Ans: Distance = 16.1
Let us consider point A with coordinates (x1,y1) and point B with coordinates (x2,y2)
The distance between the two points is given as:
d(A,B) = √(x2-x1)² + (y2-y1)²
Here: (x1,y1) = (-6,2)
(x2,y2) = (8,10)
d = √(8-(-6))² + (10-2)²
= √196+64 = √260 = 16.1
The distance between (–6, 2) and (8, 10) is 2√65
The distance formula is used to calculate the distance between two points(x₁, y₁)(x₂, y₂).
Therefore,
x₁ = -6
x₂ = 8
y₁ = 2
y₂ = 10
Therefore,
d = √(8+6)²+(10-2)²
d = √(14)² + (8)²
d = √196 + 64
d = √260
d = √4 × 65
d = 2√65
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Answer: Option (B) is the correct answer.
Explanation:
Boiling point is defined as the temperature at which vapor pressure of a liquid becomes equal to the atmospheric pressure.
Surface tension is defined as the attractive forces experienced by the surface molecules of a liquid by the molecules present beneath the surface layer of the liquid. More stronger are the intermolecular forces present within the molecules of a liquid more will be its surface tension and lower will be its vapor pressure.
Hence, higher will be the boiling point of the liquid.
And, viscosity is defined as the ability of a liquid to resist its flow. When a substance has high viscosity then it is known as a viscous substance.
Thus, we can conclude that a substance with a high boiling point will also likely have a low vapor pressure.
Answer:
a) Warmer
b) Exothermic
c) -10.71 kJ
Explanation:
The reaction:
KOH(s) → KOH(aq) + 43 kJ/mol
It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.
Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.
The enthalpy change for the dissolution of 14 g of KOH is:
Where:
m: is the mass of KOH = 14 g
M: is the molar mass = 56.1056 g/mol
The enthalpy change is:
The minus sign of 43 is because the reaction is exothermic.
I hope it helps you!
Answer:
Chloroform.
Explanation:
Given,
Solvent requires 1g of compound per 100 mL
For water,
= 1g/47ml
= 2.1
For Chloroform,
= 1 g/8.1 mL
= 12.345679
For Diethyl ether,
= 1 g/370 mL
= 0.27
For Benzene,
= 1 g/86 mL
= 1.2
Partition coefficients:
Water = -
chloroform = 5.9
Diethyl = .13
Benzene = .57
The solvent chloroform would be chosen for drawing out the compound out of an aqueous solution as it has the maximum solubility.
The solubility of a compound in different solvents will determine its concentration in each solvent. The partition coefficient represents the relative solubility of a compound in two immiscible solvents. Chloroform would be the best choice to extract the compound from an aqueous solution.
The solubility of a compound is usually expressed as grams of solute per 100 mL of solvent. To calculate the solubility, you can use the following formula:
Solubility (g/100 mL) = (mass of solute / volume of solvent) * 100
Using this formula, the solubility of the compound in water is 47 g/100 mL, in chloroform is 97.53 g/100 mL, in diethyl ether is 2.70 g/100 mL, and in benzene is 1.16 g/100 mL.
The partition coefficient is a measure of the compound's solubility in two immiscible solvents. To calculate it, divide the solubility of the compound in one solvent by its solubility in another solvent. For example, the partition coefficient between chloroform and water would be:
Partition coefficient = Solubility in chloroform / Solubility in water = 97.53 g/100 mL / 47 g/100 mL = 2.07
The larger the partition coefficient, the more soluble the compound is in the first solvent compared to the second solvent. Based on the partition coefficients, chloroform would be the best choice to extract the compound from an aqueous solution.
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Answer:
use google and use the first link
Explanation:
Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate followed by protonolysis produces dextrorotatory ketone A and a trace of isomer B. Compound A can yield optically active, dextrorotatory hydrocarbon C when treated with zinc amalgam and HCl.
Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate followed by protonolysis yields a mixture containing primarily a dextrorotatory ketone A and a trace of an optically inactive isomer B. Compound A can be treated with zinc amalgam and HCl to produce an optically active, dextrorotatory hydrocarbon C.
To provide a specific identification and stereochemical configuration of compounds A, B, and C, I would need additional information or further context about the starting components and reaction conditions, as well as any other relevant data or observations.
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b. all the chloro groups in equatorial positions.
Answer:
This is required answer.
Explanation:
Given that,
1,2,3,4,5,6-hexachlorocyclohexane
(a). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions
Using given data
We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions.
When we say that all the chloro groups in axial position that means axial bonds are vertical.
(b). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions
Using given data
We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions.
When we say that all the chloro groups in equatorial position that means axial bonds are horizontal.
Hence, This is required answer.