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dezoksy [38]
3 years ago
13

Where does the peak intensity need to fall on the spectrum for a very intense star?

Chemistry
2 answers:
umka2103 [35]3 years ago
8 0

Answer: A

Explanation:  I got it right on my test

Brainliest?

viva [34]3 years ago
7 0

Answer:

Gamma range

Explanation:

A very intense start is a bright star at very high temperature. Now, hotter the object shorter is the wavelength of peak radiation. The increasing order of wavelength of the given regions in the electromagnetic spectrum are:

gamma range < ultra violet < visible < infrared

Ideally, since the shortest wavelength based on the given options is the gamma range, the peak intensity can also be expected to fall at that wavelength. If the star spectrum was recorded in the UV-visible range of the EM spectrum, then it would fall in the UV range

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What are the similarities and differences between light microscopes used by early scientists and light microscopes used today?
Aleks [24]
Electron microscopes differ from light microscopes in that they produce an image of a specimen by using a beam of electrons rather than a beam of light. Electrons have much a shorter wavelength than visible light, and this allows electron microscopes to produce higher-resolution images than standard light microscopes
3 0
3 years ago
What is the molality (m) of a solution that contains 76.5 g of KCl dissolved in 85.0 g of<br> water?
ladessa [460]

Answer:

76.5g KCl/74.55 grams per mole Kcl = x

molality= x/.085 kg H2O

Explanation:

well remember molality is moles of solute/kilograms of solvent. So it's the moles of KCl over 85 g of h20 converted into kg. if this makes sense.

5 0
3 years ago
Which element, if combined with chlorine, will have the greatest attraction for the chlorine electrons? N P As
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3 years ago
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A substance formed when two or more elements combine and lose their distinct properties is
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7 0
3 years ago
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
inn [45]

Answer:

3.47 ×10^-10

Explanation:

The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.

E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

4 0
3 years ago
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