There are globular and open star clusters, but there are no binary, eclipsing, or wobbling ones.
Answer:
24 atm is the total pressure exerted by the gases
Explanation:
We propose this situation:
In a vessel, we have 4 gases (for example, hydrogen, Xe, methane and chlorine)
Each of the gases has the same pressure:
6 atm → hydrogen
6 atm → xenon
6 atm → methane
6 atm → chlorine
To determine the total pressure, we sum all of them:
Partial pressure H₂ + Partial pressure Xe + Partial pressure CH₄ + Partial pressure Cl₂ = Total P
6 atm + 6 atm + 6 atm + 6 atm = 24atm
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.
Using Phosphoric acid will work perfectly for producing Hydrogen halides because its not an Oxidizing agent. ...
Using an ionic chloride and Phosphoric acid
H3PO4 + NaCl ==> HCl + NaH2PO4
H3PO4 + NaI ==> HI + NaH2PO4
H2SO4 + NaCl ==> HCl + NaHSO4
This method(Using H2So4) will work for all hydrogen hydrogen halide except Hydrogen Iodide and Hydrogen Bromide.
The Sulphuric acid won't be useful for producing Hydrogen Iodide because its an OXIDIZING AGENT. Whist producing the Hydrogen Iodide... Some of the Iodide ions are oxidized to Iodine.
2I-² === I2 + 2e-
Answer:
There are three atoms in the equation