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Sonja [21]
3 years ago
10

These celestial bodies are thought to originate from two different sources. Some are called long-period and originate from the O

ort Cloud. These bodies orbit the Sun but may take as long as 200 years to complete one orbit. Others are called short-period and originate from the Kuiper Belt in our solar system.
What is this celestial body called?

A) Asteroid

B) Comet

C) Meteor

D) Planet
Chemistry
2 answers:
Whitepunk [10]3 years ago
7 0
Yes it would be metror
Alexus [3.1K]3 years ago
6 0
It’s c I believe because it says 209 years to complete an orbit so hint yourself ?.....
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The above map shows the route of a car traveling from home to the parkIf the car is traveling at a constant speed, does the car
LenaWriter [7]

Answer:

There is no map

Explanation:

Without the map, there is no way of answering this question.

8 0
2 years ago
Read 2 more answers
At 373.15K and 1 atm, the molar volume of liquid water and steamare 1.88 X 10-5 m3 and 3.06 X 10-2m3, respectively. Given that t
professor190 [17]

Answer The value of \Delta H and \Delta U is, 40.79 kJ and 37.7 kJ respectively.

Explanation :

Heat released at constant pressure is known as enthalpy.

The formula used for change in enthalpy of the gas is:

\Delta Q_p=\Delta H=40.79kJ/mol

Now we have to calculate the work done.

Formula used :

w=-P\Delta V\\\\w=-P\times (V_2-V_1)

where,

w = work done  = ?

P = external pressure of the gas = 1 atm

V_1 = initial volume = 1.88\times 10^{-5}m^3=1.88\times 10^{-5}\times 10^3L=1.88\times 10^{-2}L

V_2 = final volume = 3.06\times 10^{-2}m^3=3.06\times 10^{-2}\times 10^3L=3.06\times 10^{1}L

Now put all the given values in the above formula, we get:

w=-(1atm)\times (3.06\times 10^{1}-1.88\times 10^{-2})L

w=-30.5812L.atm=-30.5812\times 101.3J=-3097.87556J=-3.09\times 10^3J=-3.09kJ

Now we have to calculate the change in internal energy.

\Delta U=q+w

\Delta U=40.79kJ+(3.09kJ)

\Delta U=37.7kJ

Thus, the value of \Delta H and \Delta U is, 40.79 kJ and 37.7 kJ respectively.

4 0
3 years ago
Parker has a toy car he has made out of plastic building blocks. He breaks it apart so he can build something different with the
maria [59]

Answer:

C

Explanation:

3 0
3 years ago
Read 2 more answers
2C6H6 + 1502 →12CO2 + 6H20
solmaris [256]

Answer:

.926 moles

Explanation:

Rounding :

H2 0 = 18 gm/mole

   50 gm would then be   50 / 18 = 2.7777 moles of water

every two moles of 2 C6H6   produces   6 moles of water

   2.7777/6  * 2 = .926 moles

5 0
2 years ago
The photodissociation of ozone by ultraviolet light in the upper atmosphere is a first-order reaction with a rate constant of 1.
atroni [7]

Answer:

[O₃]= 8.84x10⁻⁷M  

Explanation:

<u>The photodissociation of ozone by UV light is given by:</u>

O₃ + hν → O₂ + O (1)

<u>The first-order reaction of the equation (1) is:</u>

rate = k [O_{3}] = - k \frac{\Delta [O_{3}]}{\Delta t} (2)

<em>where k: is the rate constant and Δ[O₃]/Δt: is the variation in the ozone concentration with time, and the negative sign is by the decrease in the reactant concentration </em>    

<u>We can get the following expression of the </u><u>first-order integrated law</u><u> of the reaction (1), by resolving the equation (2):</u>

[O_{3}]_{t} = [O_{3}]_{0} \cdot e^{-kt} (3)

<em>where [O₃](t): is the ozone concentration in the elapsed time and [O₃]₀: is the initial ozone concentration</em>

We can calculate the initial ozone concentration using equation (3):  

[O_{3}]_{t} = 5.0 \cdot 10^{-3}M \cdot e^{-(1.0\cdot 10^{-5}s^{-1}) (\frac{10d \cdot 24h \cdot 3600 s}{1d \cdot 1h})} = 8.84 \cdot 10^{-7}M

So, the ozone concentration after 10 days is 8.84x10⁻⁷M.

I hope it helps you!                    

3 0
3 years ago
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