Answer:
Left amount of Polonium-210 after 115 days will be 4.5927 microgram
Explanation:
We have given initial mass of Polonium-210 
Decay rate x= 0.502% = 0.00502
Time t = 115 days
We have to find the left amount of Polonium -210 after 115 days
We know that left amount is given by

So left amount of Polonium-210 after 115 days will be 4.5927 microgram
Answer:
They consist mainly of iron-nickel metal with small amounts of sulphide and carbide minerals. During the decay of radioactive elements in the early history of the solar system, many asteroids melted and the iron they contained, being dense, sank to the centre to form a metallic core.
Explanation:
Hope this helps luv
Answer:
Explanation:
From the information given:
The cell potential on mars E = + 100 mV
By using Goldman's equation:
![E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}} \Big )](https://tex.z-dn.net/?f=E_m%20%3D%20%5Cdfrac%7BRT%7D%7BzF%7DIn%20%5CBig%20%28%5Cdfrac%7BP_K%5BK%5E%2B%5D_%7Bout%7D%2BP_%7BNa%7D%5BNa%5E%2B%5D_%7Bout%7D%2BP_%7BCl%7D%5BCl%5E-%5D_%7Bout%7D%20%7D%7BP_K%5BK%5E%2B%5D_%7Bin%7D%2BP_%7BNa%7D%5BNa%5E%2B%5D_%7Bin%7D%2B%20P_%7BCl%7D%5BCl%5E-%5D_%7Bin%7D%7D%20%20%20%20%20%20%5CBig%20%29)
Let's take a look at the impermeable cell with respect to two species;
and the two species be Na⁺ and Cl⁻
![E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}](https://tex.z-dn.net/?f=E_m%20%3D%20%5Cdfrac%7BRT%7D%7BzF%7D%20In%20%5Cdfrac%7B%5BK%5E%2B%5D_%7Bout%7D%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D)
where;
z = ionic charge on the species = + 1
F = faraday constant
∴
![100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}} \Big)](https://tex.z-dn.net/?f=100%20%5Ctimes%2010%5E%7B-3%7D%20%3D%20%5CBig%20%28%5Cdfrac%7B8.314%20%5Ctimes%20298%7D%7B1%5Ctimes%2096485%7D%20%5CBig%29%20%5Cmathtt%7BIn%7D%20%20%5CBig%20%28%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}} \Big)](https://tex.z-dn.net/?f=100%20%5Ctimes%2010%5E%7B-3%7D%20%3D%200.0257%20%5CBig%20%28%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}} \Big)](https://tex.z-dn.net/?f=3.981%3D%20%5Cmathtt%7BIn%7D%20%5CBig%20%28%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\ 53.57 = \dfrac{4}{[K^+]_{in}}](https://tex.z-dn.net/?f=exp%20%28%203.981%29%20%3D%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D%20%5C%5C%20%5C%5C%20%2053.57%20%3D%20%5Cdfrac%7B4%7D%7B%5BK%5E%2B%5D_%7Bin%7D%7D)
![[K^+]_{in} = \dfrac{4}{53.57}](https://tex.z-dn.net/?f=%5BK%5E%2B%5D_%7Bin%7D%20%3D%20%5Cdfrac%7B4%7D%7B53.57%7D)
![[K^+]_{in} = 0.0476](https://tex.z-dn.net/?f=%5BK%5E%2B%5D_%7Bin%7D%20%20%3D%200.0476)
For [Cl⁻]:
![100 \times 10^{-3} = -0.0257 \ \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}} \Big)](https://tex.z-dn.net/?f=100%20%5Ctimes%2010%5E%7B-3%7D%20%3D%20-0.0257%20%5C%20%20%5Cmathtt%7BIn%7D%20%5CBig%20%28%20%5Cdfrac%7B120%7D%7B%5BCl%5E-%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![-3.981 = \ \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}} \Big)](https://tex.z-dn.net/?f=-3.981%20%3D%20%20%5C%20%20%5Cmathtt%7BIn%7D%20%5CBig%20%28%20%5Cdfrac%7B120%7D%7B%5BCl%5E-%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![0.01867 = \dfrac{120}{[Cl^-]_{in}}](https://tex.z-dn.net/?f=0.01867%20%3D%20%20%5Cdfrac%7B120%7D%7B%5BCl%5E-%5D_%7Bin%7D%7D)
![[Cl^-]_{in} = \dfrac{120}{0.01867}](https://tex.z-dn.net/?f=%5BCl%5E-%5D_%7Bin%7D%20%3D%20%5Cdfrac%7B120%7D%7B0.01867%7D)
![[Cl^-]_{in} =6427.4](https://tex.z-dn.net/?f=%5BCl%5E-%5D_%7Bin%7D%20%3D6427.4)
For [Na⁺]:
![100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}} \Big)](https://tex.z-dn.net/?f=100%20%5Ctimes%2010%5E%7B-3%7D%20%3D%200.0257%20%5CBig%20%28%20%5Cdfrac%7B145%7D%7B%5BNa%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![53.57= \Big ( \dfrac{145}{[Na^+]_{in}} \Big)](https://tex.z-dn.net/?f=53.57%3D%20%5CBig%20%28%20%5Cdfrac%7B145%7D%7B%5BNa%5E%2B%5D_%7Bin%7D%7D%20%20%20%5CBig%29)
![[Na^+]_{in}= 2.70](https://tex.z-dn.net/?f=%5BNa%5E%2B%5D_%7Bin%7D%3D%202.70)
Temperature inversion or thermal inversion is a phenomenon during which the temperature increase when there is an increase in the altitude. It is an uncommon process that happens in the troposphere-the layer of atmosphere closest to the earth.
When there is an increase in temperature with altitude that warm air acts as a blockage for the cool air near the surface of the earth. So this is just opposite to the normal behavior of the air, where it becomes cooler with an increase in altitude.
So how air circulation behaves during inversion?
So as we said that the cool air nearer to the earth will be blocked by a layer of warm air. This blockage will hinder the circulation of air upwards from the lower layers. Let’s see how:
• As we know that, warm air is less dense and more buyout than cool air and normally less dense warm air is close to the earth and cool air is in above layers. But when the process is reversed, this warm air acts like a blanket for the proper circulation of air. Due to this, many environmental pollutants which normally leave the earth during air circulation get trapped in our air.
• The trapped environmental pollutants like nitrogen oxides and hydrocarbons get convert into harmful ozone that ultimately deteriorates the quality of air and causes problems like smog.
• The improper air circulation also causes suspension of the harmful gases near the earth instead of allowing them to leave the earth atmosphere. This ultimately causes air pollution.
• When smoke will not rise, carbon dioxide will be build up in the atmosphere; this will cause greenhouse effect and global warming.
Therefore, air circulation behaves very different during temperature inversion process and poses hazardous effects on global climate.
Hope it helps! :)
Answer:
Answer D
Explanation:
improved water habitats for human recreation