1) A 75.0mL sample of water is heated to its boiling point. How much heat (in kJ
ID: 805585 • Letter: 1
Question
1)A 75.0mL sample of water is heated to its boiling point. How much heat (in kJ) is required to vaporize it? (Assume a density of 1.00 g/mL.)?
2)The density of an unknown metal is 2.64g/cm3 and its atomic radius is 0.215nm . It has a face-centered cubic lattice. Find the atomic weight of this metal.
3)How much ice (in grams) would have to melt to lower the temperature of 355mL of water from 26?C to 6?C? (Assume the density of water is 1.0 g/mL.)
4)Calculate the amount of heat required to completely sublime 81.0g of solid dry ice (CO2) at its sublimation temperature. The heat of sublimation for carbon dioxide is 32.3 kJ/mol.
Explanation / Answer
use dimensional analysis and the heat vaporization table.
Water = 40.7 kj/mol.
Start with the information that you were given and make sure that all of your units cancel out!
(70.0mL)(1.00g/1mL)(1 mole H2O/18.0g H2O)(40.7kj/1 mole H2O) = 115.820 kj.
There are 3 significant figures, so 115kj is your answer.
2)
Density = (Z * M) / ( volume * N) -----------
Z = no. of atoms in 1 unit cell. For fcc lattice Z = 4
Side = (4*radius) / (sqrt2)
so volume = (side)^3
Here convert 'nm' into 'cm' in radius.
N = avogadro no. = 6.022 * 10^23
M = atomic weight of metal.
M = {(density)*volume*N} / Z
put all the given values u will get the ans.
Ans - approx 89 g/mol (check this ans again by calculating urself,coz there may be any mistake in my calculation
3)
irst you find out the amount of energy released when the temperature of 355mL (i.e. 355 g) of water is lowered from 26?C to 6?C, using following equation
q = mcT
Then use q = ndeltaH/mol
Please be careful about the unit g and mol. Because correspondingly the value of c and deltaH will be used.
4)
moles CO2 = 81.0 g/ 44.009 g/mol=1.8405
heat required =1.8405 mol x 32.3 kJ/mol= 59.449kJ
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