I have to figure this out for chemistry, and we took all the measurements in classThe density of alumnium foil: 2.70 (of Al)Mass: .88gLength: 15.2Height: 14.5Width: xSo.2.7.88/(14.5)(15.2)(x) 2.7(14.5)(15.2)(x).88595.08x.88.001 CM^3 is the width.but then with that width how do i figure out the thickness in atoms? Someone said use angstroms but our teacher never mentioned those so i'm a little skeptical about how exactly to use them to find the thickness in ATOMs of the foil.I'm not looking for an ANSWER, i'm looking for the process so I can do it myself.thanks
You need to have a moisture barrier in that wallIt's a bad thing to have moisture build up on your interior walls, regardless of the temperature.
you can use stryofoam board, put the r value is not there you can left the sidiung on the out side walland drill holes in the joustso you can blow the inulation inthen lock the siding in and you well be done it went cost you that mucha nother way is to put a vent in the door to get warm air in the room.
Cold is cold .you have a closet on an exterior wall.heat can't get there to warm it or keep frost from buildingThe cheapest thing you can do is to keep the door open during the extreme cold or to a put louvered door on.
SureJust get the pink stuff from menards or home depotIt is denser and less likely to break apartComes in different thickness's tooWhen you open the wall, make sure there is not an open top or bottom in between the studsOlder houses didn't close these spaces and lots of cold air can come in.
Assuming that the aluminum atom is a sphere DenistyMass/Volume d3M/(4pir^3) r^33M/(d4pi) M Mass of one Aluminum atom Molar mass of Aluminum /Avogadro's number 26.9815/(6.0221410^23) pi22/7 Once you get r you can obtain the atom's length2r You've x0.001 cm So, the number of atoms representing the tickness of the aluminum foil0.001/2r Do the math yourself (As you wish)
