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Shkiper50 [21]
3 years ago
10

Cytotoxic t cells can attack target cells with which chemical weapons?

Chemistry
1 answer:
Aneli [31]3 years ago
4 0

Answer:

secrete cytotoxic substance which triggers apoptosis of target cell.

Explanation:

Cytotoxic T cells have cell surface receptor which recognizes the antigen present on the receptor of target cell. This interaction initiates the process of killing of target cell.

After interaction cytotoxic t cell release cytotoxic substance called granzyme and perforin. Granzyme triggers apoptosis through the activation of caspases or by making the release of cytochrome c and activation of the apoptosome.

Perforin make pores in the cell and its action is similar to complement membrane attack complex. Therefore cytotoxic substances are released by Tc cells which trigger apoptosis of target cell.

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Convert<br> 12 x 1025 kg/mL to fg/cm3
Anna35 [415]

Answer:

1.2x10^{38}\frac{fg}{cm^3}

Explanation:

Hello there!

In this case, for such unit conversion we need to realize that 1 kg is equal to 1000 g, 1 g is equal to 1x10⁹ and 1 mL equals 1 cm³, therefore we apply:

12x10^{25}kg*\frac{1000g}{1kg} *\frac{1x10^9fg}{1g} *\frac{1mL}{1cm^3} \\\\1.2x10^{38}\frac{fg}{cm^3}

Best regards!

5 0
4 years ago
Use the data provided to calculate benzaldehyde's heat of vaporization smartwork5
QveST [7]
This problem is incomplete. Luckily, I found a similar problem from another website shown in the attached picture. The data given can be made to use through the Clausius-Clapeyron equation:

ln(P₂/P₁) = (-ΔHvap/R)(1/T₂ - 1/T₁)

where
P₁ = 14 Torr * 101325 Pa/760 torr = 1866.51 Pa
T₁ = 345 K
P₂ = 567 Torr * 101325 Pa/760 torr = 75593.78 Pa
T₂ = 441 K

ln(75593.78 Pa/1866.51 Pa) = (-ΔHvap/8.314 J/mol·K)(1/441 K - 1/345 K)
Solving for ΔHvap,
<em>ΔHvap = 48769.82 Pa/mol or 48.77 kPa/mol</em>

6 0
3 years ago
Consider the fructose-1,6-bisphosphatase reaction. Calculate the free energy change if the ratio of the concentrations of the pr
enot [183]

Answer:

ΔG = -8.812 kJ/mol

Explanation:

To obtain the free energy of a reaction you can use the expression:

ΔG = ΔG° + RT ln Q

<em>Where: </em>

<em>ΔG° is Standard Gibbs Free energy: -16.7kJ/mol = -16700J/mol</em>

<em>R is gas constant: 8.314472 J/molK</em>

<em>T is absolute temperature (37°C + 273.15 = 310.15K)</em>

<em>And Q is reaction quotient: 21.3</em>

<em />

Replacing in the formula:

ΔG = ΔG° + RT ln Q

ΔG = -16700J/mol + 8.314472J/molK*310.15K ln 21.3

ΔG = -8812.4J/mol

<h3>ΔG = -8.812 kJ/mol</h3>

<em />

7 0
4 years ago
Why do scorpions glow on a black light?
Lana71 [14]

Answer:

All scorpions fluoresce under ultraviolet light, such as an electric black light or natural moonlight. The blue-green glow comes from a substance found in the hyaline layer, a very thin but super tough coating in a part of the scorpion's exoskeleton called the cuticle

Explanation:

6 0
3 years ago
Read 2 more answers
How many liters of 1.0 M HCl do you need to neutralize 2.0 L of 3.0 M NaOH? How many liters of 1.0 M HCl do you need to neutrali
Kruka [31]

<u>Answer: </u>

<u>For 1:</u> The volume of HCl required is 6 L.

<u>For 2:</u> The volume of HCl required is 9 L.

<u>For 3:</u> The volume of sulfuric acid required is 4.5 L.

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2        ......(1)

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base

  • <u>For 1:</u>

We are given:

n_1=1\\M_1=1M\\V_1=?L\\n_2=1\\M_2=3.0M\\V_2=2.0L

Putting values in equation 1, we get:

1\times 1\times V_1=1\times 3\times 2\\\\V_1=6L

Hence, the volume of HCl required is 6 L.

  • <u>For 2:</u>

We are given:

n_1=1\\M_1=1M\\V_1=?L\\n_2=2\\M_2=3.0M\\V_2=1.5L

Putting values in equation 1, we get:

1\times 1\times V_1=2\times 3.0\times 1.5\\\\V_1=9L

Hence, the volume of HCl required is 9 L.

  • <u>For 3:</u>

To calculate the volume of acid, we use the equation:

N_1V_1=N_2V_2

where,

N_1\text{ and }V_1 are the normality and volume of acid

N_2\text{ and }V_2 are the normality and volume of base

We are given:

N_1=1.0N\\V_1=?L\\N_2=3.0N\\V_2=1.5L

Putting values in above equation, we get:

1.0\times V_1=3.0\times 1.5\\\\V_1=4.5L

Hence, the volume of sulfuric acid required is 4.5 L.

8 0
3 years ago
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