Atoms of oxygen are electronegative and attract the shared electrons in their covalent bonds.
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
Answer:
Homogeneous
Explanation:
Homogeneous mixtures are uniform in composition. They have the same proportion of components throughout. Homogeneous mixtures are called solutions. Sugar, paint, alcohol, gold are all examples of homogeneous mixtures because they look the same throughout.
A hydrocarbon with three or more consecutive (cumulative) double bonds is known as a cumulene. They are analogous to allenes, only exhibiting a more elongated chain. The basic molecule in this category is butatriene, which is also simply known as cumulene.
In the structure of a cumulene, there are 3 double bonds and 4 single bonds. The double bond comprises 1 sigma bond, and 1 pi bond and 4 hydrogen bond produces a sigma bond with carbon. Thus, the molecule of cumulene comprises 7 sigma bonds and 3 pi bonds.
Answer:
Cyclopropane has a planar carbon back bone while propane does not
Explanation:
We have to recognize that in straight chain saturated organic compounds, carbon atoms have a tetrahedral geometry. Each carbon atom is bonded to four other atoms.
However, carbon atoms in cyclic compounds are also sp3 hybridized with each carbon bonded to only four other atoms but the ring system is highly strained.
Cyclopropane is a necessarily planar molecule with a bond angle that is far less than the expected tetrahedral bond angle due to strain in the molecule. Hence, the carbon atoms may have have a "planar backbone".
