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andrey2020 [161]
3 years ago
6

Oh no! Igor stepped on a piece of glass on the sidewalk in front of

Chemistry
1 answer:
Helen [10]3 years ago
8 0

Answer: The correct sequence for the series of event would be.

1. Igor's toe is being cut by the glass.

2. The wound surrounding the injury becomes infected  with bacteria from Igor's foot.

3. Antibodies and circulating white blood cells stick to the  bacteria creating a large complex in the lymph vessel.

4. The bacteria enter his lymph system and travel towards a lymph node.

5. The complex becomes trapped in a lymph node and is  engulfed by a phagocyte.

6. The bacteria is destroyed.

Explanation:

Whenever there is any cut or wound in the body and body encounters invasion of the foreign materials it considers it as harmful pathogen.

These pathogens when enters the body it is considered as antigen, it then travels to the lymphatic system.

These bacterial complex is then killed by the phagocytes and digested by the body.

In this way the complex is killed and the bacteria is destroyed.

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Write at least three characteristics of acids and three characteristics of bases. These characteristics should be what makes the
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1 km is equal to<br><br> 100 m<br> 10,000 dm<br> 1,000 mm<br> 1,000 cm
vovikov84 [41]
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3 0
3 years ago
A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
Igoryamba

Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

          V_{2} = 8.34 L,    V_{1} = 2.67 L

Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

    = -2818.68 J

Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

                    = -5.67 atm L

As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

            = -574.54 J

Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

3 0
3 years ago
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