Answer: The factor that lead to cyclopropane being less stable than the other cycloalkanes is the presence of a RING STRAIN.
Explanation:
In organic chemistry, the end carbon atoms of an open aliphatic chain can join together to form a closed system or ring to form cycloalkanes. Such compounds are known as cyclic compounds. Examples include cyclopropane, cyclobutane, cyclopentane and many among others.
Cyclopropane is less stable than other cycloalkanes mentioned above because of the presence of ring strain in its structural arrangement. The ring strain is the spatial orientation of atoms of the cycloalkane compounds which tend to give off a very high and non favourable energy. The release of heat energy which is stored in the bonds and molecules cause the ring to be UNSTABLE and REACTIVE.
The presence of the ring strain affects mainly the structures and the conformational function of the smaller cycloalkanes. cyclopropane, which is the smallest cycloalkane than the rest mentioned above, contains only 3 carbons with a small ring.
Answer:
= -457.9 kJ and reaction is product favored.
Explanation:
The given reaction is associated with 2 moles of 
Standard free energy change of the reaction () is given as:
, where T represents temperature in kelvin scale
So, 
So, for the reaction of 1.57 moles of , 
As, is negative therefore reaction is product favored under standard condition.
The mole fraction of KBr in the solution is 0.0001
<h3>How to determine the mole of water</h3>
We'll begin by calculating the mass of the water. This can be obtained as follow:
- Volume of water = 0.4 L = 0.4 × 1000 = 400 mL
- Density of water = 1 g/mL
- Mass of water =?
Density = mass / volume
1 = Mass of water / 400
Croiss multiply
Mass of water = 1 × 400
Mass of water = 400 g
Finally, we shall determine the mole of the water
- Mass of water = 400 g
- Molar mass of water = 18.02 g/mol
- Mole of water = ?
Mole = mass / molar mass
Mole of water = 400 / 18.02
Mole of water = 22.2 moles
<h3>How to de terminethe mole of KBr</h3>
- Mass of KBr = 0.3 g
- Molar mass of KBr = 119 g/mol
- Mole of KBr = ?
Mole = mass / molar mass
Mole of KBr = 0.3 / 119
Mole of KBr = 0.0025 mole
<h3>How to determine the mole fraction of KBr</h3>
- Mole of KBr = 0.0025 mole
- Mole of water = 22.2 moles
- Total mole = 0.0025 + 22.2 = 22.2025 moles
- Mole fraction of KBr =?
Mole fraction = mole / total mole
Mole fraction of KBr = 0.0025 / 22.2025
Mole fraction of KBr = 0.0001
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Answer:
to be honest ask your teacher
Answer:
The final and initial concentration of the acid and it's conjugate base are approximately equal, that is we use the weak acid approximation.
Explanation:
The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.
Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i
Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.
It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.
