The rising action, which follows the expositions, comes just before the climax.
Answer:
Kinetic energy: ![6.024*10^{-20}](https://tex.z-dn.net/?f=6.024%2A10%5E%7B-20%7D)
Velocity: ![3.64*10^{5}](https://tex.z-dn.net/?f=3.64%2A10%5E%7B5%7D)
Explanation:
From the equations of the photo-electric effect,
We know:
![hv=hv_0+K.E](https://tex.z-dn.net/?f=hv%3Dhv_0%2BK.E)
Where,
1.
is the Planck's constant which is ![6.626*10^{-34}](https://tex.z-dn.net/?f=6.626%2A10%5E%7B-34%7D)
2.
are the frequency of light emitted and threshold frequencies respectively.
3.
is the kinetic energy of the electrons emitted.
By fact, we come to know that the threshold frequency of Zn is 300nm
And also
Where ,
1.
is the speed of light ![=3*10^8](https://tex.z-dn.net/?f=%3D3%2A10%5E8)
2.
is the wavelength.
Thus,
![\frac{hc}{d}=\frac{hc}{d_0}+K.E\\K.E=\frac{hc}{d}-\frac{hc}{d_0}\\K.E=hc*(\frac{1}{d}-\frac{1}{d_0})\\K.E=6.626*10^{-34}*3*10^8*(\frac{1}{275*10^{-9}}-\frac{1}{300*10^{-9}})\\K.E=6.024*10^{-20}kgm^2s^{-2}](https://tex.z-dn.net/?f=%5Cfrac%7Bhc%7D%7Bd%7D%3D%5Cfrac%7Bhc%7D%7Bd_0%7D%2BK.E%5C%5CK.E%3D%5Cfrac%7Bhc%7D%7Bd%7D-%5Cfrac%7Bhc%7D%7Bd_0%7D%5C%5CK.E%3Dhc%2A%28%5Cfrac%7B1%7D%7Bd%7D-%5Cfrac%7B1%7D%7Bd_0%7D%29%5C%5CK.E%3D6.626%2A10%5E%7B-34%7D%2A3%2A10%5E8%2A%28%5Cfrac%7B1%7D%7B275%2A10%5E%7B-9%7D%7D-%5Cfrac%7B1%7D%7B300%2A10%5E%7B-9%7D%7D%29%5C%5CK.E%3D6.024%2A10%5E%7B-20%7Dkgm%5E2s%5E%7B-2%7D)
Now to find velocity:
![K.E=\frac{1}{2}mv^2\\v^2=1.324*10^{11}\\v=3.64*10^5](https://tex.z-dn.net/?f=K.E%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%5C%5Cv%5E2%3D1.324%2A10%5E%7B11%7D%5C%5Cv%3D3.64%2A10%5E5)
You can determine the mass defect when you calculate for the absolute difference of the experimentally determined molar mass and the theoretical molar mass.
64Ni means there are 28 protons, 28 electrons and (64 - 28) = 36 neutrons. The mass of each proton is 1.0072766 amu; the mass of each electron is <span>0.00054 </span>amu; the mass of each neutron is <span>1.008664 amu.
Theoretical molar mass = 28(</span>1.0072766 amu) + 28(0.00054 amu) + 36(1.008664) = 64.5308 amu
Thus,
Mass defect = |63.9280 amu - 64.5308 amu| = <em>0.6028 amu</em>
Answer:
γ−Hydrogen is easily replacable during bromination reaction in presence of light , because Allylic substitution is being preferred.
Explanation:
that's all
Answer:
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃
Step-by-step explanation:
The unbalanced equation is
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + KNO₃
Notice that the complex groups like NO₃ and CrO₄ stay the same on each side of the equation.
One way to simplify the balancing is to replace them with a single letter.
(a) For example, let <em>X = NO₃</em> and <em>Y =CrO₄</em>. Then, the equation becomes
PbX₂ + K₂Y ⟶ PbY + KX
(b) You need 2X on the right, so put a 2 in front of KX.
PbX₂ + K₂Y ⟶ PbY + 2KX
(c) Everything is balanced. Now, replace X and Y with their original meanings. The balanced equation is
Pb(NO₃)₂ + K₂CrO₄ ⟶ PbCrO₄ + 2KNO₃