Gin is uniform throughout and is a homogenous mixture. If it wasn't you would have awful lumps in your drink :). Hope I helped!
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.
Answer:
Forgive me if im wrong but i am pretty sure the answer is C
Explanation:
Answer:
1.01atm is the pressure of the gas
Explanation:
The difference in heights in the two sides is because of the difference in pressure of the enclosed gas and the atmospheric pressure. This difference is in mm of the nonvolatile liquid. The difference in mm Hg is:
32.3mm * (0.993g/mL / 13.6g/mL) = 2.36mmHg
As atmospheric pressure is 765mm Hg and assuming the gas has more pressure than the atmospheric pressure (There is no illustration), the pressure of the gas is:
765mm Hg + 2.36mm Hg = 767.36 mmHg
In atm:
767.36 mmHg * (1atm / 760 mmHg) =
1.01atm is the pressure of the gas
Answer:
520 kg
Explanation:
Let's consider the combustion of isooctane.
C₈H₁₈(l) + 12.5 O₂(g) → 8 CO₂(g) + 9 H₂O(l)
We can establish the following relations:
- 1 mL of C₈H₁₈ has a mass of 0.690 g (ρ = 0.690 g/mL).
- The molar mass of C₈H₁₈ is 114.22 g/mol.
- The molar ratio of C₈H₁₈ to O₂ is 1:12.5.
- The mole fraction of O₂ in air is 0.21.
- The molar mass of air is 28.96 g/mol.
50 L of isooctane require the following mass of air.
