statement c would be correct I think.
<u>Given:</u>
Moles of He = 15
Moles of N2 = 5
Pressure (P) = 1.01 atm
Temperature (T) = 300 K
<u>To determine:</u>
The volume (V) of the balloon
<u>Explanation:</u>
From the ideal gas law:
PV = nRT
where P = pressure of the gas
V = volume
n = number of moles of the gas
T = temperature
R = gas constant = 0.0821 L-atm/mol-K
In this case we have:-
n(total) = 15 + 5 = 20 moles
P = 1.01 atm and T = 300K
V = nRT/P = 20 moles * 0.0821 L-atm/mol-K * 300 K/1.01 atm = 487.7 L
Ans: Volume of the balloon is around 488 L
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
Answer:
The weight-average molar mass of polystyrene is 134,160 g/mol.
Explanation:
Molar mass of the monomer styrene ,
, M=104 g/mol
Given , number average molar mass of the polymer , M'= 89,440 g/mol
Degree of polymerization = n

The weight-average molar mass = 
Molar mass dispersity is ratio of weight-average molar mass to the number average molar mass of the polymer.



The weight-average molar mass of polystyrene is 134,160 g/mol.