Answer:
Main job of golgi bodies is to sort and package proteins and other substances in a plant cell.
Explanation:
Golgi bodies are also called post office of the cell because it modify and distribute proteins for the cell. First, proteins are made in the organelle of the cell i. e. endoplasmic reticulum. From here, it is send to the Golgi apparatus for modification. Golgi bodies add some special structures with the protein and this protein leaves golgi bodies which is used by the cell where it is needed.
Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.
The ruling separation is
d=1 / (470mm −1) = 2.1×10⁻³ mm
Diffraction lines occur at angles θ such that dsinθ=mλ, where λ is the wavelength and m is an integer.
Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.
We take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.
That is, find the greatest integer value of m for which mλ<d.
since d / λ = 538×10⁻⁹m / 2.1×10 −6 m ≈ 3
that value is m=3.
There are three complete orders on each side of the m=0 order.
The second and third orders overlap.
Learn more about diffraction here : brainly.com/question/16749356
#SPJ4
Isoflavones, quercetin, and anthocyanin are all Flavonoids.
Answer:
V ∝ abc
Explanation:
This task is a joint variation task involving only direct proportionality:
Direct variation is one in which two variables are in direct proportionality to each other. This means that as one increases, the other variable also increases and vice - versa.
Joint variation is one in which one variable is dependent on two or more variables and varies directly as each of them.
In this exercise:
If a ∝ b and a ∝ c, then a ∝ bc
Taking the above three proportionalities,
V ∝ a ∝ b ∝ c
V ∝ a ∝ bc
V ∝ abc
Reduction half reaction: Cu²⁺(aq) + 2e⁻ → Cu⁰(s).
Oxidation half reaction: NO₂⁻(aq) + H₂O(l) → NO₃⁻(aq) + 2H⁺(aq) + 2e⁻.
Balanced chemical reaction:
Cu²⁺(aq) + NO₂⁻(aq) + H₂O(l) → Cu(s) + NO₃⁻(aq) + 2H⁺(aq).
Copper is reduced from oxidation number +2 (Cu²⁺) to oxidation number 0 (Cu) and nitrogen is oxidized from oxidation number +3 (in NO₂⁻) to oxidation number +5 (in NO₃⁻).
