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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: matt123 on September 04, 2011, 12:53:39 pm
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hey guys , I really need help on some questions . thanks
A system delivers 1275 J of heat while the surroundings perform 855 J of work on it. Calculate ΔE in J.
Question 2 answers
-2130
-420
420
2130
I dont know if its - or + ?
Question 1 text A system that does no work but which gains heat from the surroundings has
Question 1 answers
q > 0, ΔE < 0
q > 0, ΔE > 0
q < 0, ΔE < 0
q < 0, ΔE > 0
I said D ?
Titanium tetrachloride is converted to pure titanium metal by a reduction process. The equation is
TiCl4(l) + 4Na(s) → 4NaCl(s) + Ti(s) Δ H = -840 kJ
What is the enthalpy change when 30.0 mol of titanium tetrachloride is converted to elemental titanium?
Question 4 answers
28 kJ
2.52 x 104kJ
-2.52 x 104kJ
-28 kJ
is it negetive or positive again??
Question 6 text Calculate the ΔH° rxnfor the following reaction.
ΔH° f [AsH3(g)] = 66.4 kJ/mol; ΔH° f [H3AsO4(aq)] = -904.6 kJ/mol; ΔH° f [H2O(l)] =-285.8 kJ/mol;
H3AsO4(aq) + 4H2(g) →AsH3(g) + 4H2O(l)
Again , is it neg or pos?
Which one of the following is not a correct formation reaction? (products are correct)
Question 7 answers
½H2(g) + ½Cl2(g) → HCl(g)
6C(graphite) + 3H2(g) → C6H6(l)
C(graphite) → C(diamond)
6C(graphite) + 6H2(g) + 3O2(g) → C6H12O6(s)
H2(g) + 2O(g) → H2O2(l)
Predict (do not calculate) the sign of ΔS°for the following reaction.
NaHCO3(s) → Na+(aq) + HCO3- (aq)
Question 8 answers
Δ S° ≈ 0
Δ S° > 0
Δ S°< 0
More information is needed to make a reasonable prediction.
Which of the following pairs has the member with the greater molar entropy listed first? All systems are at 25°C
Question 9 answers
KCl(s), KCl(aq)
NO(g), NO2(g)
H2S(aq), H2S(g)
CO2(g), CO2(s)
HOW DO U WORK THIS OUT? ..
Calculate ΔS° for the combustion of propane.
C3H8(g) + 5O2(g) →3CO2(g) + 4H2O(g)
Substance C3H8(g) O2(g) CO2(g) H2O(g)
S°(J/K·mol) 269.9 205.138 213.74 188.825
Question 11 answers
-100.9 J/K
-72.5 J/K
72.5 J/K
100.9 J/K
Use the given data at 298 K to calculate ΔG° for the reaction
2Cl2(g) + 2H2O(g) →4HCl(g) + O2(g)
Substance Cl2(g) H2O(g) HCl(g) O2(g)
ΔH°f (kJ/mol) 0 -242 -92 0
S°(J/K·mol) 223.0 189 187 205
Question 12 answers
11.7 x 105 kJ
78 kJ
396 kJ
11.8 kJ
Question 13 text Calculate the equilibrium constant at 800°C for the reaction of steam with hot coke to form hydrogen and carbon monoxide. The data refer to 800°C.
C(s) + H2O(g) CO(g) + H2(g)
Substance C(s) H2O(g) CO(g) H2(g)
ΔH°f (kJ/mol) 0.0 -242 -111 0.0
ΔG°f (kJ/mol) 0.0 -229 -137 0.0
S°(J/K·mol) 6.0 189 198 131
Question 13 answers
3.32 x 10-5
2.89 x 1041
1.36 x 1016
2.89 x 10-41
Question 15 text For a reaction at equilibrium, D Suniv = 0.
Question 15 answers
True
False
HELP ON ANY QUESTION IS REALLY APPRECIATED .. thanks :)
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A system delivers 1275 J of heat while the surroundings perform 855 J of work on it. Calculate ΔE in J.
Question 2 answers
-2130
-420
420
2130
ΔU = q + w
If heat is delivered, I'd assume it means absorbed? If so;
= 1275 + 855 = +2130
A system that does no work but which gains heat from the surroundings has
Question 1 answers
q > 0, ΔE < 0
q > 0, ΔE > 0
q < 0, ΔE < 0
q < 0, ΔE > 0
ΔU = q + w
If w = 0, and q is therefore positive (absorb = postive);
q > 0, ΔE > 0
Titanium tetrachloride is converted to pure titanium metal by a reduction process. The equation is
TiCl4(l) + 4Na(s) → 4NaCl(s) + Ti(s) Δ H = -840 kJ
What is the enthalpy change when 30.0 mol of titanium tetrachloride is converted to elemental titanium?
Question 4 answers
28 kJ
2.52 x 104kJ
-2.52 x 104kJ
-28 kJ
Since 840 kJ is produced per mol of TiCl4;
E = 30 x 840 = 25200 = 2.52 x 10^4
ΔH° f [AsH3(g)] = 66.4 kJ/mol; ΔH° f [H3AsO4(aq)] = -904.6 kJ/mol; ΔH° f [H2O(l)] =-285.8 kJ/mol;
H3AsO4(aq) + 4H2(g) →AsH3(g) + 4H2O(l)
ΔH = ΔH{products} - ΔH{reactants}
= [ (66.4) + (4 x -285.8) ] - [ (-904.6) + (0) ]
= -1076.8 + 904.6
= -172.2
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A system delivers 1275 J of heat while the surroundings perform 855 J of work on it. Calculate ΔE in J.
Question 2 answers
-2130
-420
420
2130
ΔU = q + w
If heat is delivered, I'd assume it means absorbed? If so;
= 1275 + 855 = +2130
A system that does no work but which gains heat from the surroundings has
Question 1 answers
q > 0, ΔE < 0
q > 0, ΔE > 0
q < 0, ΔE < 0
q < 0, ΔE > 0
ΔU = q + w
If w = 0, and q is therefore positive (absorb = postive);
q > 0, ΔE > 0
Titanium tetrachloride is converted to pure titanium metal by a reduction process. The equation is
TiCl4(l) + 4Na(s) → 4NaCl(s) + Ti(s) Δ H = -840 kJ
What is the enthalpy change when 30.0 mol of titanium tetrachloride is converted to elemental titanium?
Question 4 answers
28 kJ
2.52 x 104kJ
-2.52 x 104kJ
-28 kJ
Since 840 kJ is produced per mol of TiCl4;
E = 30 x 840 = 25200 = 2.52 x 10^4
ΔH° f [AsH3(g)] = 66.4 kJ/mol; ΔH° f [H3AsO4(aq)] = -904.6 kJ/mol; ΔH° f [H2O(l)] =-285.8 kJ/mol;
H3AsO4(aq) + 4H2(g) →AsH3(g) + 4H2O(l)
ΔH = ΔH{products} - ΔH{reactants}
= [ (66.4) + (4 x -285.8) ] - [ (-904.6) + (0) ]
= -1076.8 + 904.6
= -172.2
Thanks alot for the help mate.
but if the system delivers . .dosnt that mean getting rid of heat? e.g its exothermic? thus negative?
thanks again mate.
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luken mate can you help with these couple of questions from the wiley test,i can't do them!
For an octahedral complex of each of the following metal ions, determine the number of unpaired electrons for both high spin and low spin configurations. In some cases, both may be the same.
Zn2+
Number of unpaired electrons for high spin:
Number of unpaired electrons for low spin:
Cr2+
Number of unpaired electrons for high spin:
Number of unpaired electrons for low spin:
Co2+
Number of unpaired electrons for high spin:
Number of unpaired electrons for low spin:
Rh3+
Number of unpaired electrons for high spin:
Number of unpaired electrons for low spin:
and
f a complex appears orange-yellow, what colour light does it absorb?
orange-yellow
blue-green
blue
yellow-green
green-blue
blue-violet
red
violet-blue
red-violet
green
yellow
orange
What colour light is absorbed if the complex appears orange?
violet-blue
blue-green
orange-yellow
green
red
red-violet
green-blue
yellow-green
blue
orange
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A system delivers 1275 J of heat while the surroundings perform 855 J of work on it. Calculate ΔE in J.
Question 2 answers
-2130
-420
420
2130
ΔU = q + w
If heat is delivered, I'd assume it means absorbed? If so;
= 1275 + 855 = +2130
A system that does no work but which gains heat from the surroundings has
Question 1 answers
q > 0, ΔE < 0
q > 0, ΔE > 0
q < 0, ΔE < 0
q < 0, ΔE > 0
ΔU = q + w
If w = 0, and q is therefore positive (absorb = postive);
q > 0, ΔE > 0
Titanium tetrachloride is converted to pure titanium metal by a reduction process. The equation is
TiCl4(l) + 4Na(s) → 4NaCl(s) + Ti(s) Δ H = -840 kJ
What is the enthalpy change when 30.0 mol of titanium tetrachloride is converted to elemental titanium?
Question 4 answers
28 kJ
2.52 x 104kJ
-2.52 x 104kJ
-28 kJ
Since 840 kJ is produced per mol of TiCl4;
E = 30 x 840 = 25200 = 2.52 x 10^4
ΔH° f [AsH3(g)] = 66.4 kJ/mol; ΔH° f [H3AsO4(aq)] = -904.6 kJ/mol; ΔH° f [H2O(l)] =-285.8 kJ/mol;
H3AsO4(aq) + 4H2(g) →AsH3(g) + 4H2O(l)
ΔH = ΔH{products} - ΔH{reactants}
= [ (66.4) + (4 x -285.8) ] - [ (-904.6) + (0) ]
= -1076.8 + 904.6
= -172.2
Thanks alot for the help mate.
but if the system delivers . .dosnt that mean getting rid of heat? e.g its exothermic? thus negative?
thanks again mate.
Hmmm, yeah that kinda sounds right. Just wait for confirmation perhaps :P
Hey JDog, do you have the wiley text? All the answers are there... If not, just download the solutions from blackboard, and go to chap 13, they show all you need to know for the first question.
For the second, there's a simple graphic in wiley chap 13 where you just look at the complimentary. From here, I learnt that blue-green and green-blue are in fact different colours :P
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mm i haven't got the wiley text...
maybe ill give it a guess.
i tried the complementary for my previous two attempts and they didn't work !
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mm i haven't got the wiley text...
maybe ill give it a guess.
i tried the complementary for my previous two attempts and they didn't work !
NO DON'T, I'LL LOOK IT UP NOW!
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Zn2+
Number of unpaired electrons for high spin: 0
Number of unpaired electrons for low spin: 0
Cr2+
Number of unpaired electrons for high spin: 4
Number of unpaired electrons for low spin: 2
Co2+
Number of unpaired electrons for high spin: 3
Number of unpaired electrons for low spin: 1
Rh3+
Number of unpaired electrons for high spin: 4
Number of unpaired electrons for low spin: 0
f a complex appears orange-yellow, what colour light does it absorb? - Blue
What colour light is absorbed if the complex appears orange? - Blue Green (NOT green blue, there is a difference haha)
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btw, if you wanna know how I derived the answers for the first question, look at the solution for 13.78 in chapt 13...
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just in time!
you sir are a top bloke.
thanks!
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thanks guys so far
was wondering if anyone could help me on the remaining questions
much appreciated , Matt.
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Q7:
The general format for formation reactions is identical to addition and synthesis reactions, but the reactants must be elements and the product must have coefficient of 1: A (element) + B (element) 1 AB (more reactants can be present)
I'm gonna say H2(g) + 2O(g) → H2O2(l) , only because the O would/should perhaps be O2 rather than 2O?
Q8:
S refers to entropy, the measure of disorder in a system.
More molecules on one side implies that it has got more disordered, so I'd say Δ S° > 0
Q9:
S increases as particles move from solid -> aq -> g
CO2(g), CO2(s)
The rest are just simple calculations :)
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Q7:
The general format for formation reactions is identical to addition and synthesis reactions, but the reactants must be elements and the product must have coefficient of 1: A (element) + B (element) 1 AB (more reactants can be present)
I'm gonna say H2(g) + 2O(g) → H2O2(l) , only because the O would/should perhaps be O2 rather than 2O?
Q8:
S refers to entropy, the measure of disorder in a system.
More molecules on one side implies that it has got more disordered, so I'd say Δ S° > 0
Q9:
S increases as particles move from solid -> aq -> g
CO2(g), CO2(s)
The rest are just simple calculations :)
Cheers
LOL its the simple calculations i get mixed up about
those stupid Negetive signs! hahaha
thanks alot for the help btw.
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like how do i do these
Calculate the reaction free energy of
2NO(g) + O2(g) 2NO2(g)
when the concentrations are 0.06 mol L-1 (NO), 0.03 mol L-1 (O2), and 1.27 mol L-1 (NO2), and the temperature is 500K. For this reaction Kc= 45 at 500 K.
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http://en.wikipedia.org/wiki/Gibbs_free_energy#Useful_identities