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fenix001 [56]
3 years ago
15

Claire and Jamie are both patients at a clinic. Claire has strep throat, and Jamie has influenza. Which of the following stateme

nts is true? B cells produce different types of antibodies in each patient. T cells produce the same type of antibodies in both patients. B cells activate different types of T cells in each patient. T cells turn off to hide from pathogens in both patients.
Chemistry
2 answers:
Natali5045456 [20]3 years ago
5 0

Answer:

Option). B cells produce different types of antibodies in each patient.

Explanation:

B cells are cells of adaptive immune system that provide specific responses against pathogens. B cells have BCR (B-cell receptors0 on their surface, which are specific for different types of pathogens. So, B cells recognize each particular antigen and make specific antibodies to neutralize those pathogens.

As strep throat is generally caused by bacterial infections, Claire must has infected by bacteria. On the other hand, Jamie is infected by influenza, which represents a virus.

Hence, their B cells will produce different types of antibodies in them for different pathogens.

Thus, the correct answer is first option.

azamat3 years ago
3 0
<span>B cells produce different types of antibodies in each patient.</span>
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Answer:10.0 mL of 0.00500 M phosphoric acid

Explanation:

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Fynjy0 [20]
Q1)
the reaction that takes place is 
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balanced chemical equation for the reaction is as follows
Pb(NO₃)₂ + 2KI ----> PbI₂  + 2KNO₃

Q2)
mass of lead nitrate present - 0.600 g 
number of moles = mass present / molar mass 
number of moles - 0.600 g / 331.2 g/mol = 0.00181 mol 

Q3)
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number of moles = mass present / molar mass
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Q4)
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Q5)
next we have to calculate the number of moles of PbI₂ formed based on the amount of KI moles present , assuming all the moles of KI were used up in the reaction 
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Q6)
limting reactant is the reactant that is fully consumed during the reaction. the amount of product formed depends on the amount of limiting reactant present

if lead nitrate is the limiting reactant 
if 1 mol of Pb(NO₃)₂ reacts with 2 mol of KI 
then 0.00181 mol of Pb(NO₃)₂ reacts with - 2 x 0.00181 mol of KI = 0.00362 mol 
but 0.00512 mol of KI is present and only 0.00362 mol are required 
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Pb(NO₃)₂ is the limiting reactant 

Q7)
then the amount of PbI₂ formed depends on amount of Pb(NO₃)₂ present 
therefore number of moles of PbI₂ formed is based on number of Pb(NO₃)₂ moles present 
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number of PbI₂ moles formed - 0.00181 mol 
mass of PbI₂ formed - 461 g/mol x 0.00181 mol = 0.834 g
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Q8) 
actual yield obtained  is not always equal to the theoretical yield . therefore we have to find the percent yield. This tells us the percentage of the theoretical yield that is actually obtained after the experiment
percent yield = actual yield / theoretical yield x 100 %
percent yield = 0.475 g / 0.834 g x 100 % = 57.0 %
percent yield of lead iodide is 57.0 %
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