M=7M(H₂O)
M=7*18.015 g/mol = 126.105 g/mol
Answer:
The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
Explanation:
The given chemical reaction is as follows.
The expression for the entropy change for the reaction is as follows.
Substitute the all values in the entropy change expression.
Therefore, The standard enthalpy of formation of this isomer of is -220.1 kJ/mol.
Answer:
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.
Explanation:
Let the mass of the first solution be x and second solution be y.
Amount solution required = 1250 kg
x + y = 1250 kg....[1]
Percentage of ethanol in required solution = 12% of 1250 kg
Percentage of ethanol in solution-1 = 5% of x
Percentage of ethanol in required solution = 25% of y
5% of x + 25% of y =12% of 1250 kg
x + 5y = 3000 kg...[2]
Solving [1] and [2] we :
x = 437.5 kg , y = 812.5 kg
437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.