Chemistry, 29.01.2020 04:12 lalala1212

Relate changes in speed during a planet's orbit to the shape of the orbit and the gravitational pull of the sun

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Ok so...If the Earth were the only planet to orbit the Sun, then the Earth and Sun would each orbit the common CM, located at a point within a few hundred km of the Sun's center (the Sun is 333,000x more massive than Earth, so the CM lies this factor closer to the Sun's center than to the Earth's center) with a period of 1 year. However, in our actual solar system the Sun and Jupiter combine to have 99.95% of its mass (with the Sun having 99.85%), and their gravitational force is much stronger than that of any other planet-Sun pair. So to a good approximation the Sun and Jupiter orbit about their CM, lying at a point just outside the surface of the Sun, with Jupiter having the much larger orbit and acceleration (the Sun is 1047x more massive than Jupiter). This point lies so close to the center of the Sun that we can still refer to the planets "orbiting the Sun". The effects of the other planets are to cause the Sun to loop around a bit (i.e., yet other accelerations) about a grand solar system CM in roughly 12 years (Jupiter's orbital period being 11.86 years).

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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 :

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

Answer from: Quest

the answer is:

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).

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Relate changes in speed during a planet's orbit to the shape of the orbit and the gravitational pull...

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