Chemistry, 18.01.2020 17:31 tomasgarrido10

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i need 3 pieces of evidence supporting that the distribution of the planets in our solar system is different than commonly thought.

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Gee. I'll have to guess at what's "commonly thought".

One thing is the scale. Nobody has an accurate picture of the scale in
his head, because we never see a true-scale drawing. THAT's because
it's almost impossible to draw one on paper.

Example:
Shrink the solar system and everything in it so that the Sun
is the size of a quarter (the 25¢ coin).
Then:
-- The Earth is in orbit around the sun, 8.6 feet from it.
That's close enough that you might think you could find the
shrunken Earth. Unfortunately, it's only 0.009 inch in diameter.

-- The shrunken Jupiter is a 'huge' gas giant almost 0.1 inch in diameter.
It's orbiting the sun, about 45 feet away from it.

-- The shrunken Uranus is another gas giant, about 0.035 inch in diameter.
It's orbiting the sun, about 165 feet away from it.

-- The nearest star outside of the solar system is 441 MILES away !
On the same shrunken scale !
And there's NOTHING between here and there !

I think that's the biggest point to make about the REAL solar system ...
its utter emptiness. With the sun reduced to something you can hold
in your hand, the planets are the size of grains of sand, with hundreds
of feet of nothingness between them.

Same for its mass: The solar system is approximately nothing but a star.
That's it. A star, with some dust and some gas around it, and here and there
in the neighborhood a microscopic pebble or a chip of mineral. But mostly
it's nothing but a star ... if you went around and gathered up all that other
rubbish in the same bag and called it a part of the same solar system, the
sun would still have more than 99% of the total mass, and the bag would
hold less than 1% of it.

Book ... It's getting late, Hillary's fading, and that's all I can think of.
I hope this much is some help.

Answer from: Quest

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

40 points and brainliest to anyone who can answer these

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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