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Anna35 [415]
3 years ago
15

Could you please help

Chemistry
1 answer:
Semenov [28]3 years ago
8 0
I would say D, because you need to start with nothing to measure the different sizes as they start to grow. hope this helps!
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Help me answer this chemistry question. thank you!
Vladimir79 [104]

Answer:

Chlorine and bromine

Explanation:

I think thats the answer in your question

8 0
3 years ago
50 points!! Brainliest if correct!!
Luda [366]

The answer is C. The answer is C because if u increase the surface area, the more reactants u will get. and if u get more The reactants will move faster. Hoped that Helped!:-)

4 0
3 years ago
Which of these sentences describes a benefit of fracking?
bagirrra123 [75]
Answer is: <span>A. Fracking reduces U.S. dependency on oil and gas from other countries..
Other answers are not bad side of fracking.
</span>Fracking is technique in which rock is fractured by a pressurized liquid and is used in <span>to create cracks in the rock formations through which </span>natural gas<span>, </span>petroleum<span> will flow more freely.</span>
7 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
Suppose that on a dry, sunny day when the air temperature is near 37 ∘C,37 ∘C, a certain swimming pool would increase in tempera
ioda

Answer:

The fraction of water body necessary to keep the temperature constant is 0,0051.

Explanation:

Heat:

Q=m*Ce*ΔT

Q= heat  (unknown)

m= mass  (unknown)

Ce= especific heat (1 cal/g*°C)

ΔT= variation of temperature  (2.75 °C)  

Latent heat:

ΔE=∝mΔHvap

ΔE= latent heat

m= mass  (unknown)

∝= mass fraction (unknown)

ΔHvap= enthalpy of vaporization (539.4 cal/g)

Since Q and E are equal, we can match both equations:

m*Ce*ΔT=∝*m*ΔHvap

Mass fraction is:

∝=\frac{Ce*ΔT}{ΔHvap}

∝=\frac{(1 cal/g*°C)*2.75°C}{539.4 cal/g}

∝=0,0051

7 0
2 years ago
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