Number of electron pairs = \frac{1}{2}[V+N-C+A]
2
1
[V+N−C+A]
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
SbCl_5SbCl
5
:
In the given molecule, antimony is the central atom and there are five chlorine as monovalent atoms.
The number of electron pairs are 5 that means the hybridization will be sp^3dsp
3
B and geometry of the molecule will be trigonal bipyramidal.
The tiny holes in a sponges outer layer that take in water are called ostia, I believe. Hope it helps
Answer: -
IE 1 for X = 801
Here X is told to be in the third period.
So n = 3 for X.
For 1st ionization energy the expression is
IE1 = 13.6 x Z ^2 / n^2
Where Z =atomic number.
Thus Z =( n^2 x IE 1 / 13.6)^(1/2)
Z = ( 3^2 x 801 / 13.6 )^ (1/2)
= 23
Number of electrons = Z = 23
Nearest noble gas = Argon
Argon atomic number = 18
Number of extra electrons = 23 – 18 = 5
a) Electronic Configuration= [Ar] 3d34s2
We know that more the value of atomic radii, lower the force of attraction on the electrons by the nucleus and thus lower the first ionization energy.
So more the first ionization energy, less is the atomic radius.
X has more IE1 than Y.
b) So the atomic radius of X is lesser than that of Y.
c) After the first ionization, the atom is no longer electrically neutral. There is an extra proton in the atom.
Due to this the remaining electrons are more strongly pulled inside than before ionization. Hence after ionization, the radii of Y decreases.
There are 6.33 × 10²⁵ hydrogen atoms in this solution in total.
<h3>Explanation</h3>
- There are two hydrogen atoms in each water
molecule. - There are three hydrogen atoms in each ammonia
molecule.
2.10 × 10²⁵ water molecules and 7.10 × 10²⁴ ammonia molecules will contain
hydrogen atoms in total.
Answer:
248 mL
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.
Qw + Qc = 0
Qw = -Qc [1]
We can calculate each heat using the following expression.
Q = c × m × ΔT
where,
- ΔT: change in the temperature
163 mL of coffee with a density of 0.997 g/mL have a mass of:
163 mL × 0.997 g/mL = 163 g
From [1]
Qw = -Qc
cw × mw × ΔTw = -cc × mc × ΔTc
mw × ΔTw = -mc × ΔTc
mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)
mw × 29.0°C = 163 g × 43.9°C
mw = 247 g
The volume corresponding to 247 g of water is:
247 g × (1 mL/0.997 g) = 248 mL
