Answer: The complement components that is the direct cause of the enhanced vascular permeability and chemoattraction in the abraded skin area in this experiment is C5a. The correct option is C.
Explanation:
When the investigator gently abrades the skin from the flank of a mouse, creating a 1 × 2-cm skin window, the investigator caused a mild tissue damage which leads to a complex series of events resulting in the repair of the damaged tissue. These events are collectively called an INFLAMMATION. Inflammation can therefore be defined as the reaction of vascularized living tissue to local injury.
There are many compounds that triggers specific aspects of inflammation and they are collectively called inflammatory mediators. An example is the Complement component 5a. It has the following functions in mediating inflammatory response:
- enhances vascular permeability
- it has the ability to attract other cells to the site of the abraded skin area (a chemoattractant).
Answer:
The final temperature of the mixture is 44.9°C
Explanation:
Mass of the substance (gold) = 2.575g
Specific heat capacity of gold = 0.129J/g°C
Initial temperature (T1) = 75°C
Final temperature (T2) = ?
Energy lost = 10J
Heat energy(Q) = MC∇T
Q = heat energy (in this case lost)
M = mass of the substance
C = specific heat capacity of the substance
∇T = change in temperature of the substance = (T2 - T1)
Q = MC∇T
Q = MC(T2 - T1)
-10 = 2.575 × 0.129 × (T2 - 75) energy is -ve because it was energy lost.
-10 = 0.3321 × (T2 - 75)
-10 = 0.3321T2 - 24.9075
Collect like terms
0.3321T2 = 24.9075 - 10
0.3321T2 = 14.9075
T2 = 14.9075 / 0.3321
T2 = 44.88
T2 = 44.9°C
The final temperature of the mixture is 44.9°C
Answer:
<h2>18 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
volume = final volume of water - initial volume of water
volume = 62 - 52 = 10 mL
From the question we have
We have the final answer as
<h3>18 g/mL</h3>
Hope this helps you
Answer:
The chemist needs to react 40 g of sulfur with 60 g of oxygen to make 100 g of sulfur trioxide.
Explanation:
2S (s) + 3O₂ (g) → 2SO₃ (g)
64g + 96g → 160 g
32g + 48g → 80 g
x + y → 100 g
1 mol SO₃ ___ 80g
n _______ 100g
n = 1.25 mol SO₃
1 mol S ___ 32 g
1,25 mol S __ 40 g
1 mol O₂ ___ 32 g
1,875 mol O₂ ___ 60 g