Answer:
You may be referring to the gas that makes up 21% of the earth's atmosphere, which is oxygen.
Explanation:
According to NASA, the gases in Earth's atmosphere include:
Nitrogen — 78 percent
Oxygen — 21 percent
Argon — 0.93 percent
Carbon dioxide — 0.04 percent
(Trace amounts of neon, helium, methane, krypton and hydrogen, as well as water vapor)
<span>The solubility of KClO</span>₃ : ( 10.1 / 100 g water ) at 30ºC
10.1 g ------------ 100 g ( H₂O )
? g ------------- 100 g ( H₂O )
Mass of KClO₃ :
100 * 10.1 / 100
1010 / 100 = 10.1 g of KClO₃
hope this helps!
1. L
Number one because the lines match up
Answer:
ΔH = 
Explanation:
In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.
The heat transfer is represented by
=
where
= the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.
= the heat of combustion
Also, we know that the total heat change of the any system is
ΔH = ΔQ + ΔW
where
ΔH = the total heat absorbed by the system
ΔQ = the internal heat absorbed by the system which in this case is
ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0
substituting into the heat change equation
ΔH = + 0
==> ΔH =
C₆H₆ is benzene which has a molar mass of 78 g/mol. When benzene is burned, the reaction is called combustion. The heat produced in this reaction is called the heat of combustion. For benzene, the heat of combustion is -3271 kJ/mol.
Heat of benzene = (8.7 g)(1 mol/78 g)(-3271 kJ/mol) = -364.84 kJ
By conservation of energy,
Heat of benzene = - Heat of water
where
Heat of Water = mCp(Tf - T₀)
where Cp for water is 4.187 kJ/kg·°C
Thus,
-364.84 kJ = -(5691 g)(1 kg/1000 g)(4.187 kJ/kg·°C)(Tf - 21)
<em>Tf = 36.31°C</em>
