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Alex787 [66]
3 years ago
12

Will mark as Brainliest.

Chemistry
1 answer:
sertanlavr [38]3 years ago
8 0
<span>Atomic radius generally increases as we move 
</span><span>down a group and from right to left across a period.</span>
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Which type of wave does not require a medium to travel through?
nlexa [21]

Hello,

I'm not very sure but I think the correct answer is D) Water Wave

I really hope this helps!!!! Happy Holidays!!!! (:

8 0
3 years ago
What is the temperature of 0.5 moles of water vapor that occupies 120 dm3 and applies a pressure of 15,000 Pa to its container?
MArishka [77]
Use the ideal gas law:

PV = nRT

so, T = PV / nR

n=0.5
V= 120 dm^3 = 120 L  (1 dm^3 = 1 L)
R = 1/12
P = 15,000 Pa =  0.147 atm     (1 pa = 9.86 10^{-6} )

Put the values:

T = PV / nR
T = (0.147) (120) / (0.5) (1/12)
T= 426 K
6 0
3 years ago
Oxidation is _______?
Vika [28.1K]

Answer:

the process or result of oxidizing or being oxidized

6 0
2 years ago
What's the chemical equation for the carbon dioxide bubbles in a homemade lava lamp?
julsineya [31]

Answer:

C₆H₈O₇+ 3NaHCO₃ --› Na₃C₆H₅O₇ + 3CO2 + 3H₂O

Explanation:

The reaction occuring in lava lamp is acid base reaction.

When you drop tablet into water the citric acid reacts with sodium bicarbonate and forms water, a salt, and bubbles of carbon dioxide gas.

6 0
3 years ago
Read 2 more answers
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
3 years ago
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