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natima [27]
3 years ago
5

PLEASE PLEASE PLEASE HELP ME

Chemistry
1 answer:
andrey2020 [161]3 years ago
8 0

3. Cl₂ + 2KI --> 2KCl + I₂

Cl = 2        Cl = 2

K = 2         K = 2

I = 2           I = 2

4. 2NaCl --> 2Na + Cl₂

Na = 2      Na = 2

Cl = 2       Cl = 2

5. 4Na + O₂ --> 2Na₂O

Na = 4      Na = 4

O = 2        O = 2

6. 2Na + 2HCl --> H₂ + 2NaCl

Na = 2         Na = 2

H = 2           H = 2

Cl = 2          Cl = 2

7. 2K + Cl₂ --> 2KCl

K = 2         K = 2

Cl = 2        Cl = 2

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If a metal is less reactive than carbon, it can be extracted from its oxide by heating with carbon. The carbon displaces the metal from the compound, and removes the oxygen from the oxide. This leaves the metal.
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2 years ago
In an experiment to study the photoelectric effect, a scientist measures the kinetic energy of ejected electrons as afunction of
crimeas [40]

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!!!

5 0
4 years ago
What is the percent magnesium in a sample made of 24g Mg and 70gCl?
seropon [69]

Answer:

Total percent of magnesium in sample = 25.5%

Explanation:

Given:

Mass of magnesium = 24 gram

Mass of chlorine = 70 gram

Find:

Total percent of magnesium in sample = ?

Computation:

Total mass of sample = Mass of magnesium + Mass of chlorine

Total mass of sample = 24 gram + 70 gram

Total mass of sample = 94 gram

Total percent of magnesium in sample = [Mass of magnesium / Total mass of sample]100

Total percent of magnesium in sample = [24/94]100

Total percent of magnesium in sample = [0.255]100

Total percent of magnesium in sample = 25.5%

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