what is the oxidizing and reducing agents in Mg+CuSO4=MgSO4+Cu, 3Mg+2N=Mg3N2, ZnI2=Zn+2I?
Q. what is the oxidizing and reducing agents in Mg+CuSO4=MgSO4+Cu, 3Mg+2N=Mg3N2, ZnI2=Zn+2I?
Asked by moodle15 - Fri May 2 11:51:10 2008 - - 1 Answers - 0 Comments
A. Oxidation is 1) loss or partial loss of electrons 2) Increase in oxidation number. Mg+CuSO4=MgSO4+Cu Mg got oxidised by Cu++. Cu++ got reduced by Mg. 3Mg+2N=Mg3N2 Should be N2, usually. Like oxygen, N will act as an oxidising agent when put next to a metal. Mg gets oxidised and N gets reduced. N is the oxidising agent and Mg is the reducing agent. ZnI2=Zn+2I Zn++ becomes Zn i.e. it gets reduced. I- become I i.e. it gets oxidised. Zn++ is the oxidising agent, and I- is the reducing agent.
Answered by Sciman - Fri May 2 12:00:03 2008
Q. what is the oxidizing and reducing agents in Mg+CuSO4=MgSO4+Cu, 3Mg+2N=Mg3N2, ZnI2=Zn+2I?
Asked by moodle15 - Fri May 2 11:51:10 2008 - - 1 Answers - 0 Comments
A. Oxidation is 1) loss or partial loss of electrons 2) Increase in oxidation number. Mg+CuSO4=MgSO4+Cu Mg got oxidised by Cu++. Cu++ got reduced by Mg. 3Mg+2N=Mg3N2 Should be N2, usually. Like oxygen, N will act as an oxidising agent when put next to a metal. Mg gets oxidised and N gets reduced. N is the oxidising agent and Mg is the reducing agent. ZnI2=Zn+2I Zn++ becomes Zn i.e. it gets reduced. I- become I i.e. it gets oxidised. Zn++ is the oxidising agent, and I- is the reducing agent.
Answered by Sciman - Fri May 2 12:00:03 2008
What is the Emperial Formula of Zn+I2 --> ZnI2?
Q. What is the Emperial Formula of Zn+I2 --> ZnI2?
Asked by pvspartan99 - Wed Oct 8 01:41:22 2008 - - 1 Answers - 0 Comments
Q. What is the Emperial Formula of Zn+I2 --> ZnI2?
Asked by pvspartan99 - Wed Oct 8 01:41:22 2008 - - 1 Answers - 0 Comments
Lewis structure for ZnI2 and K2O?
Q. Checking my answers ZnI2 - Zinc has 2 valence shell electrons, Iodine has 7 * 2 = 14 Total of 16 . . : I : I Zn I : I : ' ' Potassium Oxide - (K2O) K2 has 2 valence electrons, O has 6 total of 8 k i k \ :O:
Asked by shawn s - Sun Nov 11 14:38:48 2007 - - 1 Answers - 0 Comments
A. ZnI2 is correct but the second one is wrong. The Oxygen is in the middle of two K's K--O--K Just draw 4 electrons (2 pairs) around the oxygen and you are good to go. so 4 bonded electrons: 2 per bond and 4 lone pairs around oxygen Total = electrons
Answered by Farooq - Sun Nov 11 17:17:22 2007
Q. Checking my answers ZnI2 - Zinc has 2 valence shell electrons, Iodine has 7 * 2 = 14 Total of 16 . . : I : I Zn I : I : ' ' Potassium Oxide - (K2O) K2 has 2 valence electrons, O has 6 total of 8 k i k \ :O:
Asked by shawn s - Sun Nov 11 14:38:48 2007 - - 1 Answers - 0 Comments
A. ZnI2 is correct but the second one is wrong. The Oxygen is in the middle of two K's K--O--K Just draw 4 electrons (2 pairs) around the oxygen and you are good to go. so 4 bonded electrons: 2 per bond and 4 lone pairs around oxygen Total = electrons
Answered by Farooq - Sun Nov 11 17:17:22 2007
I had to find the empirical formula for Zinc Iodide for a chem lab.?
Q. I know that the empirical formula is ZnI2, but when we did the experiment, we got 0.12g of Zn and 0.25g of I2. When calculated, the zinc to iodine ratio isn't 1:2. Did I make errors in the experiment?
Asked by mia - Sun Jan 18 23:34:30 2009 - - 1 Answers - 0 Comments
A. I think you just didn't multiply the Iodine's mass by 2. I got the right answer by doing this: Zn= 1x0.12g=0.12g I= 2x0.25g=0.50g Zn- 0.12 x 1mol/65.41g/mol= 0.002 mol I- 0.50 x 1mol/126.90= 0.004 mol Zn- 0.002/0.002= 1 I- 0.004/0.002=2 Therefore ratio is 1:2. So the formula is ZnI2
Answered by ll22 - Mon Jan 19 00:36:39 2009
Q. I know that the empirical formula is ZnI2, but when we did the experiment, we got 0.12g of Zn and 0.25g of I2. When calculated, the zinc to iodine ratio isn't 1:2. Did I make errors in the experiment?
Asked by mia - Sun Jan 18 23:34:30 2009 - - 1 Answers - 0 Comments
A. I think you just didn't multiply the Iodine's mass by 2. I got the right answer by doing this: Zn= 1x0.12g=0.12g I= 2x0.25g=0.50g Zn- 0.12 x 1mol/65.41g/mol= 0.002 mol I- 0.50 x 1mol/126.90= 0.004 mol Zn- 0.002/0.002= 1 I- 0.004/0.002=2 Therefore ratio is 1:2. So the formula is ZnI2
Answered by ll22 - Mon Jan 19 00:36:39 2009
Write balanced equation, using the smallest possible integer coefficients, for the following.?
Q. Please include physical states. ZnBr2(aq) and I2(g): (there is a reaction. The answer is NOT no reaction. Also, it is a displacement. The answer is not ZnBr2(aq)+I2(g)-->ZnI2(aq )+Br2(l) I have no idea why. I got this wrong a million times.Please help)
Asked by chemstinks926 - Mon Dec 8 04:20:15 2008 - - 1 Answers - 0 Comments
A. The only product I can think of is an ion such as (BrI2)- (IBr2)- does exist but I haven`t seen or heard of the former.
Answered by aaj - Mon Dec 8 04:49:48 2008
Q. Please include physical states. ZnBr2(aq) and I2(g): (there is a reaction. The answer is NOT no reaction. Also, it is a displacement. The answer is not ZnBr2(aq)+I2(g)-->ZnI2(aq )+Br2(l) I have no idea why. I got this wrong a million times.Please help)
Asked by chemstinks926 - Mon Dec 8 04:20:15 2008 - - 1 Answers - 0 Comments
A. The only product I can think of is an ion such as (BrI2)- (IBr2)- does exist but I haven`t seen or heard of the former.
Answered by aaj - Mon Dec 8 04:49:48 2008
Theoretical yield, percent yield, limiting reactant?
Q. You start reaction with 1.00 g of Zn and 1.00 g of I2. Calculate the theoretical yield of ZnI2 fromthe reactants. Find the limiting and excess reacant as well as the mass of excess reactant. If the actual yield is 1.12 g of ZnI2, what is the percent yield? Theoretical Yeild: ___ Limiting reactant: ___ Excess reacant: ___ grams of ___ Percent yield: ___
Asked by CHM Help - Mon Jun 15 20:55:15 2009 - - 1 Answers - 0 Comments
A. Zn + I2 = ZnI2 moles Zn = 1.00/ 65.39=0.0153 moles I2 = 1.00 / 253.808 =0.00394 ( limiting reactant) moles ZnI2 = 0.00394 mass ZnI2 = 0.00394 x 319.198=1.26 g moles Zn in excess = 0.0153 - 0.00394=0.0114 mass Zn in excess = 0.0114 x 65.39= 0.743 g % yield = 1.12 x 100/ 1.26 =88.9
Answered by Dr.A - Tue Jun 16 05:15:43 2009
Q. You start reaction with 1.00 g of Zn and 1.00 g of I2. Calculate the theoretical yield of ZnI2 fromthe reactants. Find the limiting and excess reacant as well as the mass of excess reactant. If the actual yield is 1.12 g of ZnI2, what is the percent yield? Theoretical Yeild: ___ Limiting reactant: ___ Excess reacant: ___ grams of ___ Percent yield: ___
Asked by CHM Help - Mon Jun 15 20:55:15 2009 - - 1 Answers - 0 Comments
A. Zn + I2 = ZnI2 moles Zn = 1.00/ 65.39=0.0153 moles I2 = 1.00 / 253.808 =0.00394 ( limiting reactant) moles ZnI2 = 0.00394 mass ZnI2 = 0.00394 x 319.198=1.26 g moles Zn in excess = 0.0153 - 0.00394=0.0114 mass Zn in excess = 0.0114 x 65.39= 0.743 g % yield = 1.12 x 100/ 1.26 =88.9
Answered by Dr.A - Tue Jun 16 05:15:43 2009
question about freezing point depression?
Q. with the formula deltaT = i Kf m what is the i number for Cr(CH3COO)3 and what is it for ZnI2? Because according to my homework the results i have gotten with Cr(CH3COO)3 with i as 4, and ZnI2 with i as 3 is giving me answers that dont make sense
Asked by Sean F - Tue Jul 10 18:08:08 2007 - - 2 Answers - 0 Comments
A. You know that--- i = no. of particle after dissociation / no of particle before dissociation. Cr(CH3COO)---> Cr^+3 + CH3COO^- i = 1+1 / 1 = 2 ZnI2---> Zn^+2 + I- i = 1+1/1 = 2
Answered by sillu s - Tue Jul 10 18:30:41 2007
Q. with the formula deltaT = i Kf m what is the i number for Cr(CH3COO)3 and what is it for ZnI2? Because according to my homework the results i have gotten with Cr(CH3COO)3 with i as 4, and ZnI2 with i as 3 is giving me answers that dont make sense
Asked by Sean F - Tue Jul 10 18:08:08 2007 - - 2 Answers - 0 Comments
A. You know that--- i = no. of particle after dissociation / no of particle before dissociation. Cr(CH3COO)---> Cr^+3 + CH3COO^- i = 1+1 / 1 = 2 ZnI2---> Zn^+2 + I- i = 1+1/1 = 2
Answered by sillu s - Tue Jul 10 18:30:41 2007
Checking my answer for lewis structure of:?
Q. Lewis structure for ZnI2 and K2O? Checking my answers ZnI2 - Zinc has 2 valence shell electrons, Iodine has 7 * 2 = 14 Total of 16 . . : I : I Zn I : I : ' ' Potassium Oxide - (K2O) K2 has 2 valence electrons, O has 6 total of 8 K I K \ :O: 48 minutes ago - 3 days left to answer. Report It
Asked by shawn s - Sun Nov 11 15:29:10 2007 - - 1 Answers - 0 Comments
A. your ZnI2 is correct , but the K2O isn't: K has only one valence electron and O has 6 K \ :O: / K
Answered by klimbim - Sun Nov 11 15:40:58 2007
Q. Lewis structure for ZnI2 and K2O? Checking my answers ZnI2 - Zinc has 2 valence shell electrons, Iodine has 7 * 2 = 14 Total of 16 . . : I : I Zn I : I : ' ' Potassium Oxide - (K2O) K2 has 2 valence electrons, O has 6 total of 8 K I K \ :O: 48 minutes ago - 3 days left to answer. Report It
Asked by shawn s - Sun Nov 11 15:29:10 2007 - - 1 Answers - 0 Comments
A. your ZnI2 is correct , but the K2O isn't: K has only one valence electron and O has 6 K \ :O: / K
Answered by klimbim - Sun Nov 11 15:40:58 2007
Molarity chemistry?
Q. although there are many different concentration units the one we will use is molairty. Verify numerically that all of the above are possible expression for the concentration of zinc iodide solution? a. 2.56g ZnI2/ 500 mL of solution b. 0.00512g ZnI2/ mL of solution c.0.00806 moles of ZNI2 / 500 mL of soltuion d.0.00806 moles of ZnI2/ 0.5L of solution e.0.0161 moles of ZnI2/L of solution.
Asked by Neha G - Sun Nov 25 21:15:41 2007 - - 1 Answers - 0 Comments
A. a. 2.56g ZnI2/ 500 ml (mol ZnI2/ molec mass ZnI2) (1000 ml/l) = X mol/L b. same conversion type c. .00806 mol/ 500ml (1000ml/l) = x mol/L d. exact same as C since .5L= 500ml e. Already molarity as it is mol/L you can punch the numbers into your calculator-- sorry if you were expecting someone do all your HW for you.
Answered by Abby J - Sun Nov 25 21:33:42 2007
Q. although there are many different concentration units the one we will use is molairty. Verify numerically that all of the above are possible expression for the concentration of zinc iodide solution? a. 2.56g ZnI2/ 500 mL of solution b. 0.00512g ZnI2/ mL of solution c.0.00806 moles of ZNI2 / 500 mL of soltuion d.0.00806 moles of ZnI2/ 0.5L of solution e.0.0161 moles of ZnI2/L of solution.
Asked by Neha G - Sun Nov 25 21:15:41 2007 - - 1 Answers - 0 Comments
A. a. 2.56g ZnI2/ 500 ml (mol ZnI2/ molec mass ZnI2) (1000 ml/l) = X mol/L b. same conversion type c. .00806 mol/ 500ml (1000ml/l) = x mol/L d. exact same as C since .5L= 500ml e. Already molarity as it is mol/L you can punch the numbers into your calculator-- sorry if you were expecting someone do all your HW for you.
Answered by Abby J - Sun Nov 25 21:33:42 2007
Quick zinc iodide question?
Q. How do I calculate how much the reactants cost to prepare zinc iodide? Zn + I2 ---> ZnI2 Zinc - $31.25 per 500g Iodine - $74.90 per 500g
Asked by cloud851 - Wed Mar 19 15:47:58 2008 - - 1 Answers - 0 Comments
A. Zn (65.9 grams/mol) + I2 (253.8 grams/mol) ---> ZnI2 (319.7grams) Zn (65.9 grams) @ 31.25 / 500g = $4.12 I2 (253.8 grams) @ 74.90 / 500g = $38.02 so: ZnI2 (319.7grams/mol) = $42.14 --- possible answers to make ZnI2 = $42.14 per mole of ZnI2 or @ 319.7 g/mol ... $ 0.1318 per gram ZnI2
Answered by Steve O - Wed Mar 19 18:57:35 2008
Q. How do I calculate how much the reactants cost to prepare zinc iodide? Zn + I2 ---> ZnI2 Zinc - $31.25 per 500g Iodine - $74.90 per 500g
Asked by cloud851 - Wed Mar 19 15:47:58 2008 - - 1 Answers - 0 Comments
A. Zn (65.9 grams/mol) + I2 (253.8 grams/mol) ---> ZnI2 (319.7grams) Zn (65.9 grams) @ 31.25 / 500g = $4.12 I2 (253.8 grams) @ 74.90 / 500g = $38.02 so: ZnI2 (319.7grams/mol) = $42.14 --- possible answers to make ZnI2 = $42.14 per mole of ZnI2 or @ 319.7 g/mol ... $ 0.1318 per gram ZnI2
Answered by Steve O - Wed Mar 19 18:57:35 2008
Concentration and Ksp?
Q. I just had an exam, these 2 questions I had no idea how to start or finish. The professor didnt help me much. Now i am looking on the method use so i can learn... 1.) A detergent solution has a pH of 11.63 at 25oC. What are the hydrogen and hydroxide ion concentration and dissociation constant if molarity of solution was 2.5 x 10-2 M? 2.) Which of these substance is more soluble (g/L) in water; ZnS (Ksp = 1.10 x 10-21) or Zn(OH)2 (Ksp = 2.1 x 10-16)? Calculate the solubility of each. Would the solubility increase or decrease in 0.005 M ZnI2 solution. Explain. my answer for 2.) ZnS-->Zn2++S2- x x 1.10 x 10-21 = x2x = 3.3 x 10-11 Zn(OH)2 --> Zn2++ 2OH- x 2x 2.1 x 10-16 = 4x3x = 3.7 x 10-6 Zn(OH)2 > ZnS I dont know how to… [cont.]
Asked by unknown - Wed Jun 25 01:21:03 2008 - - 1 Answers - 0 Comments
A. 1.) You can calculate [H ] and [OH ] from pH: pH = -log [H ] => [H ] = 10^(-pH ) = 10^(-11.63) = 2.34 10 M Moreover [H ] and [OH ] satisfy self-ionization equlibrium of water: [H ] [OH ] = 10 => [OH ] = 10 / [H ] = 10^(pH - 14) = 10^(-2.37) = 4.27 10 M Let the detergent dissciate as follows: roh r + OH then equilibrium constant is given by: Kb = [R ] [OH ] / [ROH] Since contribution of self dissociation of water to the formation of hydroxide ions is small compared to the detergent, it may be ignored. Hence: [R ] [OH ] The actual concentration of non-dissociated detergent is: [ROH] = [ROH] - [R ] = [ROH] - [OH ] ([ROH] is the apparent concetration of detergent) Therefore: Kb = [OH ] / ([ROH] - [OH ]) = (4.27 10 ) / (2 [cont.]
Answered by schmiso - Wed Jun 25 03:49:47 2008
Q. I just had an exam, these 2 questions I had no idea how to start or finish. The professor didnt help me much. Now i am looking on the method use so i can learn... 1.) A detergent solution has a pH of 11.63 at 25oC. What are the hydrogen and hydroxide ion concentration and dissociation constant if molarity of solution was 2.5 x 10-2 M? 2.) Which of these substance is more soluble (g/L) in water; ZnS (Ksp = 1.10 x 10-21) or Zn(OH)2 (Ksp = 2.1 x 10-16)? Calculate the solubility of each. Would the solubility increase or decrease in 0.005 M ZnI2 solution. Explain. my answer for 2.) ZnS-->Zn2++S2- x x 1.10 x 10-21 = x2x = 3.3 x 10-11 Zn(OH)2 --> Zn2++ 2OH- x 2x 2.1 x 10-16 = 4x3x = 3.7 x 10-6 Zn(OH)2 > ZnS I dont know how to… [cont.]
Asked by unknown - Wed Jun 25 01:21:03 2008 - - 1 Answers - 0 Comments
A. 1.) You can calculate [H ] and [OH ] from pH: pH = -log [H ] => [H ] = 10^(-pH ) = 10^(-11.63) = 2.34 10 M Moreover [H ] and [OH ] satisfy self-ionization equlibrium of water: [H ] [OH ] = 10 => [OH ] = 10 / [H ] = 10^(pH - 14) = 10^(-2.37) = 4.27 10 M Let the detergent dissciate as follows: roh r + OH then equilibrium constant is given by: Kb = [R ] [OH ] / [ROH] Since contribution of self dissociation of water to the formation of hydroxide ions is small compared to the detergent, it may be ignored. Hence: [R ] [OH ] The actual concentration of non-dissociated detergent is: [ROH] = [ROH] - [R ] = [ROH] - [OH ] ([ROH] is the apparent concetration of detergent) Therefore: Kb = [OH ] / ([ROH] - [OH ]) = (4.27 10 ) / (2 [cont.]
Answered by schmiso - Wed Jun 25 03:49:47 2008
Chemistry Naming Compounds!?
Q. 1. CaO 2. K2S 3. LiCl 4. BaF2 5. CuCl 6. FeO 7. Na3N 8. Al2O3 9. CrCl6 10. CuI2 11. Ca3P2 12. CoF3 13. MnO2 14. MgO 15. ZnI2 16. CuBr2 17. NiO 18. Fe2O3 19. KI 20. Cr2O3
Asked by jayme - Fri Dec 5 23:57:32 2008 - - 1 Answers - 0 Comments
Q. 1. CaO 2. K2S 3. LiCl 4. BaF2 5. CuCl 6. FeO 7. Na3N 8. Al2O3 9. CrCl6 10. CuI2 11. Ca3P2 12. CoF3 13. MnO2 14. MgO 15. ZnI2 16. CuBr2 17. NiO 18. Fe2O3 19. KI 20. Cr2O3
Asked by jayme - Fri Dec 5 23:57:32 2008 - - 1 Answers - 0 Comments
help!! stoichiometry!! 10 points?
Q. Zn + I2---> ZnI2 Determine the theoretical yield if a 126.0 g sample of zinc was used. ___ g Determine the percent yield if 510.6 g product is recovered. ___% thank you! Yeah thats what i thought. I was trying to do this problem but i thought there was supposed to be 2 quantities in order to do this problem. hm well if anyone can help me that would be really appreciated!
Asked by Bella - Mon Mar 10 02:23:56 2008 - - 2 Answers - 0 Comments
A. Well, assuming that there is an excess of Iodine, then: 126.0/63.59= 1.98 mol Zn. 1.98 mol Zn (1 mol ZnI2/ 1mol Zn)= 1.98 mol ZnI2 1.98 mol ZnI2 (379.8g/mol)=752.004g ZnI2 (theoretical (maximum) yield) 510.6/ 752.004= 67.9 % yield You're not really missing anything. if there was a mass given for iodine, then you would have to use stoichiometry to see which of the two reactants are limiting, then solve for product mass using the limiting reactant. If only on reactant mass is given, the assumption is that the other reactant is in excess.
Answered by Teancum2010 - Mon Mar 10 02:45:27 2008
Q. Zn + I2---> ZnI2 Determine the theoretical yield if a 126.0 g sample of zinc was used. ___ g Determine the percent yield if 510.6 g product is recovered. ___% thank you! Yeah thats what i thought. I was trying to do this problem but i thought there was supposed to be 2 quantities in order to do this problem. hm well if anyone can help me that would be really appreciated!
Asked by Bella - Mon Mar 10 02:23:56 2008 - - 2 Answers - 0 Comments
A. Well, assuming that there is an excess of Iodine, then: 126.0/63.59= 1.98 mol Zn. 1.98 mol Zn (1 mol ZnI2/ 1mol Zn)= 1.98 mol ZnI2 1.98 mol ZnI2 (379.8g/mol)=752.004g ZnI2 (theoretical (maximum) yield) 510.6/ 752.004= 67.9 % yield You're not really missing anything. if there was a mass given for iodine, then you would have to use stoichiometry to see which of the two reactants are limiting, then solve for product mass using the limiting reactant. If only on reactant mass is given, the assumption is that the other reactant is in excess.
Answered by Teancum2010 - Mon Mar 10 02:45:27 2008
Chem help?!? Please?
Q. The spectator ions in the reaction ZnCl2 + 2LiOH ---> LiCl + Zn(OH)2 are.. The spectator ions in the reaction KOH + HCl ---> PbI2 + KNO3 are.. The spectator ions in the reaction Pb(NO3)2 + KI --> PbI2 + KNO3 are... The spectator ions in the reaction ZnI2 +NaOH --> NaI + Zn(OH)2 are...
Asked by Samantha Jo - Tue Jan 9 20:09:51 2007 - - 2 Answers - 0 Comments
A. Spectator ions are the ones that just sit there and do nothing. 1. Good question. I would need to know if either LiCl or Zn(OH)2 is a solid or not. Do you have that info? 2. the end half of that equation should be H2O + KCl, and the answer would be K+ and Cl- because H2O is a liquid 3. I'm just gonna guess that PbI2 is a solid. In which case it's K+ and NO3- 4. Again with the phases. If it's not written in the question, try looking it up in your solubility rules. Spectator ions are the ones that stay aqueous, or are soluble.
Answered by violet - Tue Jan 9 20:20:43 2007
Q. The spectator ions in the reaction ZnCl2 + 2LiOH ---> LiCl + Zn(OH)2 are.. The spectator ions in the reaction KOH + HCl ---> PbI2 + KNO3 are.. The spectator ions in the reaction Pb(NO3)2 + KI --> PbI2 + KNO3 are... The spectator ions in the reaction ZnI2 +NaOH --> NaI + Zn(OH)2 are...
Asked by Samantha Jo - Tue Jan 9 20:09:51 2007 - - 2 Answers - 0 Comments
A. Spectator ions are the ones that just sit there and do nothing. 1. Good question. I would need to know if either LiCl or Zn(OH)2 is a solid or not. Do you have that info? 2. the end half of that equation should be H2O + KCl, and the answer would be K+ and Cl- because H2O is a liquid 3. I'm just gonna guess that PbI2 is a solid. In which case it's K+ and NO3- 4. Again with the phases. If it's not written in the question, try looking it up in your solubility rules. Spectator ions are the ones that stay aqueous, or are soluble.
Answered by violet - Tue Jan 9 20:20:43 2007
help with Chemistry hw?
Q. II. Decomposition Reactions 1. In these reactions, a single substance is broken down into two or more simpler substances. (Reverse of direct combination reactions) 2. Write the correct formula for the decomposition products on the right hand side of the arrow. Denote all gaseous products with an up arrow. KClO3 ---> KCl + O2 3. Balance the equation by inspection as in direct combination reactions. 2KClO3 ---> 2KCl + 3O2 Complete and balance the following equations: 1.MgCO3 ---> + 2.HgO ---> + 3. PCl5 ---> + 4. (NH4)2Cr2O7 ---> + + N2 5. CuO ---> + 6. Hg2CO3 ---> HgO + 7. KNO3 ---> + O2 8. H3PO4 ---> P2O5 + 9. H2SO4 ---> + 10. Ca(OH)2 ---> CaO + 11. H2SO3 ---> + 12. NH3 ---> + 13. MgSO4 7H2O ---> + 14. CuSO4 5H2O ---> + 15. Pb(NO3)2 ---> [cont.]
Asked by Kabao T - Sat Dec 6 18:00:06 2008 - - 1 Answers - 0 Comments
A. Something tells me nobody is gonna go and answer all those questions^ If you really can't work it out, there are plenty of websites that can offer solutions.
Answered by noan - Sat Dec 6 18:13:11 2008
Q. II. Decomposition Reactions 1. In these reactions, a single substance is broken down into two or more simpler substances. (Reverse of direct combination reactions) 2. Write the correct formula for the decomposition products on the right hand side of the arrow. Denote all gaseous products with an up arrow. KClO3 ---> KCl + O2 3. Balance the equation by inspection as in direct combination reactions. 2KClO3 ---> 2KCl + 3O2 Complete and balance the following equations: 1.MgCO3 ---> + 2.HgO ---> + 3. PCl5 ---> + 4. (NH4)2Cr2O7 ---> + + N2 5. CuO ---> + 6. Hg2CO3 ---> HgO + 7. KNO3 ---> + O2 8. H3PO4 ---> P2O5 + 9. H2SO4 ---> + 10. Ca(OH)2 ---> CaO + 11. H2SO3 ---> + 12. NH3 ---> + 13. MgSO4 7H2O ---> + 14. CuSO4 5H2O ---> + 15. Pb(NO3)2 ---> [cont.]
Asked by Kabao T - Sat Dec 6 18:00:06 2008 - - 1 Answers - 0 Comments
A. Something tells me nobody is gonna go and answer all those questions^ If you really can't work it out, there are plenty of websites that can offer solutions.
Answered by noan - Sat Dec 6 18:13:11 2008
Titrating the Zinc Iodide with the EDTA 4- Solution?
Q. You weigh 0.237 g of zinc iodide. You dissolve this zinc iodide in a Erlenmeyer flask and plan to titrate it with the EDTA 4- which has a molarity of 0.0189 M. You load the EDTA4- solution into the buret and take an initial volume reading of the buret. The initial volume is 0.55 ml. You then titrate the EDTA 4- into the zinc iodide solution until the endpoint is reached. The final volume reading of the buret is 42.35ml a)how many mls of EDTA 4- did you react with the zinc iodide? b)How many moles of EDTA 4- did you react with the zinc iodide? c)How many moles of zinc reacted with the EDTA 4- ? d)How many grams of zinc are in your zinc iodide sample? e)What is the experimentally determined percent ( multiply by 100 ) by mass of zinc in… [cont.]
Asked by Bla - Tue Dec 18 12:30:08 2007 - - 1 Answers - 0 Comments
A. a) 42.35 - 0.55 = 41.8 mL b) moles = 0.0189 x 41.8 / 1000 = 7.90 x 10^-4 c) 7.90 x 10^-4 d) 7.90 x 10^-4 x 65.39 g/mol =0.0517 g e) 0.0517 x 100 / 0.237 = 21.8 %
Answered by Dr.A - Tue Dec 18 13:29:58 2007
Q. You weigh 0.237 g of zinc iodide. You dissolve this zinc iodide in a Erlenmeyer flask and plan to titrate it with the EDTA 4- which has a molarity of 0.0189 M. You load the EDTA4- solution into the buret and take an initial volume reading of the buret. The initial volume is 0.55 ml. You then titrate the EDTA 4- into the zinc iodide solution until the endpoint is reached. The final volume reading of the buret is 42.35ml a)how many mls of EDTA 4- did you react with the zinc iodide? b)How many moles of EDTA 4- did you react with the zinc iodide? c)How many moles of zinc reacted with the EDTA 4- ? d)How many grams of zinc are in your zinc iodide sample? e)What is the experimentally determined percent ( multiply by 100 ) by mass of zinc in… [cont.]
Asked by Bla - Tue Dec 18 12:30:08 2007 - - 1 Answers - 0 Comments
A. a) 42.35 - 0.55 = 41.8 mL b) moles = 0.0189 x 41.8 / 1000 = 7.90 x 10^-4 c) 7.90 x 10^-4 d) 7.90 x 10^-4 x 65.39 g/mol =0.0517 g e) 0.0517 x 100 / 0.237 = 21.8 %
Answered by Dr.A - Tue Dec 18 13:29:58 2007
Due in 35 minutes! please help me! Its Urgent?
Q. 1) Classify the following reaction: Cl2 + ZnI2 --> ZnCl2 + I2 a) combustion b) formation c) decomposition d) single replacement 2) Classify the following chemical reaction: 2 Pb(NO3)2 --> 2 PbO + 4 NO2 + O2 a) combustion b) formation c) decomposition d) single replacement 3) The molecular shapes for carbon tetrachloride and hydrogen bromide are a) linear and angular b) tetrahedral and V-shaped c) tetrahedral and linear d) pyramidal and linear
Asked by Sumbal - Wed Jul 29 12:54:24 2009 - - 7 Answers - 0 Comments
A. 1) Single Replacement because I2 is replaced by Cl 2) Decomposition because the compound is breaking down 3) Tetrahedral and linear
Answered by unknown - Wed Jul 29 13:02:03 2009
Q. 1) Classify the following reaction: Cl2 + ZnI2 --> ZnCl2 + I2 a) combustion b) formation c) decomposition d) single replacement 2) Classify the following chemical reaction: 2 Pb(NO3)2 --> 2 PbO + 4 NO2 + O2 a) combustion b) formation c) decomposition d) single replacement 3) The molecular shapes for carbon tetrachloride and hydrogen bromide are a) linear and angular b) tetrahedral and V-shaped c) tetrahedral and linear d) pyramidal and linear
Asked by Sumbal - Wed Jul 29 12:54:24 2009 - - 7 Answers - 0 Comments
A. 1) Single Replacement because I2 is replaced by Cl 2) Decomposition because the compound is breaking down 3) Tetrahedral and linear
Answered by unknown - Wed Jul 29 13:02:03 2009
Chemistry: Can someone tell me if my work is right?
Q. Ok, this is for my Chemistry Lab and we are supposed to figure out the empirical formula of Zinc Iodide. If I am right I should get ZnI2 right? Well let me show the work I did and someone please tell me if I did this right or what I did wrong. Question: Determine the empirical formula of the zinc iodide product (ZnI n ), considering these two assumptions. 1. The I2 was the limiting regeant and 2. The reaction can be assumed to proceed in 100% yield. My Work: # moles of Zn = .590 g (this is how much we used in lab) / 65.38 = .009024166 mols # moles of I2 = 1.012 g ( how much we used in lab) / 253.8 = .003987392 mols So the limiting regeant is Iodide then, but the question says assume the Iodide is the limiting regeant so was I even… [cont.]
Asked by ManZ - Sun Nov 2 15:44:20 2008 - - 1 Answers - 0 Comments
A. about the limiting regeant, it's just the substance that is used up before the other...you don't do anything to it...i think you should get ZnI2
Answered by mai - Sun Nov 2 16:08:43 2008
Q. Ok, this is for my Chemistry Lab and we are supposed to figure out the empirical formula of Zinc Iodide. If I am right I should get ZnI2 right? Well let me show the work I did and someone please tell me if I did this right or what I did wrong. Question: Determine the empirical formula of the zinc iodide product (ZnI n ), considering these two assumptions. 1. The I2 was the limiting regeant and 2. The reaction can be assumed to proceed in 100% yield. My Work: # moles of Zn = .590 g (this is how much we used in lab) / 65.38 = .009024166 mols # moles of I2 = 1.012 g ( how much we used in lab) / 253.8 = .003987392 mols So the limiting regeant is Iodide then, but the question says assume the Iodide is the limiting regeant so was I even… [cont.]
Asked by ManZ - Sun Nov 2 15:44:20 2008 - - 1 Answers - 0 Comments
A. about the limiting regeant, it's just the substance that is used up before the other...you don't do anything to it...i think you should get ZnI2
Answered by mai - Sun Nov 2 16:08:43 2008
Are the following a covalent bond, ionic bond or metallic bond?
Q. 1)ZnI2 2)Ag 3)NaCl 4)C(Graphite) 5)C3H8 6)Cu
Asked by blahblahblah - Sat Dec 22 11:53:20 2007 - - 2 Answers - 0 Comments
A. ZnI2 and NaCl => ionic bond Ag and Cu => metallic bond C (graphite) and C3H8 => covalent bond
Answered by Dr.A - Sat Dec 22 12:02:52 2007
Q. 1)ZnI2 2)Ag 3)NaCl 4)C(Graphite) 5)C3H8 6)Cu
Asked by blahblahblah - Sat Dec 22 11:53:20 2007 - - 2 Answers - 0 Comments
A. ZnI2 and NaCl => ionic bond Ag and Cu => metallic bond C (graphite) and C3H8 => covalent bond
Answered by Dr.A - Sat Dec 22 12:02:52 2007
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