Iodate has a charge of -1, this means its chemical symbol
must be IO3(-1) while Lanthanum is La(+3). Therefore making the compound
La(IO3)3.
However sulfate has a chemical symbol of SO4(-2), it has
charge of -2, therefore the formula for lanthanum sulfate is:
<span>La2(SO4)<span>3</span></span>
Correct answer: Option D, <span>
K = 5.04 × 10^52</span>
Reason:
We know that,
Ecell =
,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.
Also, Ecell = <span>+1.56 v
Therefore, 1.56 = </span>
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52
Answer: 20.7 kJ
Explanation:
According to first law of thermodynamics:
=Change in internal energy
q = heat absorbed or released
w = work done or by the system
w = work done on the system=
{Work is done on the system is positive as the final volume is lesser than initial volume}
w = 4.51 kcal = 
(1kcal = 4.184kJ)
q = +1.79 kJ {Heat absorbed by the system is positive}
Thus for a system that receives 1.79 kJ of heat from surroundings and has 4.51 kcal of work done on it at the same time is 20.7 kJ
No of mole = reacting mass/ molar mass
No of mole = 0.06/56
No of mole = 0.00107mol....
Or...
1mole of iron contain 56gmol¯¹
:-x mole is contained in 0.06g
Porata
And you'll have..
x = 0.06/56
x = 0.00107mol.....
Conduction: In the conduction, the heat is transferred from the hotter body to the colder body until the temperature on both bodies are equal.
In thermal equilibrium, there is no heat transfer as the heat is transferred till the temperature on the bodies are not same.
In the given problem, an iron bar at 200°C is placed in thermal contact with an identical iron bar at 120°C in an isolated system. After 30 minutes, the thermal equilibrium is attained. Then, the temperature on both iron bars are equal.Both iron bars are at 160°C in an isolated system.
But in an open system, the temperatures of the iron bars after 30 minutes would be less than 160°C. There will be heat lost to the surrounding. The room temperature is 25°C. There will be exchange of the heat occur between the iron bars and the surrounding. But It would take more than 30 minutes for both iron bars to reach 160°C because heat would be transferred less efficiently.
