heat released Q = 749 joules
heat of fusion of silver L = 109 J/g
Here phase of silver is changing from liquid to solid
so temperature will remain same
all heat will be released due to its phase change
and in this case we use Q=mL
where m is the mass of silver in gram
Q= mL
749 = m * 109
m = 749/109
m = 6.87 gram
Answer:
As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.
Explanation:
Depression in freezing point :

where,
=depression in freezing point =
= freezing point constant
m = molality ( moles per kg of solvent) of the solution
As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:
- If molality of the solution in high the depression in freezing point of the solution will be more.
- If molality of the solution in low the depression in freezing point of teh solution will be lower .
Relative lowering in vapor pressure of the solution is given by :

= Vapor pressure of pure solvent
= Vapor pressure of solution
= Mole fraction of solute

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.
- Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
- lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.
Answer:
(D)
Explanation:
Given :
l=3.5 m


Resistance can be calculated as :


Resistance of the wire will be 1.1×
ohms
Option D is correct
Answer: -1038.8 kJ
Explanation:
From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.
from the question we were given
Mass of nitrogen, m = 7kg
initial temperature, T1 = 250k
Final temperature, T2 = 450k
Polytropic index, n = 1.4
Specific gas constant, R = 0.2968kJ/kgK
W = [p2 * v2 - p1 * v1] / 1 - n
W = [m * R * T2 - T1] / 1 - n
W = 7*0.2968*(450 - 250)] / 1 - 1.4
W = [7*0.2968*200] / -0.4
W = 415.52 / -0.4
W = -1038.8 kJ