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1 year ago

Superficial lymph nodes include all of the following, except __________.

SAT

1 answer:

Mamont248 [21]1 year ago

6 0

All of the following—aside from iliac lymph nodes—are considered superficial lymph nodes.

<h3>What is Lymph Nodes?</h3>

Usually shorter than 2.5 cm, lymph nodes are tiny bean-shaped organs. They are widely dispersed throughout the body along the lymphatic channels, where they filter the lymph before it returns to the blood.
Within the central nervous system, lymph nodes are absent. On each side of the body, there are three areas where the lymph nodes are more likely to collect.
These locations are the inguinal nodes in the groin, axillary nodes in the armpit, and cervical nodes in the neck.
A connective tissue capsule encloses the normal lymph node, which is divided into lymph nodules by this capsule. Lymph sinuses serve as the gaps that divide the dense aggregates of lymphocytes and macrophages that make up lymph nodules. The node's afferent lymphatics, which convey lymph there from the periphery and enter the node on its convex side, are at various points along its perimeter. After passing through the lymph sinuses, the lymph enters an efferent lymphatic channel, which exits the lymph node by way of the hilum, an indentation in the lymph node's surface.

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(i) The order of decomposition reaction of cyclobutane is first order.

(ii) It requires 53.15 milliseconds for 99% decomposition of cyclobutane.

(d) Rate law = k [\rm Cl_2Cl

, as chlorine is the reactant.

(e) Thus student claim is false as Chlorine gas is the reactant, not a catalyst.

(i) The order of reaction can be determined with the help of a graph given below. The graph depicts the reduction in the concentration of cyclobutane with time depicting decomposition.

The graph depicts the reaction to be of first-order kinetics. The general concentration expression can be:

\rm C_(_t_)\;=\;C_(_0_)\;e^-^k^t

Where, \rm C_(_t_) = concentration at time t

\rm C_(_0_) = initial concentration 1.60 mol/liter

k = rate constant

for first order reaction,

k = In \rm \frac{2}{Half\;time}

Halftime

From the graph, half time = 0.008 sec i.e. time when the concentration reduced to 50%.

k = In \rm \frac{2}{0.008}

0.008

k = 86.64 / sec.

(ii) The time required for 99% decomposition can be calculated, by calculating the \rm C_(_t_) for left 1 % solution.

0.01 \rm C_(_0_) = \rm e^-^k^t

-kt = -4.605

t = 53.15 miliseconds

(d) The rate law is obtained against the reaction.

In rate law, the intermediates that appear in the reaction are removed, and the equation has been formed.

The slow step is the formation of Chlorine. Thus the rate law will be:

Rate = k [\rm Cl_2Cl

(e) The catalyst function in improvising the rate of the reaction. Since, the reaction has chlorine as the part of the reactant, it is consumed in the reaction and thus does not act as a catalyst. Thus the claim of the student is false.

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