Answer:
The energy that flows into the Earth system is equal to the energy that flows out of the Earth system.
Explanation:
Because I geussed it right
Answer:
(1) Chloroplast
Explanation:
Cells of living organisms are made up of certain function-specific structures called ORGANELLES. Some organelles are present in plant cells and absent in animal cells and vice versa. In a plant cell, one notable organelle that allows it perform the photosynthetic process is the CHLOROPLAST.
However, the chloroplast is predominantly found in the LEAF part of a plant. This is because leaf cells are the site of photosynthesis. Hence, according to this question, Joe would be able to tell whether the plant cell was from the leaf or the root by looking for CHLOROPLAST as a differentiating factor in each cell.
Answer:
Density independent factor
Explanation:
in dense areas where people livein very close and tight spaces
Answer:
Option C is correct = fluorine
Explanation:
An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.
All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other. For example if neutral atom has 6 protons than it must have 6 electrons. The sum of neutrons and protons is the mass number of an atom while the number of protons are number of electrons is the atomic number of an atom.
This diagram show that there are 9 electrons and 9 protons are present. We know that the number of protons or electrons are atomic number of that element. From periodic table we know that atomic number of fluorine is 9 thus the given atom present fluorine atom.
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>