Answer:
P = 1/8
Explanation:
The wave function of a particle in a one-dimensional box is given by:
Hence, the probability of finding the particle in the one-dimensional box is:
Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:
![P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]](https://tex.z-dn.net/?f=%20P%20%3D%20%5Cfrac%7B2%7D%7BL%7D%20%5B%5Cfrac%7BL%7D%7B16%7D%20%281%20-%204%5Cfrac%7Bsin%28%5Cfrac%7Bn%20%5Cpi%7D%7B4%7D%29%7D%7Bn%20%5Cpi%7D%29%5D%20)
Solving for n=4:
I hope it helps you!
Answer:
A
Explanation:
A compound machine is a combination of two or more simple machines.
For this problem, let's use the approach of dimensional analysis. This technique is done by cancelling out like units that appear both on the numerator and the denominator side. As a result, this technique will let you know that your final answer conforms to what parameter is asked. In this case, the final answer should be in mL. The solution is as follows:
14.5 mg * 1 mL/25 mg = 0.58 mL
Answer:
C.5
Explanation:
A number of electrons present in valence shell of penultimate Shell represents the group of elements.
For s block elements: no.of group=number of valence shell electron.
p block elements: no. of group= 2 + 10 + number of valence electrons.
d block elements: no. of group= number(n-1)d electrons + number of electrons in nth shell.
Here, the differential electron is in p orbital hence, it belongs to p block
No. of group= 2+ 10 + 3=15 i.e 15th group or VA group.
Answer:
The equation is H2SO4 + 2KOH ===> K2SO4 + 2H2O
The molarity of base is 0.257 M
Explanation:
Using titration equation CAVA/CBVB = NA/NB
Where NA is the number ov mole of acid =1
Using titration equation
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 1
NB is the number of mole of base = 2
CA is the molarity of acid =1.5M
CB is the molarity of base = to be calculated
VA is the volume of acid = 13.7 mL
VB is the volume of base = 80 mL
Substituting
1.5×80/CB×13.7 = 1/2
Therefore CB = 1.5×13.7×2/80×1
CB = 0.257 M.
