Answer: "Three Electrons Are Gained"(N⁻³)
Nitrogen has an Incomplete Octet With an Electronic Configuration of
1S²2S²2P³
The P orbital takes a Maximum of 6 electrons.
When this shell is filled... The electronic Configuration of Neon is obtained.
1S²2S²2P⁶
This is also know as the Nitride Ion (N⁻³).
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)
Answer: The number of moles of
produced are, 0.287 moles.
Explanation : Given,
Mass of
= 14.9 g
Molar mass of
= 32 g/mol
First we have to calculate the moles of ![O_2](https://tex.z-dn.net/?f=O_2)
![\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DO_2%7D%7B%5Ctext%7BMolar%20mass%20%7DO_2%7D)
![\text{Moles of }O_2=\frac{14.9g}{32g/mol}=0.466mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B14.9g%7D%7B32g%2Fmol%7D%3D0.466mol)
Now we have to calculate the moles of ![CO_2](https://tex.z-dn.net/?f=CO_2)
The balanced chemical equation is:
![2C_4H_{10}+13O_2\rightarrow 10H_2O+8CO_2](https://tex.z-dn.net/?f=2C_4H_%7B10%7D%2B13O_2%5Crightarrow%2010H_2O%2B8CO_2)
From the reaction, we conclude that
As, 13 mole of
react to give 8 moles of ![CO_2](https://tex.z-dn.net/?f=CO_2)
So, 0.466 mole of
react to give
mole of ![CO_2](https://tex.z-dn.net/?f=CO_2)
Therefore, the number of moles of
produced are, 0.287 moles.
Metals that do not react with acid would be copper, platinum and mercury