Explanation:
It is known that the maximum value of ml is equal to the vale of l. But the minimum value of n is as follows.
n = l + 1
where, n = principle quantum number
l = azimuthal quantum number
Values of n can be 1, 2, 3, 4 and so on. Whereas the values of l is 0 for s, 1 for p, 2 for d, 3 for f, and so on.
Also, "m" is known as magnetic quantum number whose values can be equal to -l and +l.
- Electronic configuration of Li is
. So here, n = 2, l = 0, m = 0 and s = ± 
.
- Electronic configuration of
is ![[Ar]3d^{10}4s^{2}4p^{6} ](https://tex.z-dn.net/?f=%5BAr%5D3d%5E%7B10%7D4s%5E%7B2%7D4p%5E%7B6%7D%0A)
. So here, n = 4, l = 1, m = -1, 0, +1, and s = ± .
- Electronic configuration of
is ![[Kr]4d^{10}5s^{2}5p^{6}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B10%7D5s%5E%7B2%7D5p%5E%7B6%7D)
. So here, n = 5, l = 1, m = -1, 0, +1, and s = ± .
- Electronic configuration of B is
. So here, n = 2, l = 1, m = -1, 0, +1, and s = ± .
Answer:
Each element within a group has similar physical or chemical properties because of its atom's outermost electron shell (most chemical properties are dominated by the orbital location of the outermost electron).
Explanation:
Answer: hre
Explanation:
N2(g) + 3H2-> 2NH3(g) This is the balanced equation
Note the mole ratio between N2, H2 and NH3. It is 1 : 3 : 2 This will be important.
moles N2 present = 28.0 g N2 x 1 mole N2/28 g = 1 mole N2 present
moles H2 present = 25.0 g H2 x 1 mole H2/2 g = 12.5 moles H2 present
Based on mole ratio, N2 is limiting in this situation because there is more than enough H2 but not enough N2.
moles NH3 that can be produced = 1 mole N2 x 2 moles NH3/mole N2 = 2 moles NH3 can be produced
grams of NH3 that can be produced = 2 moles NH3 x 17 g/mole = 34 grams of NH3 can be produced
NOTE: The key to this problem is recognizing that N2 is limiting, and therefore limits how much NH3 can be produced.
Answer:
Magnitude of work done = 24.28 J
Explanation:
No. of moles = Reacting mass/ Molar mass
Reacting mass of Na = 0.45 g
Molar mass of sodium = 23g/mol
∴ No. of moles of sodium = 0.45/23 = 0.0196 mole
If 2 moles of Na react with water at 25°C and 1 atm, 24.5 L of H₂ was formed.
∴ when 0.0196 mole of Na react with water under the same conditions, (24.5×0.0196)/2 L of H₂ will be formed.
⇒ (24.5×0.0196)/2 L = 0.24 L
0.24 L × 1 atm = 0.24 L . atm
Since 1 L · atm = 101.3 J
∴ 0.24 L . atm = (0.24 L . atm ×101.3)/1 = 24.28 J
Magnitude of work done = 24.28 J
According to the kinetic theory, the mean free path is the average distance a single atom or molecule of an element or compound travels with respect with the other atoms during a collision. The greater the mean free path, the more ideal the behavior of a gas molecule is because intermolecular forces are minimum. To understand which factors affect the mean free path, the equation is written below.
l = μ/P * √(πkT/2m), where
l is the mean free path
μ is the viscosity of the fluid
P is the pressure
k is the Boltzmann's constant
T is the absolute temperature
m is the molar mass
So, here are the general effects of the factors on the mean free path:
Mean free path increases when:
1. The fluid is viscous (↑μ)
2. At low pressures (↓P)
3. At high temperatures (↑T)
4. Very light masses (↓m)
The opposite is also true for when the mean free path decreases. Factors that are not found here have little or no effect.
