The solution is:A mole of gas occupies 22.4 L A liter is 1000 cubic centimeters because 1 cubic centimeter = 1 mL
One Helium molecule (essentially one helium atom) has atomic mass 4 g/mol So for every 22400 cubic centimeters of volume, we have 4 grams of helium Density of helium = 4g / 22400 cm^3 = 1g / 5600 cm^3
Volume of a sphere = (4/3)(pi)r^3 Volume of the outside sphere (the entire sphere) is (4/3)(pi)(R+T)^3 Volume of the inside sphere (the hollow region) is (4/3)(pi)R^3
The difference for the volume of silver. (4/3)(pi)(R+T)^3 - (4/3)(pi)R^3 = (4/3)(pi)(3R^2T + 3RT^2 + T^3)
The density of silver is 10.5g/cm^3
So the mass of the silver is computed by:10.5*(4/3)(pi)(3R^2T + 3RT^2 + T^3) = (14*pi)(3R^2T + 3RT^2 + T^3) = (14pi)T(3R^2 + 3RT + T^2)
Now for the mass of helium: volume x density = (4/3)(pi)R^3 (1/5600) = (pi/4200)R^3
Set the two masses equal: (pi/4200)R^3 = (14pi)T(3R^2 + 3RT + T^2) R^3 = 58800*T(3R^2 + 3RT + T^2) R / T = 58800*(3R^2 + 3RT + T^2) / R^2 = 58800*( 3+T/R^2+(T/R)^2)
then solve for xx = T / R 1/x = 58800*( 3+x/R+x^2) 1/58800 = x (3 + x/R + x^2) 1/58800 = 3x + x^3 x^3 + 3x - 1/58800= 0 x = ~ 5.66893x10^(-6)
Ok. you're looking for the slope intercept, or how much each of them had before they started saving. Chris had $27, as that is where his line crossed the y axis, and Connie, when graphed, intercepted at $9. <span />
9514 1404 393
Answer:
11.6 cm
Step-by-step explanation:
As the page title tells you, the Pythagorean theorem must be applied more than once. As you know, it tells you the square of the hypotenuse is the sum of the squares of the two sides.
AD² = ED² +EA²
EA² = AD²-ED² = 7² -6² = 13
EA = √13 ≈ 3.606
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CD² = ED² +EC²
EC² = CD² -ED² = 10² -6² = 64
EC = √64 = 8
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The length of the horizontal diagonal is ...
AC = EA +EC = 3.6 +8 = 11.6 . . . cm
Pretty much, write two problems that equal to 0
