It is in the noble gas group which has a full valence electron shell found in group 18
I think it’s C. Stroke if not then D
Answer:
i/f = i/o + i/i f = focal, o = object, i = image
1 / i = 1 / f - 1 / o = (o - f) / o f
i = o * f / ( o - f) image distance
i = 12.5 * 22 / (12.5 - 22) = -28.9 cm
Image is real
Image is 28.9 cm to left of lens
M = - i / o = = 28.9 / 12.5 = 2.3 magnification (convex lens)
Density = (mass) divided by (volume)
We know the mass (2.5 g). We need to find the volume.
The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.
Volume = (π) (9.775 mm)² (1.55 mm)
= (π) (95.55 mm²) (1.55 mm)
= (π) (148.1 mm³)
= 465.3 mm³
We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:
mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .
Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.
(0.0054 g/mm³) x (10 mm / cm)³
= (0.0054 x 1,000) g/cm³
= 5.37 g/cm³ .
This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.
