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Dafna11 [192]
3 years ago
5

How many sites on this antibody molecule have potential to bind to a non-self molecule?

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
7 0
There are two sites on the antibody molecule that have a potential to bind to a non-self molecule. The Fab of an antibody is the region of the antibody that binds to an antigen. It consist of one constant and one variable domain from each heavy and light chain of the antibody. During immune reaction, an antigen-antibody reaction occurs between the antibodies made by the B cells and the antigens.

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Using the periodic table entry below, match the phrases with their corresponding values.
zhannawk [14.2K]

Atomic mass = 39

number of neutrons = 20

Number of protons= 19

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

The number of electron in first shell = 2

number of electrons in n=3 shell = 8


5 0
3 years ago
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How are compounds created?
Flauer [41]

The answer is that compounds are formed when elements are joined and held together by strong forces called chemical bonds.Covalent bonds share electrons between atoms in order to fill their electron shells. In the compound, molecules are held together by the attraction between the nucleus and the shared electrons. (I got this from the internet so credits to the internet)

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Which option lists the layers of the rainforest in the correct order from top to bottom?
andrey2020 [161]

Answer: B.

Explanation:

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8 0
3 years ago
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Label the liquids from least to most dense, Vegetable oil, water, Dish soap, Corn syrup, Honey
masya89 [10]

Answer:

1. Vegetable oil

2. Water

3. Dish soap

4. Corn syrup

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8 0
3 years ago
If 255 J of heat is applied to 20.0 g of
Arisa [49]

Answer:

ΔT  = 98.84 °C

Explanation:

Given data:

Heat absorbed = 255 J

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Specific heat capacity of gold = 0.129 J/g.°C

Temperature change = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

by putting values,

255 J = 20.0 g × 0.129 J/g.°C × ΔT

255 J  = 2.58 J / °C × ΔT

ΔT  = 255 J  / 2.58 J / °C

ΔT  = 98.84 °C

8 0
3 years ago
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