Iwill give brainliest to best and most thorough answer! we know that standard pressure is one atmosphere, or 760 millimeters of mercury. this pressure results from the weight of gas molecules in the atmosphere. as a diver enters the water, he is subject to both water pressure and air pressure. because water is much denser than air, the pressure increases rapidly as the diver descends. at the depth of 34 feet in fresh water, the diver is experiencing 2 atmospheres of pressure (one from air pressure and one from the 34 feet of water). for every additional 34 feet the diver descends he will be under an additional atmosphere of pressure. since water pressure is proportional to depth, how many atmospheres of pressure would a diver experience at 102 feet? why wouldn't this pressure squash the diver? answering this second question may be easier if you think of the reason a person on land is not squashed by one atmosphere of pressure. explain your answer in detail.

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

The second and last option are the answers

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is  2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry  problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that     10 dm = 1 m  ⇔ 10³ dm³ = 1 m³then we use the fact that     10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting:     now we have to find the moles of hydrochloric  acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by    molar mass = (mass ca) + (mass c) + 3(mass o)    = 40 + 12 + 3(16)    = 100 g/molconvert the given 5 g to moles:     mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis    mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃)    = 0.05 mol hcl * 2    = 0.1 mol hcl  we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get     volume = moles / concentration    volume = 0.1 mol / (0.002 mol/cm³)    volume = 50 cm³