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Rom4ik [11]
3 years ago
15

In the 1990s scientists were studying the comet named Shoemaker-Levy. Write a short report about this comet.

Chemistry
1 answer:
Korolek [52]3 years ago
7 0

Answer:

Please find the report on the comet below

Explanation:

Specifically in the year 1993, a comet was discovered by three individuals namely: Eugene Shoemaker, Carolyn Shoemaker, and David Levy. The comet was named after the first discoverers as; Shoemaker-Levy 9.

Shoemaker-Levy 9 was reported to have already been in pieces as at the time of discovery. It was believed to have been torn apart by the gravitational force emanating from the biggest of the planets called JUPITER.

In July 1994, Shoemaker-Levy 9 came in contact with Jupiter in what seemed to be the end of the discovered comet. This collision of Shoemaker-Levy 9 with planet Jupiter was said to be the first ever solar systems collision observed.

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If a container were to have 24 molecules of C5H12 and 24 molecules of O2 initially, how many total molecules (reactants plus pro
frutty [35]

Answer:

81 molecules

Explanation:

The reaction between C5H12 and O2 is a combustion reaction and is represented by the following equation;

C5H12 + 8O2 --> 5CO2 + 6H2O

The ratio of C5H12 to O2 from the above equation is 1 : 8.

Aplying the conditins of the question; 24 molecules each of C5H12 and O2 we have;

3C5H12 + 24O2 --> 15CO2 + 18H2O

This means we have 24 - 3 = 21 molecules of C5H12 that are unreacted.

Total molecules is given as;

3(C5H12) + 24(O2) + 15(CO2) + 18(H2O) + 21(Unreacted C5H12) = 81 molecules

5 0
3 years ago
The amount of kinetic energy an object has depends on its:
UkoKoshka [18]

Answer:

the 4th one

Explanation:

kinitc energy formula is1/2mv^

there is mass ,and velocity (speed)

I hope it help

3 0
2 years ago
Read 2 more answers
Maltase is an enzyme that breaks down maltose. If a maltase enzyme has just completed catalyzing the decomposition of maltose, t
netineya [11]
Yes it is available. It will continue catalyzing the reactions until it becomes completely consumed. That's how enzymes work. They work and are eventually consumed in the process completely without altering the reaction in any way other than speeding it up.
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7 0
3 years ago
A 100 g sample of potassium chlorate, KCIO3(s), is completely decomposed by heating:
Mama L [17]
Explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
. Under these conditions for pressure and temperature, one mole of any ideal gas occupies
22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas.
This tells you that the reaction will always produce
3
2
times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
1 mole KClO
3
122.55
g
=
1.885 moles KClO
3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L
6 0
3 years ago
what method can be used to seperate parts of liquid mixture wwhen the entire mixture can pass through a filter
QveST [7]
Distillation is the <span>method that can be used to seperate parts of liquid mixture when the entire mixture can pass through a filter.</span>
6 0
4 years ago
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