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Setler [38]
2 years ago
10

. Each antibody is designed to fight off how many pathogens

Chemistry
1 answer:
Alex_Xolod [135]2 years ago
4 0
Antibodies can destroy pathogens by (i) binding to and blocking the pathogen's receptors, thus causing neutralization of the pathogen, (ii) binding to the pathogen and activating complement, and (iii) binding to the pathogen and facilitating its opsonization and uptake by macrophages, which utilize their Fc receptors ...
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Which of these elements is the most reactive?<br> potassium<br> calcium<br> titanium<br> scandium
vladimir1956 [14]

Answer:

Scandium is the most reactive

3 0
3 years ago
Please help
Colt1911 [192]
A i belive is the correct answer

3 0
2 years ago
Can someone help me please?
morpeh [17]

Answer:

metal ball that moves at high speed

Explanation:

4 0
3 years ago
Bromine reacts with nitric oxide to form nitrosyl bromide as shown in this reaction: br2(g) + 2 no(g) → 2 nobr(g) a possible mec
Svetach [21]

The overall reaction is given by:

Br_{2}(g) + 2 NO(g) \rightarrow  2 NOBr(g)

The fast step reaction is given as:

NO(g) + Br_{2}(g) \rightleftharpoons NOBr_{2}(g) (fast; k_{eq}= \frac{k_{1}}{k_{-1}})

The slow step reaction is given as:

NOBr_{2}(g) + NO(g) \rightarrow  2 NOBr(g) (slow step k_{2})

Now, the expression for the rate of reaction of fast reaction is:

r_{1}=k_{1}[NO][Br_{2}]-k_{-1}[NOBr_{2}]

The expression for the rate of reaction of slow reaction is:

r_{2}=k_{2}[NOBr_{2}] [NO]

Slow step is the rate determining step. Thus, the overall rate of formation is the rate of formation of slow reaction as [NOBr_{2}] takes place in this reaction.

The expression of rate of formation is:

\frac{d(NOBr)}{dt}=r_{2}

= k_{2}[NOBr_{2}][NO]    (1)

Now, consider that the fast step is always is in equilibrium. Therefore, r_{1}=0

k_{1}[NO][Br_{2}]= k_{-1}[NOBr_{2}]

[NOBr_{2}] = \frac{k_{1}}{k_{-1}}[NO][Br_{2}]

Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









4 0
3 years ago
How many L of a 7.5 H2SO4 stock solution would you need to prepare a dilute solution 100 L of 0.25M H2SO4
blsea [12.9K]

Answer:

3.33 L

Explanation:

We can solve this problem by using the equation:

  • C₁V₁=C₂V₂

Where the subscript 1 refers to one solution and subscript 2 to the another solution, meaning that in this case:

  • C₁ = 0.25 M
  • V₁ = 100 L
  • C₂ = 7.5 M
  • V₂ = ?

We input the data:

  • 0.25 M * 100 L = 7.5 M * V₂
  • V₂ = 3.33 L

Thus the answer is 3.33 liters.

3 0
2 years ago
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