1.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that reacts with water to produce a lilac flame
2.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element used as an inert atmosphere
3.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element that has a valency of 3
4.Write a balanced chemical equation for the reaction between potassium and water. (Non-anonymous question). .
(1 Point)
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5.For each of the following, give the name of an element from Period 4 (potassium to krypton), which matches the description.
Elements may be used once, more than once or not all.. Single line text.
(1 Point)
an element with a fixed valency of 2 that not is not in group 2
HELP
6 + 3q + 306 + 153q + 15
156q + 327
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
Answer:
w > 7
Step-by-step explanation:
L = 3w
2L + 2w > 56
sub for L
2(3w) + 2w > 56
6w + 2w > 56
8w > 56
w > 7
Check: L = 3(8)
L = 24
2(24) + 2(8) > 56
48 + 16 > 56
64 > 56
