<span>At 100 feet, the diver is under about 4 atmospheres pressure. If she is free diving, her lungs will be compressed to about 1/4 their size on the surface (with some movement of the major abdominal organs). If she is scuba diving, the air which she is breathing is also at 4 atmospheres and there is no problem. (The non-gas spaces in the body are not-compressible and are unaffected.) The only problems she has to concern herself with are the beginnings to nitrogen narcosis and the nitrogen which is dissolving (Henry's law) into her body tissues. On the way up, she also has to remember that the air in her lungs will expand by a factor of 4 and she better exhale! Hope this helps you</span>
Answer:
.✓is related to the solute content
✓gives information about potential changes in cell volume when cells are placed in that solution
√is related to membrane permeability to solutes.
Explanation:
Tonicity of a solution can be explained as how an extracellular solution can give room for the liquid to move in and out of the cell through osmosis.
It should be noted that Tonicity of a solution is
.✓is related to the solute content
✓gives information about potential
changes in cell volume when cells are placed in that solution
√is related to membrane permeability to solutes.
Answer:
The human heart is an organ that pumps blood throughout the body via the circulatory system, supplying oxygen and nutrients to the tissues and removing carbon dioxide and other wastes.
Explanation:
Hope this helped! Good luck on whatever it is you're working on!
Diamond and graphite are made of carbon. So is most of charcoal.
Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>
