The effects on the concentration of SO3 gas when the
following changes occur after initial equilibrium has been established in this
system (N.C. = no change) by adding a catalyst.
2SO2(g) + O2(g) 2SO3(g) + 46.8 kcal
So that even though there is an addition of catalyst, no
change in reactants or products has occurred because catalyst only provides a
faster pathway for the reaction to occur.
Answer:
Kc → 5.58×10⁻⁴
Explanation:
Equilibrium reaction is:
2NOCl (g) ⇄ 2NO (g) + Cl₂(g)
Initially we have 1.25 moles of NOCl
After the equilibrium, we have 1.10 moles. So, during the process:
(1.25 mol - 1.1 mol) = 0.15 moles have reacted.
As ratio are 2:2, and 2:1, 0.15 moles of NO and (0.15 /2) = 0.075 moles of chlorine, were produced in the equilibrium.
Finally in equilibrium we have: 1.10 moles of NOCl, 0.15 moles of NO and 0.075 moles of Cl₂. But these amount are not molar, so we need molar concentration in order to determine Kc:
1.10 mol /2.50L = 0.44 M
0.15 mol / /2.50L = 0.06 M
0.075 mol /2.50L = 0.03 M
Let's make expression for Kc → [Cl₂] . [NO]² / [NOCl]²
Kc = (0.03 . 0.06²) / 0.44² → 5.58×10⁻⁴
Answer:
7.5 L of the 10% and 22.5 L of the 30% acid solution, she should mix.
Explanation:
Let the volume of 10% acid solution used to make the mixture = x L
So, the volume of 30% acid solution used to make the mixture = y L
Total volume of the mixture = <u>x + y = 30 L .................. (1)
</u>
For 10% acid solution:
C₁ = 10% , V₁ = x L
For 30% acid solution :
C₂ = 30% , V₂ = y L
For the resultant solution of sulfuric acid:
C₃ = 25% , V₃ = 30 L
Using
C₁V₁ + C₂V₂ = C₃V₃
10×x + 30×y = 25×30
So,
<u>x + 3y = 75 .................. (2)
</u>
Solving 1 and 2 we get,
<u>x = 7.5 L
</u>
<u>y = 22.5 L</u>
This would be c as for the amswer
A solid has definite volume and shape, a liquid has a definite volume but no definite shape, and a gas has neither a definite volume nor shape. ... (a) Solid O2 has a fixed volume and shape, and the molecules are packed tightly together.
Also I need brainlest please
