Answer:- solution boiling point = 102.23 degree C (102 degree C with three sig figs).
Solution:- When a non volatile solute is added to a solvent then boiling point increases. Elevation in boiling point is directly proportional to the molality of the solution.
The equation is:
where, 
is the elevation in boiling point, i is the Van't hoff factor, 
is the molal elevation constant and m is the molality.
Value of i is 1 as ethylene glycol is a covalent molecule that does not break to give ions. for water is 
.
We can calculate the molality from the given grams of ethylene glycol and liters of water as molality is moles of solute per kg of solvent.
Molar mass of ethylene glycol is 62 gram per mol and density of water is 1.00 kg per liter.
= 2.50 kg
Let's calculate the moles of ethylene glycol.
= 10.9 mol
molality of the solution = 
= 4.36m
Let's plug in the values in the equation we have on the top for elevation in boiling point.
= 
Boiling point of pure water is 100 degree C. So, the boiling point of the solution = 100 + 2.23 = 102.23 degree C
(If we fix the three sig figs then it could be written as 102 degree C.)
For this case, we use the equation for an ideal gas which is expressed as PV=nRT where P is the pressure, V is the volume, n is the number of moles and T is the temperature. We calculate as follows:
PV = nRT
T = PV / nR
T = 20 kPa (100 L) / 1 mol (8.314)
T = 240.56 K
Answer:
Explanation:
Froath flotaion process is based on thie wttebality properteis like hydrophillicity and hydrophobicity.
Incase of sulphide ore processing,
The metallic sulphide particles of ore are preferentially wetted by oil and the gangue particles by water. By doing thise you are sperating required sulphide ore particels from unwanted gaunge , by doing this you are incresing the concentation of Sulfide in ore by removing the not required gaunge particles.
Hope this helps you to decide your annswer
Bc an MRI is just a scanner it's not and never will be meant for treatment
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
