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
Here, we are required to determine which element has atoms with Valence electrons in a higher energy level than those of calcium.
The Valence electrons of the Caesium, Cs atoms has higher energy level than those of Calcium.
First, we need to determine the electronic configuration of each of the elements
- Therefore the Calcium electron configuration will be:
- 1s²2s²2p⁶3s²3p⁶4s²
- Therefore, for Cs Electron Configuration:
- 1s² 2s²2p⁶3s²3p⁶3d¹⁰ 4s²4p⁶4d¹⁰ 5s²5p⁶ 6s¹
- Therefore, For lithium, The electron configuration of lithium is :
- 1s²2s¹
- For oxygen, Therefore the O electron configuration will be:
- 1s²2s²2p⁴
- For bromine, the electronic configuration is 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁵
Among the elements in the option, the Valence electrons of the Caesium, Cs atoms has higher energy level than those of Calcium.
This is so because the only Valence electron of Cs is located in the 6s orbital and as such has the highest energy level amongst all of the elements.
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Answer:
given function is
formula for newton's method is
so derivative of function is
now put values and solve
or you can also use MATLAB code to solve
i.e
function p= newton(x)
e=0.001;
for i=1:100
if abs(d(x))>e
if abs(k(x))>0
xm=x-(k(x)/d(x));
x=xm;
else
end
break;
end
end
disp(x)
disp(k(x))
return;
Answer:
A) A warm wire
Explanation:
A warm wire has the most resistance. Heating the metal wire causes atoms to vibrate more, which in turn makes it more difficult for the electrons to flow, increasing resistance. Heating the wire increases resistivity.
Answer:
m = 4.5021 kg
Explanation:
given,
Apparent mass of aluminium = 4.5 kg
density of air = 1.29 kg/m³
density of aluminium = 2.7 x 10⁷ kg/m³
true mass of the aluminium = ?
Weight in Vacuum
W = m g
W = ρV g
Air buoyancy acting on aluminium
B = ρ₀V g
Volume is the same in both cases since the volume of the aluminum
displaces an equal amount of volume air.
Apparent weight:
ρV g − ρ₀V g = 4.5 g
ρV − ρ₀V = 4.5
m = ρV
m = 4.5021 kg
