The correct answer is letter (A) Acetylene. Acetylene is the most simplest form of alkyne and at the same time a hydrocarbon. It is unsaturated because of the presence of only two carbon atoms that are bonded together in a triple bond. I<span>n its pure, it is unstable and thus, it is usually held and handled as a solution.</span>
Answer:
A. is the correct point.
Explanation:
This is true because no matter how many mL of water is added, the solution only gets more height; the concentration in everything else stays the same, and water doesn't have any concentration. Very confusing, I know. Good luck!
Test tube of ammonium chloride (NH4Cl) being heated over a bunsen burner flame. Ammonium chloride decomposes readily when heated, but condenses in the cooler area at the top of the test tube. This is a reversible reaction, where the ammonium chloride decomposes into the gases ammonia (NH3) and hydrogen chloride (HCl).
LiF or lithium fluoride is the non covalent molecule or ionic compound.
Option 3.
<h3><u>Explanation:</u></h3>
Covalent molecules are those molecules which do have actual bonds between the atoms present in the molecule by sharing of the electrons. But in ionic molecules, there's no actual bonds between the atoms, but the oppositely charged ions are attracted towards each other by means of electrostatic force of attraction.
The molecules that are formed by the atoms with high electronegativity and electropositivity are actually ionic because the atoms with high electronegativity are able to actually gain electron readily and the atoms with high electropositivity are actually ready to give the electrons to the electronegative elements.
Lithium is highly electropositive and fluoride is highly electronegative. So they establish an ionic bond. But other molecules like fluorine molecule has both the electronegative elements, Carbon monoxide has carbon which isn't electropositive highly, and ammonia has hydrogen which isn't electropositive.
So lithium fluoride is the ionic compound.
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
