In: Chemistry
1. An electrode of solid chromium is immersed in a 1M solution of Cr3+; In a second beaker, a tin electrode is immersed in a 1 M solution of Sn2+. The two electrodes are connected by a wire through a voltmeter, and the two beakers are connected via a salt bridge. The entire system is maintained at a T = 298K. When the connection is made between the two cells, what is the Eo for the cell, in volts, if the reduction potential for Cr(III) to Cr(0) is -0.74 V and the reduction potential for Sn(II) to Sn(0) is -0.14V? (a) –0.78V (b) 0.78V (c) –0.60V (d) 0.60V (e) none of these
2. Using the data in the table provided, calculate the approximate standard state Gibbs free energy change at 298 K for the reaction Al2O3(s) + 3H2(g) => 2Al(s) + 3H2O(g)
(a) 236 (b) 886 (c) –344 (d) –1021 (e) none of these
3. The Ksp for lead (II) fluoride in water is 3.6 x 10-8. What is its solubility in g/L? (a) 3.3 x 10-3 (b) 1.3 x 10-4 (c) 0.051 (d) 2.1 (e) none of these
4. The lower the vapor pressure of a liquid, the (a) higher the boiling point (b) the lower the boiling point (c) the higher the melting point (d) the lower the melting point (e) none of these relationships are true.
5. Elemental phosphorous, which exists as P4 molecules, has the following solubility properties: Water – insoluble; CCl4 – soluble. What may be ascertained about the type of intermolecular forces which are operative when P4 forms solutions? (a) they are ion-dipole in nature (b) they are dipole-dipole in nature (c) they are London force in nature (d) they are ionic in nature (e) they are covalent in nature
6. What is the order of the reaction in NO? (a) -1 (b) 0 (c) 1 (d) 2 (e) none of these
7. What is the order of the reaction in H2? (a) -1 (b) 0 (c) 1 (d) 2 (e) none of these
2Cr ---------------> 2Cr+3 + 6e- E0 = 0.74V
3Sn+2 + 6e- ---------> 3Sn E0 = -0.14V
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2Cr(s) + 3Sn+2 (aq) ---------> 2Cr+3 + 3Sn E0cell = 0.60V
(d) 0.60V
Al2O3(s) + 3H2(g) => 2Al(s) + 3H2O(g)
G0rex = G0 f products - Gf reactants
= -3*228.4+0 - (-1576.5 +0 ) = 891.3KJ >>>> answer
3. PbF2(s) ---------> Pb+2 (aq) + 2F-
S 2S
Ksp = [Pb+][F-]2
3.6*10-8 = s*(2s)2
4s3 = 3.6*10-8
S3 = 9*10-9
S = 2.08*10-3
solubility is 2.08*10-3 M >>> answer