A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.
<h3>What is Combined Gas Law ?</h3>
This law combined the three gas laws that is (i) Charle's Law (ii) Gay-Lussac's Law and (iii) Boyle's law.
It is expressed as
where,
P₁ = first pressure
P₂ = second pressure
V₁ = first volume
V₂ = second volume
T₁ = first temperature
T₂ = second temperature
Now put the values in above expression we get
P₂ = 1.76 atm
Thus from the above conclusion we can say that A sample of an ideal gas has a volume of 2.30 L at 281 K and 1.02 atm. 1.76 atm is the pressure when the volume is 1.41 L and the temperature is 298 K.
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Answer:
Empirical formula is Cr₂O₃.
Explanation:
Given data:
Percentage of Cr = 68.4%
Percentage of O = 31.6%
Empirical formula = ?
Solution:
Number of gram atoms of Cr = 68.4 / 52 = 1.3
2
Number of gram atoms of O = 31.6 / 16 = 1.98
Atomic ratio:
Cr : O
1.32/1.32 : 1.98/1.32
1 : 1.5
Cr : O = 1 : 1.5
Cr : O = 2(1 : 1.5)
Empirical formula is Cr₂O₃.
I believe that the answer is 12 because there is already 3 O molecules and since its in parentheses with 3 outside it that means that there are 3 of those CO molecules meaning that for every 1 CO there will be 3 O’s so 3, four times Is 12
Answer:
It's well Explained below.
Explanation:
First of Excess product of CaCO_3 would be produced due to the fact that there would not be enough CaCl_2 to react with Na_2•CO_3. The main purpose of having stoichiometric quantities is for us to know the correct amount or near the correct amount of each reactant in order to create a product that will be close to the theoretical amount and thus have a higher percent yield.
Answer:
24.03 J/mol.ºC
Explanation:
For a calorimeter, the heat lost must be equal to the heat gained from water plus the heat gained from calorimeter, which has the same initial temperature as the water.
-Qal = Qw + Qc (minus signal represents that the heat is lost)
-mal*Cal*ΔTal = mw*Cw*ΔTw + Cc*ΔTc
Where m is the mass, C is the specific heat, ΔT is the temperature variation, al is from aluminum. w from water and c from the calorimeter. Cw = 4.186 J/gºC
-25.5*Cal*(22.7 - 100) = 99.0*4.186*(22.7 - 18.6) + 14.2*(22.7 - 18.6)
1971.15Cal = 1699.10 + 58.22
1971.15Cal = 1757.32
Cal = 0.89 J/g.ºC
The molar mass of Al is 27 g/mol
Cal = 0.89 J/g.ºC * 27 g/mol
Cal = 24.03 J/mol.ºC
