The first element to have 4d electrons in its electron configuration is Yttrium with an atomic number of 39 and the chemical symbol Y. The electron configuration of an atom refers to the distribution of its electrons in the atomic orbitals. These orbitals are represented by letters s,p,d,f and so on. The electron configuration of Yttrium is [Kr] 4d¹ 5s², wherein [Kr] refers to the electron configuration of the noble gas, Krypton.
Answer: The correct answer is "B" two bonding domains(or bonding pairs) or two non bonding domains(or lone pairs)
Explanation:
Molecular geometry/structure is a three dimensional shape of a molecule. It is basically an arrangement of atoms in a molecule.It is determined by the central atom, its surrounding atoms and electron pairs.According to VSEPR theory, there are 5 basic shapes of a molecule: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.
A)Four bonding domains and zero non bonding domains: shape is tetrahedral and bond angle is 109.5°
B)Two bonding domains and two non bonding domains(lone pairs): shape is bent and bond angle is 104.5°
C)Three bonding domains and one non bonding domain: shape is trigonal pyramidal and bond angle is 107°
D)Two bonding domain and zero non bonding domain: shape is linear and bond angle is 107°
E)Two bonding domain and one non bonding domain: bent shape and bond angle is 120°
F)Three bonding domains and zero nonbonding domain: shape is trigonal planar and bond angle is 120°
Hence Two bonding domains and two non bonding domains have the smallest bond angle.
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Answer:
At the burner temp. and pressure, 18.85 litres of air is needed to completely combust each gram of propane
Explanation:
The combustion stoichiometry is as follows:
C₃H₈ + 5O₂ = 4 H₂O + 3CO₂ The molecular weights (g/mol) are:
MW 44 5x32 4x18 3x44
So each gram of propane is 1/44 = 0.02272 mol propane
and will need 5 x 0.02272 = 0.1136 mol oxygen
At 0.21 mol fraction oxygen in air, 0.1136 / 0.21 = 0.54 mol air is needed to burn the propane.
At the low pressure in the burner we can use the Ideal Gas Law
PV=nRT, or V = nRT/P
P = 1.1 x 101325 Pa = 111457 Pa
T = 195°C + 273 = 468 K
R = 8.314
and we calculated n = number of moles air = 0.54 mol
So V m³ = 0.54 x 8.314 x 468 / 111457 = 0.0188 m³ = 18.85 litres air.
Answer:
Mass = 6.183 g
Solution:
Step 1: Calculate number of moles of Boric acid using following formula,
Molarity = Moles ÷ Volume
Solving for Moles,
Moles = Molarity × Volume
Putting Values,
Moles = 0.05 mol.L⁻¹ × 2.0 L
Moles = 0.1 mol
Step 2: Calculate Mass of Boric Acid using following formula,
Moles = Mass ÷ M.mass
Solving for Mass,
Mass = Moles × M.mass
Putting values,
Mass = 0.1 mol × 61.83 g.mol⁻¹
Mass = 6.183 g
Flask used to prepare this solution is called as Volumetric flask. Take 2 L volumetric flask, add 6.183 g of Boric acid and fill it to the mark with distilled water.
