The correct answer from the choices given is the last option. The can from the <span> car will lose the carbon more quickly because there are fewer solute–solvent collisions. The can in the car has a lower temperature than the one in the refrigerator. At low temperature, the solubility of carbon dioxide in the liquid decrease therefore particles would tend to be in the vapor phase and escape from the liquid.</span>
Explanation:
Since, it is given that critical temperature of Argon is 150.9 K and critical pressure of Argon is 48.0 atm.
It is known that gas phase of neon occurs at 50 K. As the boiling point of Ar is more than the boiling point of neon which means that there is strong intermolecular force of attraction between argon molecules as compared to neon molecules.
This is also because argon is larger in size. As a result, induced dipole-induced dipole forces leads to more strength in Ar as compared to Ne.
Answer:
A: 20 km/h, B: 10 km/h
Explanation:
Speed of Vehicle A = 40-20/ 2-1
=20 km/h
If A is 20km/h, B is 10km/h as given in option
Answer:
[COF₂] = 0.346M
Explanation:
For the reaction:
2COF₂(g) ⇌ CO₂(g) + CF₄(g)
Kc = 5.70 is defined as:
Kc = [CO₂] [CF₄] / [COF₂]² = 5.70 <em>(1)</em>
Equilibrium concentrations of each compound after addition of 2.00M COF₂ will be:
[COF₂] : 2.00M - 2x
[CO₂] : x
[CF₄] : x
Replacing in (1):
5.70 = [X] [X] / [2-2x]²
22.8 - 45.6x + 22.8x² = x²
0 = -21.8x² + 45.6x - 22.8
Solving for x:
X = 1.265 <em>-False answer, will produce negative concentrations-</em>
<em>X = 0.827.</em>
Replaing, molar concentration of COF₂ is:
[COF₂] : 2.00M - 2×0.827 = <em>0.346M</em>
I hope it helps!
