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3 years ago

In ancient times, what would the Moon have looked like from the Earth?

Chemistry

2 answers:

ss7ja [257]3 years ago

6 0

The moon would have appeared more vividly and brightly due to the absence of light pollution and smog

Ipatiy [6.2K]3 years ago

6 0

Answer:

It would have looked bigger. Active volcanoes might have been visible on it.

Explanation:

The most widely accepted theory of the Moon formation is Giant Impact hypothesis which states that a giant Mars sized celestial body: Theia hit planet Earth and the material that ejected out of the Earth coalesced and formed Moon. The Moon as observed today is moving away at a rate of around 4 cm per year which means that in ancient times it would have looked much bigger in the sky.

Frozen lava has been seen on the Lunar surface which are called as Maria (dark spots). In ancient times we could have witnessed those active volcanoes.

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Why do substances with different structures have similar conductivity values

weeeeeb [17]

Answer:

because of the material, they are made of

Explanation:

like a paperclip and a car door maybe they both are metal or steel whatever they both can conduct energy. because of what they are made of.

i hope this helps and sorry i am rushing, i also just woke up

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3 years ago

_____ are isotopes that emit radiation because they have unstable nuclei.

alexandr1967 [171]

The correct answer among the choices given is option B. Radioisotopes are isotopes that emit radiation because they have unstable nuclei. These are radioactive isotopes of an element. They are defined as atoms that contain an unstable combination of neutrons and protons.

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3 years ago

Read 2 more answers

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

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3 years ago

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dmitriy555 [2]

A.electrons are shared between two different nuclei

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in their mechanism of action, a difference between lipid-soluble and water-soluble hormones is that _____.

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3 years ago