Answer:
36.55kJ/mol
Explanation:
The heat of solution is the change in heat when the KNO3 dissolves in water:
KNO3(aq) → K+(aq) + NO3-(aq)
As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.
To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:
<em>Moles KNO3 -Molar mass: 101.1032g/mol-</em>
10.6g * (1mol/101.1032g) = 0.1048 moles KNO3
<em>Change in heat:</em>
q = m*S*ΔT
<em>Where q is heat in J,</em>
<em>m is the mass of the solution: 10.6g + 251.0g = 261.6g</em>
S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-
And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C
q = 261.6g*4.184J/g°C*3.5°C
q = 3830.87J
<em>Molar heat of solution:</em>
3830.87J/0.1048 moles KNO3 =
36554J/mol =
<h3>36.55kJ/mol</h3>
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Hello there.
Thomson's atomic model is best described by which of the following statements?
A nucleus with electrons moving around it like planets.
Explanation:
A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.
The given reaction is 
Now, we balance the mass on both reactant and product side as follows.
235 + 1 = 
236 = 234 + x
x = 236 -234
= 2
So, now we balance the charge on both reactant and product side as follows.
92 + 0 = 
92 = 96 - y
y = 4
Thus, we can conclude that there are 2 neutrons and 4 beta-particles are produced in the given reaction.
Therefore, reaction equation will be as follows.
As I am reading the problem, I see they gave you two pressures, one volume and they are asking for another volume. this should give you a hint that you need to use the following formula.
P1V1= P2V2
P1= 1.00 atm
V1= 0.50 ft³
P2= 3.00 atm
V2= ?
Now we plug the values
(1.00 x 0.50)= (3.00 x V2)
V2= 0.17 ft³
Answer: Option (E) is the correct answer.
Explanation:
Since, the conductor is hollow which means that it is opened on both the ends. Hence, when a small uncharged metal ball is passed through it with the help of a silk thread then due to the presence of this insulating thread the ball will not come directly in contact with the charged rod.
As a result, there will occur no formation of opposite charge on the metal ball. Therefore, the ball will remain uncharged in nature.
Thus, we can conclude that after the given ball is removed, it will have no appreciable charge.
