The perimeter of ΔWXY is : ( D ) 14.5 cm
<u>Calculating the </u><u>perimeter </u><u>of ΔWXY</u>
QR = WY / 2
RS = XW / 2
QS = XY / 2
Given that : QR = 2.93 cm , RS = 2.04 cm, QS = 2.28 cm
Therefore
Perimeter of ΔWXY = ∑ WY + XW + XY
= 2SR + 2QS + 2QR
= 2(2.04) + 2(2.28) + 2(2.93)
= 14.5 cm
Hence we can conclude that the perimeter of ΔWXY = 14.5 cm
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The equation to be used is written as:
ρ = nA/VcNₐ
where
ρ is the density
n is the number of atoms in unit cell (for FCC, n=4)
A is the atomic weight
Vc is the volume of the cubic cell which is equal to a³, such that a is the side length (for FCC, a = 4r/√2, where r is the radis)\
Nₐ is Avogradro's constant equal to 6.022×10²³ atoms/mol
r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³
ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]
ρ = 21.46 g/cm³