Delta S= Delta H
It boils
The new volume is 544,5 l.
The solubility increase from 38,7 g KCl/100g water to 40,7 g KCl/100 g water.
Melting point 1132 degC, boiling point 4131 degC
For an approximate calculation: specific heat capacity for water = 4.18 J/(g*degC) (how much energy is required per gram per change in degrees C) mass = 25g Change in temperature = 60-10 = 50 degC energy required = mass * change in temperature * specific heat capacity = 25g * 50 degC * 4.18 J/(g*degC)
A change of 100 degrees K is equal to a change of 100 degrees Celsius.
A PWR has an inlet water temperature of 275 degC and outlet 325 degC
dT = (i)(Kf)(m) is the equation that can be used to model the freezing point depression. dT represents the change change in temperature. i represents the amount of ions formed from the dissolution of the solute. Kf represents the constant of the freezing point depression of water. m is the molality of the solute in solution. dT = (3)(1.86 degC/m)(2.65m) dT = 14.8 degC 0-14.8 degC = -14.8 degC So the freezing point is -14.8 degrees C. The reason why this value was subtracted from zero is because the presence of a solute lowers the freezing point.
It boils
Water is transformed in vapors.
The energy is 103,6 kcal.
Liquid
It would be approx 9042 litres.
(4.184 J/g*degC)(400g)(40.0*degC-80.0*degC)+(200g)
The new volume is 544,5 l.
How can the temperature DECREASE from 30.5ºC to 35.6ºC? That's an INCREASE in temperature. So, assuming you meant the temperature INCREASED to 35.6ºC, then it is an endothermic reaction.q = mC∆T = (1000 g)(4.184 J/g/deg)(5.1 deg) = 20,920 J = 20.9 kJ. This is the heat change for this reaction.
Sulphur dioxide is a colourless liquid or pungent gas, which is the product of the combustion of sulphur on air. Its melting point is -72.7 degC, its boiling point is -10 degC and its relative density is 1.43.