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4 years ago

What would happen if IgE were produced against a self molecule?

Chemistry

2 answers:

Lisa [10]4 years ago

8 0

Answer: Option A

Explanation:

If the IgE were produced against a self disorder then it will be considered as autoimmune disorders in which the body's own molecules or substances are considered as antigen.

The immune system of the body starts producing the antibodies against these antigens.This can be considered as hypersensitive reaction which would lead to anaphylatic shocks.

Lady_Fox [76]4 years ago

6 0

If IgE were produced against a self molecule:
The hypersensitive reaction would lead to anaphylactic shock.

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Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank

marysya [2.9K]

The rate of Formation of Carbocation mainly depends on two factors'

1) Stability of Carbocation:
The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

2) Ease of detaching of Leaving Group:
The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

R-I > R-Cl > R-F

B > C > A

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3 years ago

Wha is boiling point

Troyanec [42]

The boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure surrounding the liquid. Therefore, the boiling point of a liquid depends on atmospheric pressure.

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3 years ago

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

Phoenix [80]

Answer:

pH = 8.0

Explanation:

First, we have to calculate the moles of NaOH.

$35.8 \times 10^{-3}L.\frac{0.020mol}{L} =7.2\times 10^{-4}mol$

Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

$\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M$

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb) where Cb is the initial concentration of the base

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

pH + pOH = 14

pH = 14 - pOH = 14 - 6.0 = 8.0

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3 years ago

What role does sea grass play in this food chain?

rjkz [21]

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3 years ago

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abruzzese [7]

Answer:

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3 years ago

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