According to the fact that t<span>hey both include the reaction change of elements through different processes, there should be a nice connection between them. In math, numbers pose as an element, and in chem - </span><span>elements are physical things. Both of them can change due to reaction or a formula.</span>
Answer:
Gravity pulls all sides equally forming a ball
Answer:
Cell Type I: Animal cell surrounded by a plasma membrane only.
Cell Type II: Plant cell surrounded by a plasma membrane and a cell wall.
Explanation:
Hello,
In this case, we must remember that animal cells are covered only by a plasma membrane, which is not enough to keep the cell from bursting as no limit for the inlet salt is established, causing it both to swell and burst. On the other hand, plant cells are covered by both a plasma membrane and a cell wall which contributive effect allow the inlet of the salt but prevent the cell to burst as the cell wall is rigid. In such a way, based on the described situation, one infers that the cell type I is an animal cell surrounded by a plasma membrane only the cell type II is a plant cell surrounded by a plasma membrane and a cell wall.
Best regards.
Answer:
Option A.
Explanation:
1 mol of anything contains 6.02×10²³ particles.
We know that 1 mol of oxygen gas contains 2 moles of O.
1 mol of oxygen weighs 16 g/mol, the mass for 1 molecule of O.
By the way, the mass for 1 mol of O₂ may be:
Option A → 16 g/mol . 2 mol
32 g
Oyxgen is a dyatomic molecule, that's why we have 2 moles of O.
Another example can be:
1 mol of water (H₂O) contains 2 moles of H and 1 mol of O.
Explanation:
It is given that,
The electron in a hydrogen atom, originally in level n = 8, undergoes a transition to a lower level by emitting a photon of wavelength 3745 nm. It means that,
The amount of energy change during the transition is given by :
![\Delta E=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]](https://tex.z-dn.net/?f=%5CDelta%20E%3DR_H%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7Bn_i%5E2%7D%5D)
And
![\dfrac{hc}{\lambda}=R_H[\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}]](https://tex.z-dn.net/?f=%5Cdfrac%7Bhc%7D%7B%5Clambda%7D%3DR_H%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7Bn_i%5E2%7D%5D)
Plugging all the values we get :
![\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{3745\times 10^{-9}}=2.179\times 10^{-18}[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\\dfrac{5.31\times 10^{-20}}{2.179\times 10^{-18}}=[\dfrac{1}{n_f^2}-\dfrac{1}{8^2}]\\\\0.0243=[\dfrac{1}{n_f^2}-\dfrac{1}{64}]\\\\0.0243+\dfrac{1}{64}=\dfrac{1}{n_f^2}\\\\0.039925=\dfrac{1}{n_f^2}\\\\n_f^2=25\\\\n_f=5](https://tex.z-dn.net/?f=%5Cdfrac%7B6.63%5Ctimes%2010%5E%7B-34%7D%5Ctimes%203%5Ctimes%2010%5E8%7D%7B3745%5Ctimes%2010%5E%7B-9%7D%7D%3D2.179%5Ctimes%2010%5E%7B-18%7D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B8%5E2%7D%5D%5C%5C%5C%5C%5Cdfrac%7B5.31%5Ctimes%2010%5E%7B-20%7D%7D%7B2.179%5Ctimes%2010%5E%7B-18%7D%7D%3D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B8%5E2%7D%5D%5C%5C%5C%5C0.0243%3D%5B%5Cdfrac%7B1%7D%7Bn_f%5E2%7D-%5Cdfrac%7B1%7D%7B64%7D%5D%5C%5C%5C%5C0.0243%2B%5Cdfrac%7B1%7D%7B64%7D%3D%5Cdfrac%7B1%7D%7Bn_f%5E2%7D%5C%5C%5C%5C0.039925%3D%5Cdfrac%7B1%7D%7Bn_f%5E2%7D%5C%5C%5C%5Cn_f%5E2%3D25%5C%5C%5C%5Cn_f%3D5)
So, the final level of the electron is 5.
