Explanation:
Newton's second law states the "net force on a body is the product of its mass and acceleration".
When a body has a net force acting on it, it will change position and begin to move.
Ordinarily, a car will not move until it is towed when the engine is faulty. The weight of the car and the frictional force of the tire acts to prevent its motion.
Now, the towing van supplies a force that is greater than the weight of the car and the frictional force between the tires. This way the towing van is able to move the faulty car.
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Answer: The change in entropy change when 207 g of toluene boils at 
is 223 
.
Explanation:
It is given that mass of toulene is 207 g and its molar mass is 92.14 g/mol. So, its moles will be calculated as follows.
No. of moles = 
= 
= 2.25 moles
As we are given that heat of vaporization for 1 mol is 38.1 kJ/mol. So, heat of vaporization for 2.25 moles will be calculated as follows.
= 85.725 KJ
Now, we know that the relation between enthalpy change and entropy change is as follows.
= 
= 0.223 
or, = 223 (as 1 kJ = 1000 J)
Thus, we can conclude that the change in entropy change when 207 g of toluene boils at is 223 .
Answer:
Because the optimal range of buffering for a formic acid potassium formate buffer is 2.74 ≤ pH ≤ 4.74.
Explanation:
Every buffer solution has an optimal effective range due to pH = pKa ± 1. Outside this range, there is not enough acid molecules or conjugate base molecules to sustain the pH without variation. There is a certain amount of both molecules that has to be in the solution to maintain a pH controlled.
Being for the formic acid the pKa 3.74, the optimal effective range is between 2.74 and 4.74. Upper or lower these range a formic acid/potassium formate buffer does not work.
I would say that the first answer choice makes sense since the charge of Oxygen is +2. But I may be wrong, try it out!
