The given question is incomplete. The complete question is:
The change in entropy is related to the change in the number of moles of gas molecules. Determine the change in moles of gas for each of the reactions and decide if the entropy increases decreases or has little to no change:
A. 
B. 
C. 
D. 
Answer: A. : decreases
B. 
: decreases
C. : no change
D. : increases
Explanation:
Entropy is defined as the randomness of the system.
Entropy is said to increase when the randomness of the system increase, is said to decrease when the randomness of the system decrease and is said to have no change when the randomness remains same.
In reaction , as gaseous reactant is changed to solid product, entropy decreases.
In reaction , as 4 moles of gaseous reactants is changed to 2 moles of gaseous product, entropy decreases.
In reaction , as 3 moles of gaseous reactants is changed to 3 moles of gaseous product, entropy has no change.
In reaction , as 1 mole of gaseous reactant is changed to 3 moles of gaseous product, entropy increases.
Because when you take the sand out it doesn't stay the same shape
The answer is D. Use the equation PV=nRT
P=(.567mol)(.0821)(300K)/4.5L
Answer:
1.089%
Explanation:
From;
ν =1/2πc(k/meff)^1/2
Where;
ν = wave number
meff = reduced mass or effective mass
k = force constant
c= speed of light
Let
ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl
ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl
The between the two is obtained from;
ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2
Therefore;
ν' - ν /ν = [meff/m'eff]^1/2 - 1
Substituting values, we have;
ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1
ν' - ν /ν = -0.01089
percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;
ν' - ν /ν * 100
|(-0.01089)| × 100 = 1.089%
I'm guessing you mean O2 gas? This is an elemental compound. It can't be a mixture because it is composed of only 1 element. Hope I helped!
