Answer : The partial pressure of
is, 67.009 atm
Solution : Given,
Partial pressure of
at equilibrium = 30.6 atm
Partial pressure of
at equilibrium = 13.9 atm
Equilibrium constant = 
The given balanced equilibrium reaction is,

The expression of
will be,

Now put all the values of partial pressure, we get


Therefore, the partial pressure of
is, 67.009 atm
Answer:
you absolute buffoon Use Ohms' Law: V = RI
V = (1x10^3)(5x10^-3) = 5 volts
Yes, this is in the range of normal household voltages.
Explanation:
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,



So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
no not always sometimes they react at all so false I hope I helped :)
Answer:
Explanation:
This is a circular motion questions
Where the oscillation is 27.3days
Given radius (r)=3.84×10^8m
Circular motion formulas
V=wr
a=v^2/r
w=θ/t
Now, the moon makes one complete oscillation for 27.3days
Then, one complete oscillation is 2πrad
Therefore, θ=2πrad
Then 27.3 days to secs
1day=24hrs
1hrs=3600sec
Therefore, 1day=24×3600secs
Now, 27.3days= 27.3×24×3600=2358720secs
t=2358720secs
Now,
w=θ/t
w=2π/2358720 rad/secs
Now,
V=wr
V=2π/2358720 ×3.84×10^8
V=1022.9m/s
Then,
a=v^2/r
a=1022.9^2/×3.84×10^8
a=0.0027m/s^2