Answer:
529.2 N
Explanation:
As we have studied the first law of motion, which states that every action has some reaction, equal in magnitude but having an opposite direction.
The force that is acting on the student will be due to gravitational force, that is equal to his weight.
F=mg: 54kg x 9.8m/s^2 =529.2 N
So the weight of student is exerting downwards towards the stool and land. The stool will also exert a force on the student that will be equal in magnitude but opposite in direction, then it will be 529.2 N.
This is because the student is sitting in a constant state and all the weight is exerted on the stool.
Note: This answer is very generic supposing that all the weight of the student is on stool. But, if we suppose that student's legs are on floor so it means the force of gravity acting on the stool has become less because student's mass on stool is less. So the answer would be a force somehow less than 529.2 N. However, since the question asked normal force, it would be weight of student in general terms.
Hope it helps!
Answer:
B. They are dimensionless ratios of the actual concentration or pressure divided by standard state concentration, which is 1 M for solutions and 1 bar for gases.
Explanation:
Activity of a substance is defined as the ratio of an effective concentration or an effective pressure to a standard state pressure or a standard state pressure. It is usually a unit less ratio.
Concentrations in an equilibrium constant are really dimensionless ratios of actual concentrations divided by standard state concentrations. Since standard states are 1 M for solutes, 1 bar for gases, and pure substances for solids and liquids, these are the units to be used.
Hence, activity is a fudge factor to ideal solutions that correct the true concentration. Activity of a gas and solute concentration is a ratio with no unit.
Answer : The vapor pressure of bromine at 
is 0.1448 atm.
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of bromine at = ?
= vapor pressure of propane at normal boiling point = 1 atm
= temperature of propane = 
= normal boiling point of bromine = 
= heat of vaporization = 30.91 kJ/mole = 30910 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the vapor pressure of bromine at is 0.1448 atm.
Answer:
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