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| | (1 recommendation so far) | Message 1 of 22 in Discussion |
| From:  Abalo3 (Original Message) | Sent: 1/22/2008 12:09 AM |
I am preparing to take a last test and have questions regarding some exercises. You can give some suggestions.
Thank you.
1. How many electrons are in the outermost shell of the In3+ ion in its ground state?
a.2
b.3
c.6
d.18
2. Which of these elements has the highest first electron affinity?
a.Ca
b. N
c. Ne
d. S
3. What are the coefficients in front of NO3 (-1)(aq) and Zn(s) when the following equation is balanced in a basic solution?
----NO3(-1)(aq) + ----Zn(s)------>----Zn(+2)(aq) + ---NO(g) |
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| | | From: ·Steve· | Sent: 1/29/2008 3:28 AM |
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2. You are given ΔH for the reaction
I2 (g) ––�?gt; I2 (s)
and you want ΔH for the sublimation reaction
I2 (s) ––�?gt; I2 (g)
3. The molecular geometry of SF6 is _________. (It's octahedral.) You can look at an illustration of this geometry to see what the F-S-F angles are (and memorize!).
4. This is a "products minus reactants" calculation in which you are given ΔH for the reaction and are solving for ΔHf of one of the substances in the reaction.
ΔHrxn = [ 2(ΔHf of CHCl3 (l) + 3(ΔHf of H2 (g) ] �? [ 2(ΔHf of CH4 (g) + 3(ΔHf of Cl2 (g) ]
Plug in all the numbers you are given and solve for ΔHf of CH4 (g). Be careful about the signs (easy to make mistakes!).
Are you sure this is the correct reaction? If you actually do it, you would get HCl (g) instead of H2 (g). But your reaction works "on paper!"
Steve |
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| | | From: Abalo3 | Sent: 1/29/2008 4:59 PM |
I would like to see if I made a mistake.
1. Step# 1
CH4 + 2O2---------> CO2 + 2H2O
Step#2
Moles of CH4
=3.2 g * ( 1mole/16.0107 g CH4)
= 0.1998663 moles
Moles of Oxygene
=12.8g O2 * ( 1 mole/32g O2)
= 0.4 mole O2
Mole of CO2
= 8.8 g CO2 * ( 1 mole/44.0107g CO2)
= 0.19995137 mole CO2
Step#3
Mole of desired substance H2O
Mole of H2O(Based on CH4) = (0.1998663 mole CH4/1)* (2 moles H2)/1 mole Ch4)
= 0.3997326 mole H2O
Mole of H2O (Based on O2) = (0.4 mole O2 /1)* (2 moles H2O)/2 mole O2)
= 0.4 mole H2O
Mole of H2O(Based on CO2) = (0.19995137 mole CO2/1)* (2 moles H2O)/ 1 mole CO2)
= 0.39990274 mole H2O
Conversion into gram
=0.3997326 mole H20 * 18g/ mole H20
=7.1951868g
=7.2g
4. This is the correct problem:
For the reaction
2CH4 (g) + 3Cl2 (g)-------> 2CHCl3 (l) + 3H2(g)
DeltaH degree= -118.6kJ . DeltaH degree f = -134.1 kJ/mol for CHCl3 (l). Find DeltaH degree f for CH4 (g).
But ΔHf of H2 (g) was not given.How to do the calculation???
Mole of H2O (Based on O2) = (0.4 mole O2 /1 ) * (2moles/1mole O2)
= 0.4 mole H20 |
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| | | From: Abalo3 | Sent: 1/29/2008 5:00 PM |
Disregard this message "Mole of H2O (Based on O2) = (0.4 mole O2 /1 ) * (2moles/1mole O2)
= 0.4 mole H20" on the bottom of my previous note sent.
Thank you. |
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| | | From: ·Steve· | Sent: 1/29/2008 5:10 PM |
Ooops, I see we're posting at the same time! I have to get to class right now, but I will check these later today.  |
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| | | From: ·Steve· | Sent: 1/29/2008 8:10 PM |
Nice job on #1, 7.2 g of H2O is correct. Neither reactant was limiting. Starting with the grams of CO2 also gives the correct answer. Total mass of reactants = 3.2 g + 12.8 g = 16.0 g Total mass of products = 8.8 g + 7.2 g = 16.0 g   >> But ΔHf of H2 (g) was not given. How to do the calculation??? << ΔHf of an element in its natural state at 25° and 1 atm is.................. ZERO! |
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| | | From: Abalo3 | Sent: 1/30/2008 4:22 AM |
I am still struggling with:
4. Find DeltaH degree f for CH4 (g)
ΔHrxn = [ 2(ΔHf of CHCl3 (l) + 3(ΔHf of H2 (g) ] �? [ 2(ΔHf of CH4 (g) + 3(ΔHf of Cl2 (g) ]
= [2(-134.1kJ/molCHCl3 (l) + 3(0) ΔHf of H2 (g) ] - [ 2x of CH4 (g) + 3(-118.6 of Cl2 (g) ]
= -268.2kJ/mol CHCl3(l) -2xof CH4 (g) + 355.8of Cl2 (g)
= 43.8 kJ/mol
5. Which statement about elemental analysis by combustion is not correct?
a. carbon is determined from teh amount of CO2 formed
b. hydrogen is determined from the amount of H2O formed
c. Oxygen is determined from the amount of H2O formed
d. Only carbon can be determined directly from CO2 and H2O
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| | | From: ·Steve· | Sent: 1/30/2008 5:09 AM |
4. You omitted ΔHrxn, �?18.6 kJ. The answer from a table of standard enthalpies of formations is �?4.6 kJ/mol for the ΔHf° of CH4 (g), so you should get something close to that. 5. C is not correct. Oxygen is usually determined by subtracting the percentages of the other elements from 100% if no other percentages are missing. Or, calculate the grams of C from the CO2 and the grams of H from the H2O and subtract these weights from the sample weight to get the grams of O in the sample, again if there are no other elements that were undetermined. The percentage of oxygen can be determined, but not from the amount of H2O formed. Answer D is not quite correct either. Carbon is determined from the amount of CO2 formed, but not from the amount of H2O. |
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| | | From: Abalo3 | Sent: 1/30/2008 2:50 PM |
4.Find DeltaH degree f for CH4 (g)
ΔHrxn = [ 2(ΔHf of CHCl3 (l) + 3(ΔHf of H2 (g) ] �?[ 2(ΔHf of CH4 (g) + 3(ΔHf of Cl2 (g) ]
-118.6kJ = [2(-134.1kJ/molCHCl3 (l) + 3(0) ΔHf of H2 (g) ] - [ 2x of CH4 (g) + 3(-118.6 of Cl2 (g) ]
-118.6kJ = -268.2kJ/mol CHCl3(l) -2xof CH4 (g) + 355.8of Cl2 (g)
-118.6kJ/mol = 87.6 kJ/mol - 2x
-206.2kJ/mol/2= x
103.1kJ/mol=x
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| | | From: Abalo3 | Sent: 1/30/2008 6:48 PM |
4.Find DeltaH degree f for CH4 (g)
ΔHrxn = [ 2(ΔHf of CHCl3 (l) + 3(ΔHf of H2 (g) ] �?[ 2(ΔHf of CH4 (g) + 3(ΔHf of Cl2 (g) ]
-118.6kJ = [2(-134.1kJ/molCHCl3 (l) + 3(0) ΔHf of H2 (g) ] - [ 2x of CH4 (g) + 3(-92.3 of Cl2 (g) ]
-118.6kJ = -268.2kJ/mol CHCl3(l) -2xof CH4 (g) - 276.9 of Cl2 (g)
-118.6kJ/mol = -545.1 kJ/mol - 2x
426.5 kJ/mol/- 2 = x
-213.25 kJ/mol =x
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| | | From: ·Steve· | Sent: 1/30/2008 7:41 PM |
You're not using the correct ΔHf for Cl2 (g). Look it up in the table of thermodynamic data in your textbook. Hint: Cl2 (g) is an element in its natural state at 25°C and 1 atm, just like H2 (g).  Let me know that you found the table! |
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| | | From: Abalo3 | Sent: 1/30/2008 10:34 PM |
According to the book the value of ΔHf for Cl2 (g)=0.
Therefore,
-118.6kJ = [2(-134.1kJ/molCHCl3 (l) + 3(0) ΔHf of H2 (g) ] - [ 2x of CH4 (g) + 3(0) of Cl2 (g) ]
-118.6kJ/mol = [268.2 kJ/mol CHCl3 (l) ]- [2x CH4
59.3kJ/mol= x
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| | | From: ·Steve· | Sent: 1/30/2008 10:45 PM |
Right-O. The rule is, ΔHf of an element in its natural state at 25° and 1 atm is zero (Message 13). You have all the right numbers there, but you are making a mistake in solving for x. Remember that the correct value for ΔHf of CH4 (g) is �?4.6 kJ/mol, which you can also verify from the table if that entry is in it. So you know that is about what x should come out to be. |
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| | | From: Abalo3 | Sent: 1/30/2008 11:14 PM |
| Thank you for the clarification on that.. |
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