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2 years ago

What is the difference between B-cell lymphocytes and T-cell lymphocytes?

2 answers:

3 0

There are different types of B cell and T cell. both are lymphocytes, a subclass of white blood cell. the T cells are mainly used in identifying antigens and releasing chemicals which attack macrophages (big immune cells which 'eat' antigens), to destroy the antigen. B cells are used in the antibodies. When they encounter a new antigen, plasma cells and memory cells are antibodies. Formed from the division of a B cell. the memory cell remembers the antigen and which antibody to use, while the plasma cell makes the antibodies to fight a particular antigen or class of antigens.
 Hope this could help you with your question. : )

Thepotemich [5.8K]2 years ago

3 0

Answer:

Explanation:

Lymphocytes are a class of white blood cell in the body's immune system. B-cell and T-cell lymphocytes are types of lymphocytes.

There are differences between the T-cell lymphocytes and the B-cell lymphocytes. The major differences are

T-cell lymphocytes are produced in the thymus while B-cell lymphocytes are produced in the bone marrow
T-cell lymphocytes are involved in cell-mediated immunity while B-cell lymphocytes are involved in humoral immunity

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What role do guanacos play in the Andes Mountains ecosystem?.

Nikitich [7]

Grasses and plants are producers, guanacos are herbivores and are primary consumers and pumas are predators and secondary consumers

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2 years ago

Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec

Bess [88]

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0$

Let energy change be E for 1 atom.

$E=E_{\infty}-E_1=0-(-13.6 eV)=13.6 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol$

$E'=1,312.17 kJ/mol$

The energy required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ , $B^{4+}$ atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-340eV)=340 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 340eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol$

$E'=32,804.31 kJ/mol$

The energy required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ , $Li^{2+}$ atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol$

$E'=11,809.55 kJ/mol$

The energy required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ , $Mn^{24+}$ atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

$E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV$

Energy of n = ∞ in an hydrogen like atom:

$E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0$

Let energy change be E.

$E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV$

$1 mole = 6.022\times 10^{-23}$

Energy for 1 mole = E'

$E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV$

$1 eV=1.60218\times 10^{-22} kJ$

$E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol$

$E'=820,107.88 kJ/mol$

The energy required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0

3 years ago

Akosua has observed that anytime she drinks sobolo, her stool appears dark green. On one occasion

Anna007 [38]

Answer:

i. Sobolo is a Ghanian drink that is produced from red hibiscus flower that has an average pH of 6.7

It contains cyanidin and anthocyanins, which is a red pigment that is red in an acidic medium and changes green when introduced in a basic medium that has a high pH

The pH at the rectum of the digestive system = 5 to 8 (Slightly basic)

Therefore, what made the stool of Akosua green is that the sobolo drink changes to green in basic solution

ii. The stool which appeared green because she took sobolo turn into bright red upon mixing with the acidic WC water because of the presence of anthocyanins in sobolo, it turns red in an acidic medium

iii. Sobolo which turns green, or blue in a basic medium and red in an acidic medium can be used as a litmus solution to test the pH of a given substance

Explanation:

Sobolo or soobolo in Ghana is a name for the Hibiscus tea or tisane, which is made from calyces of the hibiscus plant, and has a sour (tangy) taste and appears bright red in color

8 0

3 years ago

I WILL GIVE BRAINIEST AND MORE !!

icang [17]

Composting is a technique that allows biodegradable materials of urban, domestic and industrial origin to be degraded to form humus, a type of soil that is very fertile and extremely rich in micronutrients.
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Hope this helps :)

3 0

2 years ago

How much heat is gained when a 50.32g piece of aluminum is heated from 9.0°c to 16°c

Rashid [163]

Answer: 317 joules

Explanation:

The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

In this case,

Q = ?

Mass of aluminium = 50.32g

C = 0.90J/g°C

Φ = (Final temperature - Initial temperature)

= 16°C - 9°C = 7°C

Then, Q = MCΦ

Q = 50.32g x 0.90J/g°C x 7°C

Q = 317 joules

Thus, 317 joules of heat is gained.

5 0

3 years ago

Read 2 more answers