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GrogVix [38]
2 years ago
8

Which of Hubble's findings supported the big bang theory? Check all that apply.

Chemistry
2 answers:
Stella [2.4K]2 years ago
8 0

Answer:

C,D,E

Explanation:

JUST TOOK THE TEST

grigory [225]2 years ago
3 0

<u>Out of all given options, Hubble's findings supported the big bang theory are as follows: </u>

  • The universe formed from a central point.
  • Most galaxies in the universe are moving away from Earth.
  • There are billions of galaxies in the universe, not just one.

Option C, D, and E

<u>Explanation: </u>

According to big bang theory, the universe is increasing its region of occupancy in free space. As it is known that the universe is composed of several galaxies. So this spreading of the universe is nothing but spreading of galaxies.

And this is confirmed from Hubble’s findings using Hubble’s telescope.  As they have observed that the speed at which the galaxies are travelling in free space has been increased at a faster rate.

So, the galaxies are used to move away from Earth with time at a faster rate which confirms the big bang theory of universe formation. Also, in simple words, universe started from big bang according to cosmology' standard theory, it’s just moves from everything else.

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How much heat is required to convert 20.0 g of ice at 50.0⁰C to liquid water at 0.0⁰C? The specific heat of ice is 2.06 J/(g∙⁰C)
Alex777 [14]

Answer:

8740 joules are required to convert 20 grams of ice to liquid water.

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The amount of heat required (Q), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:

Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}] (1)

Where:

m - Mass, measured in grams.

c - Specific heat of ice, measured in joules per gram-degree Celsius.

T_{o}, T_{f} - Temperature, measured in degrees Celsius.

L_{f} - Latent heat of fussion, measured in joules per gram.

If we know that m = 20\,g, c = 2.06\,\frac{J}{g\cdot ^{\circ}C}, T_{f} = 0\,^{\circ}C, T_{o} = -50\,^{\circ}C and L_{f} = 334\,\frac{J}{g }, then the amount of heat is:

Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}

Q = 8740\,J

8740 joules are required to convert 20 grams of ice to liquid water.

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