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Alona [7]
3 years ago
8

Are the orbits of the planets could you pls help!!!! Pls don't send scam links !! in the solar system as they revolve around the

Sun perfectly
circular?
Explain.
Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0

Answer: Orbits are elliptical, Sun is in focal point of the ellipse.

Explanation:

Orbits are not perfect circles. Distance of Earth from Sun is between

147 million kilometres - 152 million kilometres. When Earth is in closest point of the Sun it moves faster tant in the point which is most far away from Sun.

Kepler´s Laws give more information.

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Convert 213.62 grams of xenon gas at STP to L.
jolli1 [7]

Answer: The coefficient is 3.645

The exponent is 1

There are 4 significant digits

The rightmost significant figure is 5

Explanation:

Scientific notation is defined as the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example : 5000 is written as 5.0\times 10^3

According to avogadro's law, 1 mole of every gas contains avogadro's number 6.023\times 10^{23} of particles, occupy 22.4 L at STP and weighs equal to its molecular mass.

131.29 g of Xe occupy = 22.4 L at STP.

Thus 213.62 g of Xe occupy = \frac{22.4}{131.29}\times 213.62=36.45L at STP.

Scientific notation = 3.645\times 10^1L

The coefficient is 3.645

The exponent is 1

There are 4 significant digits

The rightmost significant figure is 5

4 0
3 years ago
The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t
11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:                                                        

k = Ae^{-\frac{E_{a}}{RT}}

For the reaction without catalyst we have:

k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}   (1)

And for the reaction with the catalyst:

k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}   (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:                      

\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}

\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}    

\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}    

Since the reaction rate is related to the time as follow:

k = \frac{\Delta [R]}{t}

And assuming that the initial concentrations ([R]) are the same, we have:

\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}

\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}

t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!                            

4 0
3 years ago
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