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.
Answer:
124.91mL
Explanation:
Given parameters:
P₁ = 1.08atm
V₁ = 250mL
T₁ = 24°C
P₂ = 2.25atm
T₂ = 37.2°C
V₂ = ?
Solution:
To solve this problem, we are going to apply the combined gas law;
P, V and T represents pressure, volume and temperature
1 and 2 delineates initial and final states
Convert the temperature to kelvin;
T₁ = 24°C, T₁ = 24 + 273 = 297K
T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K
Input the variables and solve for V₂
V₂ = 124.91mL
This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:
Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g
% O = 48/84 * 100 = <em>57.14%</em>
ΔS =S(products) -S(reactants)
Where ΔS is the change of entropy in a reactions
a. ΔS = (2) - (2+1) = -1
b. ΔS = (1+1) -(1) = 1
c. ΔS = (1+2) - (1) = 2
d. ΔS = (2) - (2+1) = -1
e. ΔS = (1) - (1) = 0
ΔS is negative for reaction a. and d.
Answer:
Explanation:
Hello!
In this case, since the dissolution of copper (I) chloride is:
And its equilibrium expression is:
![Ksp=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
We can represent the molar solubility via the reaction extent as 
, however, since there is 0.050 M KCl we immediately add such amount to the chloride ion concentration since KCl is readily ionized; therefore we write:
Thus, solving for , we obtain:
By using the quadratic equation, we obtain:
Clearly, the solution is 
because no negative results are
allowed. Therefore, the molar solubility is:
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