Density
=mass÷volume
=277÷38
=7.29 g/cm3
Answer:
Kp = 1.41 x 10⁻⁶
Explanation:
We have the chemical equation:
2 A(g) + 3 B(g)⇌ C(g)
In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):
dn= (sum moles products - sum moles reactants)
= (moles C - (moles A + moles B))
= (1 - (2+3))
= 1 - 5
= -4
We have also the following data:
Kc = 63.2
T= 81∘C + 273 = 354 K
R = 0.082 L.atm/K.mol (it is a constant)
Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:
= (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶
Answer:
Keep temperature constant and increase the pressure of the reaction. The rate of reaction increases.
Explanation:
First of all, the question is asking us to design an experiment to investigate the effect of pressure on the rate of reaction hence the pressure can not be held constant since it is the variable under investigation. This eliminates the first option.
Secondly, increasing the pressure of the reaction means that particles of the gas collide more frequently leading to a greater number of effective collisions and a consequent increase in the rate of reaction according to the collision theory.
Hence the answer above.
This is a one-step unit analysis problem. Since we are staying in moles, grams of our compound, and thus molar mass, is not needed.
1 mole is equal to 6.022x10²³ particles as given, so:

<h3>
Answer:</h3>
2.49 mol
Let me know if you have any questions.