Answer:
0.67mol/Kg
Explanation:
The following were obtained from the question:
Mole of solute = 0.50mol
Mass of solvent = 750g = 750/1000 = 0.75Kg
Molality =?
Molality = mole of solute /mass of solvent
Molality = 0.5/0.75
Molality = 0.67mol/Kg
Answer:
See explanation
Explanation:
The third law of thermodynamics states that "the entropy of a perfect crystal of a pure substance approaches zero as the temperature approaches zero" (Wikipedia).
One example of the third law of thermodynamics has to do with steam. Steam is gaseous water. Since it is a gas, its molecules are free to move around therefore its entropy is high. When the temperature of the steam is decreased below 100 degrees, the molecules of steam loose energy and turn into liquid water and do not move as freely as they did in the gaseous state. If the temperature is further decreased to yield ice at zero degrees, the molecules of water are "frozen" in their positions and the entropy of the system decreases to zero.
Also, the ions in ionic crystal solids move around when the substance is in solution or in molten state hence the substance conducts electricity. When the ionic substance is in solid state, the ions do not move about and the entropy of the solid system tends towards zero.
Corrosion of aluminium metal is a chemical process. Corrosion happens when a certain chemical reaction takes place.
Answer:
B. potential energy
Explanation:
Potential energy ( Mechanical in this case ) is energy stored by force. so If you are compressing a spring you are increasing more potential to it by force.
Answer:
0.36 M
Explanation:
There is some info missing. I think this is the complete question.
<em>Suppose a 250 mL flask is filled with 0.30 mol of N₂ and 0.70 mol of NO. The following reaction becomes possible:
</em>
<em>N₂(g) +O₂(g) ⇄ 2 NO(g)
</em>
<em>The equilibrium constant K for this reaction is 7.70 at the temperature of the flask. Calculate the equilibrium molarity of O₂. Round your answer to two decimal places.</em>
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Initially, there is no O₂, so the reaction can only proceed to the left to attain equilibrium. The initial concentrations of the other substances are:
[N₂] = 0.30 mol / 0.250 L = 1.2 M
[NO] = 0.70 mol / 0.250 L = 2.8 M
We can find the concentrations at equilibrium using an ICE Chart. We recognize 3 stages (Initial, Change, and Equilibrium) and complete each row with the concentration or change in the concentration.
N₂(g) +O₂(g) ⇄ 2 NO(g)
I 1.2 0 2.8
C +x +x -2x
E 1.2+x x 2.8 - 2x
The equilibrium constant (K) is:
![K=7.70=\frac{[NO]^{2}}{[N_{2}][O_{2}]} =\frac{(2.8-2x)^{2} }{(1.2+x).x}](https://tex.z-dn.net/?f=K%3D7.70%3D%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BN_%7B2%7D%5D%5BO_%7B2%7D%5D%7D%20%3D%5Cfrac%7B%282.8-2x%29%5E%7B2%7D%20%7D%7B%281.2%2Bx%29.x%7D)
Solving for x, the positive one is x = 0.3601 M
[O₂] = 0.3601 M ≈ 0.36 M
