Answer:
Final temperature of the gas is 576 
.
Explanation:
As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:
where 
and 
are volume of gas at 
and 
temperature (in kelvin scale) respectively.
Here 
, 
and 
So 
849 K = (849-273) = 576
So final temperature of the gas is 576 .
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
Answer: scientific explanation
Explanation:
A really good scientific explanation should do two main things: It should explain all the observations and data we have. It should allow us to make testable predictions that we can check using future experiments.
Of course, at STP, dioxygen is a gas, but 10.0 g is still 10.0 g. We could calculate its volume at STP, which is 22.4 L × its molar quantity, approx. 8⋅L . There are 1.51×1023molecules O2 in 10.0 g O2 .
Answer:
D
We don't have a way to contain the radioactive waste produced.
Explanation
Nuclear waste is non disposable and takes millions of years to biodegrade and lose it radio activeness
