂ can produce 2 moles of N₂O

Then, the 0.00547 will produce (0.00547 . 2) /1 = 0.01095 moles

We convert to mass: 0.01095 mol . 44 g /mol = 0.482 g

That's the answer.

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In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
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Answer:

a) v₀ = 4.41 × 10¹⁴ s⁻¹

b) W₀ = 176 KJ/mol of ejected electrons

c) From the graph, light of frequency less than v₀ will not cause electrons to break free from the surface of the metal. Electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. Explanation is in the section below.

Explanation:

The plot for this question which is attached to this solution has Electron kinetic energy on the y-axis and frequency of incident light on the x-axis.

a) Wavelength, λ = 680 nm = 680 × 10⁻⁹ m

Speed of light = 3 × 10⁸ m/s

The frequency of the light, v₀ = ?

Frequency = speed of light/wavelength

v₀ = (3 × 10⁸)/(680 × 10⁻⁹) = 4.41 × 10¹⁴ s⁻¹

b) Work function, W₀ = energy of the light photons with the wavelength of v₀ = E = hv₀

h = Planck's constant = 6.63 × 10⁻³⁴ J.s

E = 6.63 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹J

E in J/mol of ejected electrons

Ecalculated × Avogadros constant

= 2.92 × 10⁻¹⁹ × 6.023 × 10²³

= 1.76 × 10⁵ J/mol of ejected electrons = 176 KJ/mol of ejected electrons

c) Light of frequency less than v₀ does not possess enough energy to cause electrons to break free from the metal surface. The energy of light with frequency less than v₀ is less than the work function of the metal (which is the minimum amount of energy of light required to excite electrons on metal surface enough to break free).

As evident from the graph, electron kinetic energy remains at zero as long as the frequency of incident light is less than v₀.

d) When frequency of the light exceeds v₀, there is an increase of electron kinetic energy from zero steadily upwards with a constant slope. This is because, once light frequency exceeds, v₀, its energy too exceeds the work function of the metal and the electrons instantaneously gain the energy of incident light and convert this energy to kinetic energy by breaking free and going into motion. The energy keeps increasing as the energy and frequency of incident light increases and electrons gain more speed.

e) The slope of the line segment gives the Planck's constant. From the mathematical relationship, E = hv₀,

And the slope of the line segment is Energy of ejected electrons/frequency of incident light, E/v₀, which adequately matches the Planck's constant, h = 6.63 × 10⁻³⁴ J.s

Hope this Helps!!!

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Answer:

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Explanation:

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A center of symmetry is said to exist in a molecule when reflection of all parts of the molecule through the center of symmetry produces an indistinguishable configuration(Housecroeft and Sharpe,2012)

Obviously, the Cl2 molecule has a center of symmetry, hence it is symmetrical. Reflection of the molecules through its center of symmetry produces an indistinguishable configuration.

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