In: Physics
a) Determine the amount of heat per gram which must be removed to convert 110 OC steam (H2O(g)) into –10 OC ice(H2O(s)), then find the total heat removed if there is 9.50 g of steam.
b) For acetic acid (CH3COOH, with one O bonded to an H), acetone (CH3COCH3, with the O bonded only to a C), and carbon tetrachloride (CCl4, see non polar molecules in section 12.10), determine the predominant types of intermolecular attraction for the pure liquids, and predict whether the vapor pressures, boiling points, viscosity, and surface tensions will be low or high.
c)For tantalum (Ta), calcium chloride (CaCl2), fructose (C6H12O6), and diamond (C), determine the type of solid and describe the expected physical properties.
I have solved your problem which is shown below:
Amount of heat per gram : As we know Know that we have to bring to so firstly we bring it to 110 to 0 degree which is known as specific heat. And then from 0 to -14 we need to find the latent heat of fusion to remove energy.
Some known values:
Specific heat of ice = 2.1 Jg-1 0C-1
Latent heat of Steam = 2268 Jg-1
Latent heat of ice = 336 Jg-1
Temperature of Steam = 1100C
Temperature of ice = -100C
Specific heat of water = 4.2 Jg-1 0C-1
Specific heat of steam = 1.996 Jg-1 0C-1
Heat removed when water temperature comes from 110 to 100 for 1 gram
heat removed = m CsT
=1.99610 = 19.96 J
Now to convert steam to water for 1 gram
Latent heat of steam = 2268 J
Now changing to water the amount of heat released
Heat removed = 4.2 100 = 420 J
Now changing Water to ice latent heat of fusion releases
Heat removed of latent heat of fusion = 336 J for 1 gram
Now to convert to so that comes out to be specific heat of ice * change in temperature
Heat removed = 2.1 * 14 = 29.4 J
So total amount of heat released to change steam to ice for 1 gram
Total heat released = 19.96 + 2268+420+336+29.4 = 3073.36 J For 1 gram
Now given mass is 9.5 gram
So for 9.5 gram amount of heat released = 3073.36 * 9.5 = 29196.92 J
Which is our required result
Part b
For Acetic acid (CH3COOH) O Bonded with H In this there is Hydrogen bonding between O and H is present and also the Dipole Dipole interactions are also present in it.
Vapor Pressure = It has Low Vapour pressure comparitive to water.
Boiling Point = It has 17.9 degree high boiling point comparitive to water.
Viscosity = Viscosity of acetic acid is 1.222mPa.s but at same temperature viscosity of water is 0.89 mPa.s Which shows it has high viscosity at 25 degree celcius.
Surface Tension = Surface Tension of acetic acid - air interface is 26.6 dyne/cm while surface tension of water - air interface is 72 dyne /cm at C for both liquid . This shows acetic acid has Low surface tension.
Acetone (CH3COCH3)
Types of interactions = Dipole Dipole interactions
Vapor pressure = Acetone has 30 KPa while water has 2.4KPa at C temperature which shows it has very high vapor pressure.
Boiling Point = shows it has Low boiling point
Viscosity = 0.295 mPa.s Which shows it has Low viscosity as compare to water i.e 0.89 mPa.s
Surface Tension = 25.2 dyne/cm which shows it has low surface tension comparitive to water.
Carbon Tetrachloride ()
Types of interaction = It has london dispersive forces because it is non polar in nature.
Vapor Pressure = CCl4 has 5.2 KPa while water has 2.4KPa at C temperature which shows it has very high vapor pressure compare to water.
Boiling Point = which shows low boiling point comparitve to water
Viscosity = 0.901 mPa.s Shows has high viscosity comparitive to water.
Surface Tension = 26.8 dyn/cm while surface tension of water - air interface is 72 dyne /cm at C for both liquid . This shows CCl4 has Low surface tension.
Part c
Tantalum(Ta)
Structure = Solid structure having BCC (body cubic crystal ) structure
If you have any query please write in a comment. I hope you like it. Thank you