[tex]1s^2 2s^2 2p^6 3s^2 3p^3[/tex] is the correct electron configuration for an element with five electrons in the third energy level.
What are elements?Elements are the simplest substances which cannot be broken down using chemical methods.
The shell nearest to the nucleus, 1n, can carry two electrons, while the next shell, 2n, can carry eight, and the third shell, 3n, can carry up to eighteen.
The third shell carries 18 electrons; 2 in a 3s orbital; 6 in three 3p orbitals; and 10 in five 3d orbitals. The fourth shell carries 32 electrons; 2 in a 4s orbital; 6 in three 4p orbitals; 10 in five 4d orbitals; and 14 in seven 4f orbitals.
The element would be phosphorus. Its electron configuration is [tex]1s^2 2s^2 2p^6 3s^2 3p^3[/tex]
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Cubes are three-dimensional square shapes that have equal sides. What is the density of a cube that has a mass of 12.6 g and a measured side length of 4.1 cm? (Density: D = )
Answer:
0.1828g/cm³
Explanation:
density= mass÷volume
m= 12.6g
v= 4.1×4.1×4.1 = 68.921cm³
•
density= 12.6÷68.921 = 0.1828g/cm³
Calculate the pressure drop over a 100m length due to friction when a slurry made from 1.0-mm silica particles is pumped through a horizontal 6-cm diameter pipeline (smooth pipe) at 2.5 m/s. The slurry contains 25 per cent silica by volume. The density of silica is 2700 kg/m3, rhow = 1000 kg/m3, μw = 0.001 kg/ms. Use a value of 82 for Ω. Drag coefficient for these particles,CD, may be taken as 0.44 .
For the slurry/Pipe system in question 1, estimate the deposition velocity.
Answer:
The answer is "2.78".
Explanation:
Given values:
CD= 0.44
Formula:
[tex]\bold{f_r= \frac{v^{2}}{ g(s-1)D}}[/tex]
g=9.8
s= 2.7
D= 0.06
[tex]\to f_r=\frac{2.5^2}{9.8(2.7-1)0.06}\\\\[/tex]
[tex]=\frac{2.5 \times 2.5}{9.8 \times 1.7 \times 0.06 }\\\\=\frac{6.25}{.9996 }\\\\=6.252501[/tex]
[tex]\phi = \frac{82\times v}{ \sqrt{cD} \times f_r^{-1.5}}\\\\[/tex]
[tex]= \frac{82 \times 0.25 }{ \sqrt{0.44} \times 6.25^{1.5}}\\\\=2.4285\\[/tex]
[tex]\frac{\bigtriangleup P f_1 s_1}{L} = \frac{\bigtriangleup Pf_w}{L}(1+\phi)\\[/tex]
[tex]=\frac{2fwSwv^2 (1+2.4285)}{D}\\\\[/tex]
[tex]Re= \frac{D \bar v Sw}{M_w}\\[/tex]
[tex]=\frac{0.06 \times 2.5 \times 1000 }{0.001}\\\\=\frac{150 }{0.001}\\\\= 150 \times 10^{3}\\\\= 1.50 \times 10^{5}\\\\[/tex]
[tex]fw= 0.00389[/tex]
[tex]\to \frac{\bigtriangleup P f_1 s_1}{L}[/tex]
[tex]\to \frac{2 \times 0.00389 \times 1000 \times2.5^2 \times 3.4265}{0.06}\\\\\to 2.78[/tex]