the production cost is proportional to the area of the aluminium sheet used.the volume that each tin can hold is 1000cm3(1litter)1)Determine the value of h, r and hence calculate the ratio of h/r when the total surface area of each tin is minimumhere h cm denotes the height and r cm the radius of the tin.2)the top and bottom pieces of the tin of height h cm are cut from square-shaped aluminium sheetsdetermine the value for r, h and hence calculate the ratio h/r so that the total area of the aluminium sheets used for making the tin is minimum.(refer to the diagram below)3)investigate cause where the top and buttom surface are cut from 1equllateral triangle 2.regular hexagon find the ratio of h/r for each case
Non-metals can only have 8 electrons, which is called an octet, when it receives the number of electrons it needsFor example, Chlorine needs only one more electron, since it already has 7When it has received one from a metal, such as Sodium, then the octet will be completeHowever, it becomes a compound with Sodium - called NaClMetals can only have 8 electrons when it gives up the extra electrons it hasSodium (Na) in the given example above is the metal, and the extra electron it has was given to ChlorineBecause of this, both elements have 8 electrons eachBoth are also bonded because of the giving of electronsTo find the number of electrons an element needs / has to give, you can check their group numberElements on the left, known as the metals, always give up their electronsOn the other hand, the elements on the right, the non-metals, accept the electrons the metals have to give up.
No, it depends on the non-metal and how negatively it is charged, and valence electrons are the number of electrons in the outer most shell, not the total numberFor the most part everything but Hydrogen will always have some electronsChlorine gaining one and Aluminum losing three are attempts to get eight electrons in their outer most shell because that is the most stable confirmationThey are trying to be like the noble gases (Helium, Neon, Argon etc.) Thats a really simplified version and there is much more on this topicAlso remember that the A columns in the periodic table tell you how many valence electrons all the atoms in that column have, IA have 1 IIA have 2 and so on.
1) A can with constant volume could be tall and narrow, or short and wide, and both of these would have a lot of areaThe one with the least area is in between these two, and has a nearly square profile; meaning h ~ 2r (but not exactly it turns out)You can imagine that if you plotted area vsh or r, the curve would start high, and as h or r increased, it would slope down to a minimum area, then go back up againYou know you're at the minimum area when the slope of that curve is zeroAssume the volume is constant 1000 cm^3 (for consistancy use units of cm) Volume of a cylinder area of the circular base times the height V pi r^2 h 1000 cm^3 Area area of top + bottom + side A 2 (pi r^2) + 2 pi r h You have 2 equations (for A and V), and 2 unknowns (r and h)solve this in the usual way, by solving one equation for one variable (the easiest one) and substituting into the other equationUsing: V pi r^2 h 1000 cm^3 h 1000 / (pi r^2) Now substitute this eqn for h into the eqn for A, and you get: A 2 pi r^2 + 2000/r now take the derivative w.r.tr dA / dr 4 pi r - 2000 / r^2 set this equal to zero, and solve for rr cube root of (500/pi) ~ 5.419 solve for h 10.8 check that h ~ 2r h/r ~ 1/2 2 3 ) Use the same method, but adjust the equation for the Area so that instead of the area of the can, the eqn is for the original stock material that the can is made of.
