Explanation:
It is known that equation for ideal gas is as follows.
PV = nRT
The given data is as follows.
Pressure, P = 1500 psia, Temperature, T = 
= 104 + 460 = 564 R
Volume, V = 2.4 cubic ft, R = 10.73 
Also, we know that number of moles is equal to mass divided by molar mass of the gas.
n = 
m = 
= 
= 9.54 lb
Hence, molecular weight of the gas is 9.54 lb.
- We will calculate the density as follows.
d = 
= 
= 3.975 
- Now, calculate the specific gravity of the gas as follows.
Specific gravity relative to air = 
= 
= 51.96
Answer:
Explanation:
From the given information:
The equation for the reaction can be represented as:
The I.C.E table can be represented as:
2SO₂ O₂ 2SO₃
Initial: 14 2.6 0
Change: -2x -x +2x
Equilibrium: 14 - 2x 2.6 - x 2x
However, Since the amount of sulfur trioxide gas to be 1.6 mol.
SO₃ = 2x,
then x = 1.6/2
x = 0.8 mol
For 2SO₂; we have 14 - 2x
= 14 - 2(0.8)
= 14 - 1.6
= 12.4 mol
For O₂; we have 2.6 - x
= 2.6 - 1.6
= 1.0 mol
Thus;
[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,
[O₂] = 1/50 = 0.02 M ,
[SO₃] = 1.6/50 = 0.032 M
Kc = [SO₃]² / [SO₂]² [O₂]
= ( 0.032²) / ( 0.248² x 0.02)
= 0.8325
Recall that; the equilibrium constant for the reaction = 0.8325;
If we want to find:
Then:
Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.
The name of the covalent compound CCl4 is 'Carbon tetrachloride'.
Hope this helps!
A. The heat is needed to melt 100.0 grams of ice that is already at 0°C is +33,400 J.
<h3>What is Specific heat capacity?</h3>
Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass.
<h3>
Heat needed to melt the cube of ice</h3>
The heat is needed to melt 100.0 grams of ice that is already at 0°C is calculated as follows;
Q = mL
where;
- m is mass of the ice
- L is latent heat of fusion of ice = 334 J/g
Q = 100 x 334
Q = 33,400 J
Thus, the heat is needed to melt 100.0 grams of ice that is already at 0°C is +33,400 J.
Learn more about heat capacity here: brainly.com/question/16559442
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2.91 mol Al * ( 26.982 g Al / 1 mol Al) = 78.518 grams
