The reaction is 2 NO (g) <----> N2(g) + O2
partial pressures
Initial 37.30 0 0
Change -2p +p +p
Equilibrium 37.30-p p p
Kp = pN2 X pO2 / (pNO)^2
2400 = p^2 / (37.30-p)^2
3339096 - 179040p + 2400p^2 = p^2
2399p^2 + 3339096 -179040 p = 0
On solving
p = 36.55atm
Thus partial pressure of N2 and O2 = 36. 55 atm
There are 4 quantum numbers that can be used to describe the space of highest probability an electron resides in.
First quantum number is the principal quantum number- n , states the energy level.
Second quantum number states the angular momentum quantum number - l,
states the subshell and the shape of the orbital
values of l for n energy shells are from 0 to n-1
third is magnetic quantum number - m, which tells the specific orbital.
fourth is spin quantum number - s - gives the spin of the electron in the orbital
here we are asked to find l for 3p1
n = 3
and values of l are 0,1 and 2
for p orbitals , l = 1
therefore second orbital for 3p¹ is 1.
Which two solutions, when mixed together, will undergo a double replacement reaction and form a white, solid substance?
1. NaCl(aq) and LiNO3(aq)
2. KCl(aq) and AgNO3(aq) answer
3. KCl(aq) and LiCL(aq)
4. NaNO3(aq) and AgNO3(aq)
2 is the answer because AgCl is formed and that is a white ppt.
Just use the Heisenberg Uncertainty principle:
<span>ΔpΔx = h/2*pi </span>
<span>Δp = the uncertainty in momentum </span>
<span>Δx = the uncertainty in position </span>
<span>h = 6.626e-34 J s (plank's constant) </span>
<span>Hint: </span>
<span>to calculate Δp use the fact that the uncertainty in the momentum is 1% (0.01) so that </span>
<span>Δp = mv*(0.01) </span>
<span>m = mass of electron </span>
<span>v = velocity of electron </span>
<span>Solve for Δx </span>
<span>Δx = h/(2*pi*Δp) </span>
<span>And that is the uncertainty in position. </span>
