95) Mercury and bromine will react with each other to produce mercury(II) bromide: Hg + Br2----->HgBr2...?
Q. 95) Mercury and bromine will react with each other to produce mercury(II) bromide: Hg + Br2--->HgBr2 a) What mass of Hgr2 can be produced from the reaction of 10 g Hg and 9 g Br2? What mass of which reagent is left unreacted? b) What mass of HgBr2 can be produced from the reaction of 5 mL mercury (density=13.6 g/mL) and 5 mL bromine (density=3.10 g/mL)?
Asked by Fantasy - Fri Aug 14 03:09:39 2009 - - 1 Answers - 0 Comments
A. moles Hg = 10 g/ 200.59 g/mol=0.0499 moles Br2 = 9 / 159.808 g/mol=0.0563 ( in excess) moles HgBr2 = 0.0499 mass HgBr2 = 0.0499 x 360.398 g/mol=18.0 g moles Br2 in excess = 0.0563 - 0.0499=0.0064 mass Br2 in excess = 0.0064 x 159.808 =1.0 g mass Hg = 5 x 13.6 =68 g moles Hg = 0.339 mass Br2 = 5 x 3.10= 15.5 moles Br2 = 0.0970 ( limiting reactant) moles HgBr2 = 0.0970 mass HgBr2 = 0.0970 x 360.398 =35 g
Answered by Dr.A - Fri Aug 14 10:00:10 2009
Q. 95) Mercury and bromine will react with each other to produce mercury(II) bromide: Hg + Br2--->HgBr2 a) What mass of Hgr2 can be produced from the reaction of 10 g Hg and 9 g Br2? What mass of which reagent is left unreacted? b) What mass of HgBr2 can be produced from the reaction of 5 mL mercury (density=13.6 g/mL) and 5 mL bromine (density=3.10 g/mL)?
Asked by Fantasy - Fri Aug 14 03:09:39 2009 - - 1 Answers - 0 Comments
A. moles Hg = 10 g/ 200.59 g/mol=0.0499 moles Br2 = 9 / 159.808 g/mol=0.0563 ( in excess) moles HgBr2 = 0.0499 mass HgBr2 = 0.0499 x 360.398 g/mol=18.0 g moles Br2 in excess = 0.0563 - 0.0499=0.0064 mass Br2 in excess = 0.0064 x 159.808 =1.0 g mass Hg = 5 x 13.6 =68 g moles Hg = 0.339 mass Br2 = 5 x 3.10= 15.5 moles Br2 = 0.0970 ( limiting reactant) moles HgBr2 = 0.0970 mass HgBr2 = 0.0970 x 360.398 =35 g
Answered by Dr.A - Fri Aug 14 10:00:10 2009
Name this ionic compount. Hg2Br2!!! it says it is mercury bromide everywhere.. but i got it wrong :(?
Q. Name this ionic compount. Hg2Br2!!! it says it is mercury bromide everywhere.. but i got it wrong :(?
Asked by chemistry - Wed Jan 30 17:46:55 2008 - - 4 Answers - 0 Comments
A. Guys, the prefixes to molecular names only apply to molecules that are covalently bonded, not ionic bonded. For example, H2O (water) which is convalentely bonded, could be called dihydrogen monooxide. But Something like MgFl_2 will still be called Magnesium Fluoride because it is ionically bonded. As far as I can tell, Mercury Bromide should be the right name. Why exactly did your teacher tell you it's wrong? Teachers should be willing to explain why you got an answer wrong if you ask them.
Answered by math geek - Wed Jan 30 18:25:16 2008
Q. Name this ionic compount. Hg2Br2!!! it says it is mercury bromide everywhere.. but i got it wrong :(?
Asked by chemistry - Wed Jan 30 17:46:55 2008 - - 4 Answers - 0 Comments
A. Guys, the prefixes to molecular names only apply to molecules that are covalently bonded, not ionic bonded. For example, H2O (water) which is convalentely bonded, could be called dihydrogen monooxide. But Something like MgFl_2 will still be called Magnesium Fluoride because it is ionically bonded. As far as I can tell, Mercury Bromide should be the right name. Why exactly did your teacher tell you it's wrong? Teachers should be willing to explain why you got an answer wrong if you ask them.
Answered by math geek - Wed Jan 30 18:25:16 2008
Mercury and bromine will react with each other to produce mercury(II) bromide.Hg(l) + Br2(l) HgBr2(s)?
Q. (a) What mass of HgBr2 is produced from the reaction of 9.23 g Hg and 11.6 g Br2? What mass of which reactant is left unreacted? (b) What mass of HgBr2 is produced from the reaction of 5.30 mL of mercury (density = 13.6 g/mL) and 5.30 mL bromine (density = 3.10 g/mL)?
Asked by DaniK - Sun Oct 12 01:36:01 2008 - - 1 Answers - 0 Comments
A. I believe this is just simple stoichiometry. a) First find the moles of each reactant: n(mercury) = m / M = 9.23 / 200.59 = 0.04601 moles of mercury n(bromine) = m / M = 11.6 / 2(79.904) = 0.07259 moles of bromine One mole of mercury combines with one mole of bromine so mercury is the limiting reagent. Therefore 0.04601 moles of mercury bromide is produced. This has a mass of: n = m / M 0.04601 = m / (200.59 + 2(79.904)) m = 16.58 g 0.07259 - 0.04601 = 0.02658 moles of bromine not reacted. This has a mass of 2(79.904) x 0.02658 = 4.25 g b) This is basically the same question as above but you need to find the mass of mercury and bromine first by multiplying the volume by the density. Example: 5.30 mL x 13.6g/mL = 72.08 g
Answered by Goddord - Sun Oct 12 04:32:36 2008
Q. (a) What mass of HgBr2 is produced from the reaction of 9.23 g Hg and 11.6 g Br2? What mass of which reactant is left unreacted? (b) What mass of HgBr2 is produced from the reaction of 5.30 mL of mercury (density = 13.6 g/mL) and 5.30 mL bromine (density = 3.10 g/mL)?
Asked by DaniK - Sun Oct 12 01:36:01 2008 - - 1 Answers - 0 Comments
A. I believe this is just simple stoichiometry. a) First find the moles of each reactant: n(mercury) = m / M = 9.23 / 200.59 = 0.04601 moles of mercury n(bromine) = m / M = 11.6 / 2(79.904) = 0.07259 moles of bromine One mole of mercury combines with one mole of bromine so mercury is the limiting reagent. Therefore 0.04601 moles of mercury bromide is produced. This has a mass of: n = m / M 0.04601 = m / (200.59 + 2(79.904)) m = 16.58 g 0.07259 - 0.04601 = 0.02658 moles of bromine not reacted. This has a mass of 2(79.904) x 0.02658 = 4.25 g b) This is basically the same question as above but you need to find the mass of mercury and bromine first by multiplying the volume by the density. Example: 5.30 mL x 13.6g/mL = 72.08 g
Answered by Goddord - Sun Oct 12 04:32:36 2008
How do you find out if a chemical reaction occurs based on solubility?
Q. I have a chemistry lab that i must do, and the prelab questions are "using the solubility rules, determine whether the following reactions occur or not Lithium chloride + sodium nitrate sodium bromide + Mercury (11) nitrate strontium acetate + sodium phosphate silver nitrate + lithium bromide Potassium chloride+ silver sulfate
Asked by thatguy - Thu Mar 8 21:19:06 2007 - - 1 Answers - 0 Comments
A. Do you have solubility charts? If so, it will not be too difficult. Double replacement reactions (that's what each of your given reactions can be classified as) are characterized by the presence of a precipitate (insoluble compound) or a pure gas or liquid (usually water).
Answered by Shanny - Thu Mar 8 21:26:03 2007
Q. I have a chemistry lab that i must do, and the prelab questions are "using the solubility rules, determine whether the following reactions occur or not Lithium chloride + sodium nitrate sodium bromide + Mercury (11) nitrate strontium acetate + sodium phosphate silver nitrate + lithium bromide Potassium chloride+ silver sulfate
Asked by thatguy - Thu Mar 8 21:19:06 2007 - - 1 Answers - 0 Comments
A. Do you have solubility charts? If so, it will not be too difficult. Double replacement reactions (that's what each of your given reactions can be classified as) are characterized by the presence of a precipitate (insoluble compound) or a pure gas or liquid (usually water).
Answered by Shanny - Thu Mar 8 21:26:03 2007
What is the name of the compound HgBr2?
Q. What is the name of the compound HgBr2? A. Mercuric bromate B. Mercury(II) bromide C. Mercurium brominium D. Mercury dibromine E. Hygronium bromide
Asked by Drew - Thu Jan 24 21:53:08 2008 - - 2 Answers - 0 Comments
A. B Just like CaCl2, you'll call it calcium chloride, so Hg is still called mercury (eliminate A, C and E) and Br2 you'll call bromide (eliminate D).
Answered by eugenics23 - Thu Jan 24 21:58:31 2008
Q. What is the name of the compound HgBr2? A. Mercuric bromate B. Mercury(II) bromide C. Mercurium brominium D. Mercury dibromine E. Hygronium bromide
Asked by Drew - Thu Jan 24 21:53:08 2008 - - 2 Answers - 0 Comments
A. B Just like CaCl2, you'll call it calcium chloride, so Hg is still called mercury (eliminate A, C and E) and Br2 you'll call bromide (eliminate D).
Answered by eugenics23 - Thu Jan 24 21:58:31 2008
Can someone please help me write these chemical formulas?
Q. a) copper(I)oxide b) potassium peroxide c) aluminum hydroxide d) zinc nitrate e) mercury(I)bromide f) iron(III)carbonate g) sodium hypobromite h) magnesium nitride i) potassium hypochlorite j) iron(II) sulfite
Asked by HELLO! - Thu Aug 28 13:32:50 2008 - - 1 Answers - 0 Comments
A. a) Cu2O b) K2O2 c) AlOH3 d) Zn(NO3)2 e) HgBr2 f) Fe2(CO3)3 g) NaBrO h) Mg3N2 i) KClO2 j) FeSO2
Answered by Ravz - Thu Aug 28 13:58:44 2008
Q. a) copper(I)oxide b) potassium peroxide c) aluminum hydroxide d) zinc nitrate e) mercury(I)bromide f) iron(III)carbonate g) sodium hypobromite h) magnesium nitride i) potassium hypochlorite j) iron(II) sulfite
Asked by HELLO! - Thu Aug 28 13:32:50 2008 - - 1 Answers - 0 Comments
A. a) Cu2O b) K2O2 c) AlOH3 d) Zn(NO3)2 e) HgBr2 f) Fe2(CO3)3 g) NaBrO h) Mg3N2 i) KClO2 j) FeSO2
Answered by Ravz - Thu Aug 28 13:58:44 2008
Write the formulas for the following compounds:?
Q. Lead(IV) Oxide Chromuim(III) Chloride Copper(II) Chloride Iron(III) Bromide Mercury(I) Sulphide Also, Name the following compounds. Include the numeral: Cu2O MnCl2 PbO Fel3 AuBr3
Asked by K3L53Y 15 TH3 53X. - Tue Aug 12 21:04:02 2008 - - 2 Answers - 0 Comments
A. Since each of these species is uncharged, the oxidation numbers of each of the atoms that make up the molecule must add to equal zero. The Roman Numeral given after the name of the metal is equal to the oxidation state of the metal. Also, the oxidation states of the ions attached to the metal are equal to the charge on that ion if it was isolated from the metal (example: sulfide has a charge of 2-, so the oxidation state is -2. Note the difference in notation between a charge and an oxidation state.). With these ideas in mind, let's work on your list. Lead(IV) oxide. Oxide (O) has a charge of 2- and the oxidation state on lead in this compound is +4. To have a total oxidation state of zero, this compound must have 2 oxide ions to… [cont.]
Answered by Cytosis - Tue Aug 12 21:35:43 2008
Q. Lead(IV) Oxide Chromuim(III) Chloride Copper(II) Chloride Iron(III) Bromide Mercury(I) Sulphide Also, Name the following compounds. Include the numeral: Cu2O MnCl2 PbO Fel3 AuBr3
Asked by K3L53Y 15 TH3 53X. - Tue Aug 12 21:04:02 2008 - - 2 Answers - 0 Comments
A. Since each of these species is uncharged, the oxidation numbers of each of the atoms that make up the molecule must add to equal zero. The Roman Numeral given after the name of the metal is equal to the oxidation state of the metal. Also, the oxidation states of the ions attached to the metal are equal to the charge on that ion if it was isolated from the metal (example: sulfide has a charge of 2-, so the oxidation state is -2. Note the difference in notation between a charge and an oxidation state.). With these ideas in mind, let's work on your list. Lead(IV) oxide. Oxide (O) has a charge of 2- and the oxidation state on lead in this compound is +4. To have a total oxidation state of zero, this compound must have 2 oxide ions to… [cont.]
Answered by Cytosis - Tue Aug 12 21:35:43 2008
chemical reactions with limiting reactants?
Q. So i am confused with this question and i need a place to start the problem. Mercury and bromine will react with each other to produce mercury (II) bromide: Hg (l) + Br2 (l) => HgBr2 (s) A. What is the mass of HgBr2 can be produced from the reaction of 10.0 g Hg and 9.00 g Br2? B. What mass of which reagent is left unreacted? You don't have to give the answers, but can someone walk me through or help me start the question? :] thanks!
Asked by sooper mouse! - Tue Aug 15 14:57:05 2006 - - 4 Answers - 0 Comments
A. Firstly, the reagents react in the 1:1 proportion. Now, with the help of the periodic table and a calculator we find that Hg is the limiting reagent. mol Hg = 10.0 g / 200.59 (g/mol) = 0.0499 mol mol Br2 = 9.00 g /159.8 (g/mol) = 0.0563 mol Only 0.0499 of Br2 will react and you can only get this quantity of the product. (0.0499 HgBr2) x (360.39 g/mol) = 17.98 g HgBr2 This leaves an excess of 0.0064 mol of Br2 (0.0563 - 0.0499) unreacted or 1.02 g.
Answered by DONN - Tue Aug 15 15:46:58 2006
Q. So i am confused with this question and i need a place to start the problem. Mercury and bromine will react with each other to produce mercury (II) bromide: Hg (l) + Br2 (l) => HgBr2 (s) A. What is the mass of HgBr2 can be produced from the reaction of 10.0 g Hg and 9.00 g Br2? B. What mass of which reagent is left unreacted? You don't have to give the answers, but can someone walk me through or help me start the question? :] thanks!
Asked by sooper mouse! - Tue Aug 15 14:57:05 2006 - - 4 Answers - 0 Comments
A. Firstly, the reagents react in the 1:1 proportion. Now, with the help of the periodic table and a calculator we find that Hg is the limiting reagent. mol Hg = 10.0 g / 200.59 (g/mol) = 0.0499 mol mol Br2 = 9.00 g /159.8 (g/mol) = 0.0563 mol Only 0.0499 of Br2 will react and you can only get this quantity of the product. (0.0499 HgBr2) x (360.39 g/mol) = 17.98 g HgBr2 This leaves an excess of 0.0064 mol of Br2 (0.0563 - 0.0499) unreacted or 1.02 g.
Answered by DONN - Tue Aug 15 15:46:58 2006
Write the formulas or the compound?
Q. Write the formulas or the compound a. mercury(II) bromide b. ammonium dichromate c. Lithium hydrogen sulfate d. chromium(III) nitrite
Asked by d marques - Thu Nov 30 20:48:14 2006 - - 1 Answers - 0 Comments
A. a. HgBr2 b. (NH4)2Cr207 c. LiHSO4 (bisulfate is another name for hydrogen sulfate) d. Cr(NO3)3
Answered by youresohxc56 - Thu Nov 30 20:58:39 2006
Q. Write the formulas or the compound a. mercury(II) bromide b. ammonium dichromate c. Lithium hydrogen sulfate d. chromium(III) nitrite
Asked by d marques - Thu Nov 30 20:48:14 2006 - - 1 Answers - 0 Comments
A. a. HgBr2 b. (NH4)2Cr207 c. LiHSO4 (bisulfate is another name for hydrogen sulfate) d. Cr(NO3)3
Answered by youresohxc56 - Thu Nov 30 20:58:39 2006
Stoichiometry help?
Q. I need someone to help me double check my work please. I am not lazy or a cheater so please spare me the lecture, I did my work but would like to know it is correct, thank you!!! 1) Hg + Br2 ---> HgBr2 What mass of mercury (II) bromide can be produced from the reaction if 10 g of mercury are used? 18G HgBr2 2) Fe2O3 + 2Al ---> 2Fe + Al2O3 How many grams of iron (III) oxide are required to produce 15.0g of iron ore? 21.4 g Fe2O3 3) Which elements are reduced and which oxidized? Zn + 2HCl ---> ZnCl2 + H2 Cl- reduced Zn- Oxidized H- Oxidized (I could use some pointers on how to identify each type of agent THANKS in advance!) 4) calc. molarity of solution of 5.25 g of HCl in 500 mL of water? .288 M HCl 5) How many grams of… [cont.]
Asked by j h - Sun Nov 12 21:25:10 2006 - - 1 Answers - 0 Comments
A. I see you are studying the basic course... Ask me for help when you get to *something* a little more difficult. The stuff you are working on is mere child's play for a man of my stature and intelligence. John Zonk Genius
Answered by johnny_zonker - Sun Nov 12 21:36:45 2006
Q. I need someone to help me double check my work please. I am not lazy or a cheater so please spare me the lecture, I did my work but would like to know it is correct, thank you!!! 1) Hg + Br2 ---> HgBr2 What mass of mercury (II) bromide can be produced from the reaction if 10 g of mercury are used? 18G HgBr2 2) Fe2O3 + 2Al ---> 2Fe + Al2O3 How many grams of iron (III) oxide are required to produce 15.0g of iron ore? 21.4 g Fe2O3 3) Which elements are reduced and which oxidized? Zn + 2HCl ---> ZnCl2 + H2 Cl- reduced Zn- Oxidized H- Oxidized (I could use some pointers on how to identify each type of agent THANKS in advance!) 4) calc. molarity of solution of 5.25 g of HCl in 500 mL of water? .288 M HCl 5) How many grams of… [cont.]
Asked by j h - Sun Nov 12 21:25:10 2006 - - 1 Answers - 0 Comments
A. I see you are studying the basic course... Ask me for help when you get to *something* a little more difficult. The stuff you are working on is mere child's play for a man of my stature and intelligence. John Zonk Genius
Answered by johnny_zonker - Sun Nov 12 21:36:45 2006
Ionic Compounds?
Q. I did these questions for homework, and I was wondering if you can tell me if I made any mistakes in these answers. Writing formulas for Ionic Compounds. 1. Sodium carbonate - Na*2CO*3 2. Potassium sulfite - K*2SO*3 3. Cadmium sulfate - Cd*2 (CO*3)*2 4. Mangnesium nitrate - Mg (NO*3)*2 5. Barium bromide - BaBr*2 6. Barium nitrate - Ba (NO*3)*2 7. Iron (11) oxide - FeO 8. Iron (11) sulfate - FeSO*4 9. Strontium hydroxide - Sr*3 (PO*4)*2 10. Strontium sulfide - SrS 11. Mercury (11) bromide - Hg (Br)*2 12. Calcium hypochorite - ClClO 13. Manganese (11) sulfate - MnSO*4 14. Chromium (111) sulfite - CrSO*3 15. Iron (111) oxide - FeO*2 16. Ammonium sulfite - (NH*4) Cl*3 17. Barium cyanide - Ba(CN)*2 18. Zinc thiosulfate - Zn*2(S*2O*3)*2
Asked by (: Muffin Man :) - Mon Oct 1 22:08:45 2007 - - 1 Answers - 0 Comments
A. 3. Cadmium Sulfate - CdSO4 9. Strontium Hydroxide - Sr(OH)2 12. Calcium Hypochlorite - Ca(ClO)2 14. Chromium (III) sulfite - Cr2(SO3)3 15. Iron (III) Oxide - Fe2O3 16. Ammonium Sulfite - (NH4)2SO3 18. Zinc Thiosulfate - ZnS2O3 There ya go
Answered by Skeez - Mon Oct 1 22:16:18 2007
Q. I did these questions for homework, and I was wondering if you can tell me if I made any mistakes in these answers. Writing formulas for Ionic Compounds. 1. Sodium carbonate - Na*2CO*3 2. Potassium sulfite - K*2SO*3 3. Cadmium sulfate - Cd*2 (CO*3)*2 4. Mangnesium nitrate - Mg (NO*3)*2 5. Barium bromide - BaBr*2 6. Barium nitrate - Ba (NO*3)*2 7. Iron (11) oxide - FeO 8. Iron (11) sulfate - FeSO*4 9. Strontium hydroxide - Sr*3 (PO*4)*2 10. Strontium sulfide - SrS 11. Mercury (11) bromide - Hg (Br)*2 12. Calcium hypochorite - ClClO 13. Manganese (11) sulfate - MnSO*4 14. Chromium (111) sulfite - CrSO*3 15. Iron (111) oxide - FeO*2 16. Ammonium sulfite - (NH*4) Cl*3 17. Barium cyanide - Ba(CN)*2 18. Zinc thiosulfate - Zn*2(S*2O*3)*2
Asked by (: Muffin Man :) - Mon Oct 1 22:08:45 2007 - - 1 Answers - 0 Comments
A. 3. Cadmium Sulfate - CdSO4 9. Strontium Hydroxide - Sr(OH)2 12. Calcium Hypochlorite - Ca(ClO)2 14. Chromium (III) sulfite - Cr2(SO3)3 15. Iron (III) Oxide - Fe2O3 16. Ammonium Sulfite - (NH4)2SO3 18. Zinc Thiosulfate - ZnS2O3 There ya go
Answered by Skeez - Mon Oct 1 22:16:18 2007
writing formulas, several problems.?
Q. 1) magnesium oxide 2) lithium bromide 3) Calcium nitrate 4) aluminum sulfide 5) potassium iodide 6) chromium(III) oxide 7) copper(I) sulfide 8) lead(IV) iodide 9) gold(I) flouride 10) mercury(I) bromide 11) silver carbonate 12) aluminum hydroxide 13) calcium acetate 14) lithium carbonate 15) tin(IV) chlorite 16) mercury(II) phosphate 17) tin(II) carbonate
Asked by hugemarkus - Mon Dec 17 22:51:32 2007 - - 1 Answers - 0 Comments
A. 1. Mg(OH)2 2. LiBr 3. Ca(NO3)2 4. Al2(SO4)3 5. KI 6. Cr2O3 7. Cu2S 8. PbI4 9. AuF 10. HgBr 11. Ag2CO3 12. Al(OH)3 13. Ca(CH3COO)2 14. Li2CO3 15. Sn(ClO3)4 16. Hg3(PO4)2 17. SnCO3 I hope these help.
Answered by o-O - Mon Dec 17 22:56:48 2007
Q. 1) magnesium oxide 2) lithium bromide 3) Calcium nitrate 4) aluminum sulfide 5) potassium iodide 6) chromium(III) oxide 7) copper(I) sulfide 8) lead(IV) iodide 9) gold(I) flouride 10) mercury(I) bromide 11) silver carbonate 12) aluminum hydroxide 13) calcium acetate 14) lithium carbonate 15) tin(IV) chlorite 16) mercury(II) phosphate 17) tin(II) carbonate
Asked by hugemarkus - Mon Dec 17 22:51:32 2007 - - 1 Answers - 0 Comments
A. 1. Mg(OH)2 2. LiBr 3. Ca(NO3)2 4. Al2(SO4)3 5. KI 6. Cr2O3 7. Cu2S 8. PbI4 9. AuF 10. HgBr 11. Ag2CO3 12. Al(OH)3 13. Ca(CH3COO)2 14. Li2CO3 15. Sn(ClO3)4 16. Hg3(PO4)2 17. SnCO3 I hope these help.
Answered by o-O - Mon Dec 17 22:56:48 2007
Taking AP Chem but forgot how to answer certain problems please help?
Q. 1)Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane. 2 NH3(g) + 3 O2(g) + 2 CH4(g) 2 HCN(g) + 6 H2O(g) If 6.00 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield? 2)Mercury and bromine will react with each other to produce mercury(II) bromide. Hg(l) + Br2(l) HgBr2(s)What mass of reagent is left unreacted? What mass of HgBr2 can be produced from the reaction of 2.68 mL mercury (density = 13.6 g/mL) and 2.68 mL bromine (density = 3.10 g/mL)? 3)A student prepared aspirin in a laboratory experiment using the following reaction. C7H6O3 + C4H6O3 C9H8O4 + HC2H3O2 salicylic acid acetic anhydride aspirin The student… [cont.]
Asked by Abby M - Thu Aug 14 21:43:10 2008 - - 1 Answers - 0 Comments
A. Moles NH3 = 6000 g / 17 g/mol = 353 moles O2 = 6000 g / 32 = 187.5 Moles CH4 = 6000 g / 16 = 375 ratio = 2 :3 : 2 O2 is the limiting reactant we get 187.5 x 2 / 3 = 125 moles HCN => 125 x 27 g/mol = 3375 g and 187.5 x 6 / 3 =375 moles water => 375 x 18 =6750 g Mass Hg = 2,68 x 13.6 = 36.4 g moles Hg = 36.4 / 200.6 = 0.182 mass Br2 = 2.68 x 3.10 = 8.31 Moles Br2 = 8.31 / 159.8 = 0.05200 Moles Hg in excess = 0.182 - 0.0520 =0.130 Mass Hg in excess = 0.130 x 200.6 = 26.1 g we get 0.0520 moles HgBr2 => 0.0520 x 360.4 = 18.7 g I hope this can help you
Answered by Dr.A - Fri Aug 15 11:39:56 2008
Q. 1)Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane. 2 NH3(g) + 3 O2(g) + 2 CH4(g) 2 HCN(g) + 6 H2O(g) If 6.00 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield? 2)Mercury and bromine will react with each other to produce mercury(II) bromide. Hg(l) + Br2(l) HgBr2(s)What mass of reagent is left unreacted? What mass of HgBr2 can be produced from the reaction of 2.68 mL mercury (density = 13.6 g/mL) and 2.68 mL bromine (density = 3.10 g/mL)? 3)A student prepared aspirin in a laboratory experiment using the following reaction. C7H6O3 + C4H6O3 C9H8O4 + HC2H3O2 salicylic acid acetic anhydride aspirin The student… [cont.]
Asked by Abby M - Thu Aug 14 21:43:10 2008 - - 1 Answers - 0 Comments
A. Moles NH3 = 6000 g / 17 g/mol = 353 moles O2 = 6000 g / 32 = 187.5 Moles CH4 = 6000 g / 16 = 375 ratio = 2 :3 : 2 O2 is the limiting reactant we get 187.5 x 2 / 3 = 125 moles HCN => 125 x 27 g/mol = 3375 g and 187.5 x 6 / 3 =375 moles water => 375 x 18 =6750 g Mass Hg = 2,68 x 13.6 = 36.4 g moles Hg = 36.4 / 200.6 = 0.182 mass Br2 = 2.68 x 3.10 = 8.31 Moles Br2 = 8.31 / 159.8 = 0.05200 Moles Hg in excess = 0.182 - 0.0520 =0.130 Mass Hg in excess = 0.130 x 200.6 = 26.1 g we get 0.0520 moles HgBr2 => 0.0520 x 360.4 = 18.7 g I hope this can help you
Answered by Dr.A - Fri Aug 15 11:39:56 2008
limiting reactant?
Q. Mercury and bromine will react with each other to produce mercury(II) bromide. Hg(l) + Br2(l) HgBr2(s) (a) What mass of HgBr2 is produced from the reaction of 9.78 g Hg and 10.7 g Br2? What mass of which reactant is left unreacted? (b) What mass of HgBr2 is produced from the reaction of 6.84 mL of mercury (density = 13.6 g/mL) and 6.84 mL bromine (density = 3.10 g/mL)?
Asked by misstalkalot12345 - Wed Jul 16 18:52:04 2008 - - 1 Answers - 0 Comments
A. Hg(l) + Br2(l) = HgBr2(s) Hg = 201g/mol Br2 = 79.9*2 = 159.8g/mol HgBr2 = 360.8g/mol 9.78g (Hg) : X (HgBr2) = 201g (Hg) : 360.8 (HgBr2) X = 17.15g *** 6.84mL * 13.6g/mL = 93.054g 93.024g / 201g/mol = 0.463mol 6.84mL*3.10g/mL = 21.204g 21.204g / 159.8g/mol =0.133mol 1mol(H2) : 1mol(Br2) = 0.463 : x x = 0.463 moles of Br2 > 0.133 then Br2 is the limiting reactant: 21.204g (Br2) : Xg(HgBr2) = 159.8g (Br2) : 360.8g (HgBr2) X = 47.87g
Answered by il Visitatore - Thu Jul 17 07:41:19 2008
Q. Mercury and bromine will react with each other to produce mercury(II) bromide. Hg(l) + Br2(l) HgBr2(s) (a) What mass of HgBr2 is produced from the reaction of 9.78 g Hg and 10.7 g Br2? What mass of which reactant is left unreacted? (b) What mass of HgBr2 is produced from the reaction of 6.84 mL of mercury (density = 13.6 g/mL) and 6.84 mL bromine (density = 3.10 g/mL)?
Asked by misstalkalot12345 - Wed Jul 16 18:52:04 2008 - - 1 Answers - 0 Comments
A. Hg(l) + Br2(l) = HgBr2(s) Hg = 201g/mol Br2 = 79.9*2 = 159.8g/mol HgBr2 = 360.8g/mol 9.78g (Hg) : X (HgBr2) = 201g (Hg) : 360.8 (HgBr2) X = 17.15g *** 6.84mL * 13.6g/mL = 93.054g 93.024g / 201g/mol = 0.463mol 6.84mL*3.10g/mL = 21.204g 21.204g / 159.8g/mol =0.133mol 1mol(H2) : 1mol(Br2) = 0.463 : x x = 0.463 moles of Br2 > 0.133 then Br2 is the limiting reactant: 21.204g (Br2) : Xg(HgBr2) = 159.8g (Br2) : 360.8g (HgBr2) X = 47.87g
Answered by il Visitatore - Thu Jul 17 07:41:19 2008
PLEASE HELP ME! ! ! (CHEMISTRY SOLVING)?
Q. Question in Chemistry (SOLVING)? 1. Two different compounds are formed by the elements carbon and oxygen. The first compound contains 42.9% carbon by mass and 57.1% oxygen by mass. The second compound contains 27.3% carbon by mass and 72.7% oxygen by mass. Show that the data are consistent with the law of multiple of proportions. 2. A sample of mercury (II) bromide, HgBr2 , weighs 4.26g . How many moles are in this sample? 3.How many molecules are contained in 5.22 mol of nitrogen gas, N2? 4.A sample of oxygen gas, O2, weighs 28.4g .How many molecules of O2 and how many atoms of O are present in this sample? 5. What is the mass of 2.6x 10^23 molecules of ammonia, NH3? please help!. i really need this. IT's ok if only one number you… [cont.]
Asked by sacred f - Tue Feb 3 05:25:14 2009 - - 2 Answers - 0 Comments
A. You have to take the atomic weights of each of the elements, e.g. ammonia =the weight of nitrogen and the weight of 3 hydrogens, add it all up and that's how you get the number of grams per mole of substance, then you figure the rest out by calculations.
Answered by Anagram Lacking - Tue Feb 3 05:33:44 2009
Q. Question in Chemistry (SOLVING)? 1. Two different compounds are formed by the elements carbon and oxygen. The first compound contains 42.9% carbon by mass and 57.1% oxygen by mass. The second compound contains 27.3% carbon by mass and 72.7% oxygen by mass. Show that the data are consistent with the law of multiple of proportions. 2. A sample of mercury (II) bromide, HgBr2 , weighs 4.26g . How many moles are in this sample? 3.How many molecules are contained in 5.22 mol of nitrogen gas, N2? 4.A sample of oxygen gas, O2, weighs 28.4g .How many molecules of O2 and how many atoms of O are present in this sample? 5. What is the mass of 2.6x 10^23 molecules of ammonia, NH3? please help!. i really need this. IT's ok if only one number you… [cont.]
Asked by sacred f - Tue Feb 3 05:25:14 2009 - - 2 Answers - 0 Comments
A. You have to take the atomic weights of each of the elements, e.g. ammonia =the weight of nitrogen and the weight of 3 hydrogens, add it all up and that's how you get the number of grams per mole of substance, then you figure the rest out by calculations.
Answered by Anagram Lacking - Tue Feb 3 05:33:44 2009
Give the chemical formula for each of the following ionic compounds.?
Q. chromium(III) carbonate potassium chromate sodium dihydrogen phosphate iron(III) cloride mercury(I) bromide Name the following compounds. Cu(ClO4)2 The radius of the chromium (Cr) atom is about 1.9 A (a) Express this distance in nanometers (nm). (b) How many chromium (Cr) atoms would have to be lined up to span 1.0 mm? (c) If the atom is assumed to be a sphere, what is the volume in cm3 of a single chromium (Cr) atom?
Asked by Cynthia - Sun May 25 19:34:57 2008 - - 3 Answers - 0 Comments
A. Cr2(CO3)3 K2CrO4 NaH2PO4 FeCl3 HgBr copper (II) perchlorate
Answered by science teacher - Sun May 25 19:44:41 2008
Q. chromium(III) carbonate potassium chromate sodium dihydrogen phosphate iron(III) cloride mercury(I) bromide Name the following compounds. Cu(ClO4)2 The radius of the chromium (Cr) atom is about 1.9 A (a) Express this distance in nanometers (nm). (b) How many chromium (Cr) atoms would have to be lined up to span 1.0 mm? (c) If the atom is assumed to be a sphere, what is the volume in cm3 of a single chromium (Cr) atom?
Asked by Cynthia - Sun May 25 19:34:57 2008 - - 3 Answers - 0 Comments
A. Cr2(CO3)3 K2CrO4 NaH2PO4 FeCl3 HgBr copper (II) perchlorate
Answered by science teacher - Sun May 25 19:44:41 2008
Stable Ionic Compounds... are they right?
Q. sodium chloride- NaCl beryllium bromide- BeBr2 sodium sulfide- Na2S barium phosphide- Ba3P2 potassium oxide- K2O aluminum oxide- Al2O3 cesium nitride- Cs3N aluminum nitride- AlN magnesium selenide- MgSe manganese phosphide- Mn3P2 lead (IV) oxide- PbO2 tin (II) phosphide- SnP2 copper (I) chloride- CuCl copper (I) selenide- CuSe chromium (III) fluoride- CrF3 iron (II) sulfide-FeS iron (II) nitride- Fe2 N tin (IV) hydride- SnH4 cobalt (II) bromide- CoBr2 mercury (II) hydride-HgH2
Asked by dyb - Mon Aug 3 01:55:21 2009 - - 1 Answers - 0 Comments
A. All right
Answered by Dr.A - Mon Aug 3 08:28:01 2009
Q. sodium chloride- NaCl beryllium bromide- BeBr2 sodium sulfide- Na2S barium phosphide- Ba3P2 potassium oxide- K2O aluminum oxide- Al2O3 cesium nitride- Cs3N aluminum nitride- AlN magnesium selenide- MgSe manganese phosphide- Mn3P2 lead (IV) oxide- PbO2 tin (II) phosphide- SnP2 copper (I) chloride- CuCl copper (I) selenide- CuSe chromium (III) fluoride- CrF3 iron (II) sulfide-FeS iron (II) nitride- Fe2 N tin (IV) hydride- SnH4 cobalt (II) bromide- CoBr2 mercury (II) hydride-HgH2
Asked by dyb - Mon Aug 3 01:55:21 2009 - - 1 Answers - 0 Comments
A. All right
Answered by Dr.A - Mon Aug 3 08:28:01 2009
what are the following chemicals types of reaction? (please o please help)?
Q. ex(synthesis, decomposition, single replacement, double replacement, neutralization, combustion) 1) sodium + bromine = sodium bromide 2)magnesium chlorate = magnesium chloride + oxygen 3)potassium oxide + water = potassium hydroxide 4)hydrochloric acid + sodium hydroxide = sodium chloride + water 5)gold + magnesium bromide = gold (III) bromide + magnesium 6)mercury (II) oxide = mercury + oxygen 7)diphosphorous pentoxide + water = phosphoric acid 8)propane (c3h8) + oxygen = carbon dioxide + water 9)nickel + lead (IV) nitrate = lead + nickel (II) nitrate 10)copper (II) carbonate = copper (II) oxide + carbon dioxide 11)barium chloride + copper (II) sulfate = barium sulfate + copper (II) chloride 12)aluminum + sulfur = aluminum sulfide 13)ethan [cont.]
Asked by JDSmooooth - Fri Feb 27 17:33:17 2009 - - 1 Answers - 0 Comments
A. 1) sodium + bromine = sodium bromide synthesis 2)magnesium chlorate = magnesium chloride + oxygen decomposition 3)potassium oxide + water = potassium hydroxide synthesis 4)hydrochloric acid + sodium hydroxide = sodium chloride + water double replacement 5)gold + magnesium bromide = gold (III) bromide + magnesium single replacement 6)mercury (II) oxide = mercury + oxygen decomposition 7)diphosphorous pentoxide + water = phosphoric acid synthesis 8)propane (c3h8) + oxygen = carbon dioxide + water combustion 9)nickel + lead (IV) nitrate = lead + nickel (II) nitrate single replacement 10)copper (II) carbonate = copper (II) oxide + carbon dioxide decomposition 11)barium chloride + copper (II) sulfate = barium sulfate + copper (II)… [cont.]
Answered by kodie - Fri Feb 27 17:43:31 2009
Q. ex(synthesis, decomposition, single replacement, double replacement, neutralization, combustion) 1) sodium + bromine = sodium bromide 2)magnesium chlorate = magnesium chloride + oxygen 3)potassium oxide + water = potassium hydroxide 4)hydrochloric acid + sodium hydroxide = sodium chloride + water 5)gold + magnesium bromide = gold (III) bromide + magnesium 6)mercury (II) oxide = mercury + oxygen 7)diphosphorous pentoxide + water = phosphoric acid 8)propane (c3h8) + oxygen = carbon dioxide + water 9)nickel + lead (IV) nitrate = lead + nickel (II) nitrate 10)copper (II) carbonate = copper (II) oxide + carbon dioxide 11)barium chloride + copper (II) sulfate = barium sulfate + copper (II) chloride 12)aluminum + sulfur = aluminum sulfide 13)ethan [cont.]
Asked by JDSmooooth - Fri Feb 27 17:33:17 2009 - - 1 Answers - 0 Comments
A. 1) sodium + bromine = sodium bromide synthesis 2)magnesium chlorate = magnesium chloride + oxygen decomposition 3)potassium oxide + water = potassium hydroxide synthesis 4)hydrochloric acid + sodium hydroxide = sodium chloride + water double replacement 5)gold + magnesium bromide = gold (III) bromide + magnesium single replacement 6)mercury (II) oxide = mercury + oxygen decomposition 7)diphosphorous pentoxide + water = phosphoric acid synthesis 8)propane (c3h8) + oxygen = carbon dioxide + water combustion 9)nickel + lead (IV) nitrate = lead + nickel (II) nitrate single replacement 10)copper (II) carbonate = copper (II) oxide + carbon dioxide decomposition 11)barium chloride + copper (II) sulfate = barium sulfate + copper (II)… [cont.]
Answered by kodie - Fri Feb 27 17:43:31 2009
Chemistry Homework! Please help.?
Q. i need some help with my chemistry, if you could help it will be greatly appreciated. some problems go from Name--> formula & formula --> Name. thanks very much. NO2 magnesium chloride Al2 (SO4)3 cabonic acid aluminum oxalate CoI3 BF3 NaNO3 lead (II) phosphite calcium hydroxide SrCo3 Be (ClO)2 sodium chlorite SO2 Sn (HCO3)2 Carbon Monoxide Copper (II) phosphate Mg So4 mercury (I) sulfide barium bromide Hg (HSO4)2 magnesium phosphate Ca C2 O4 magnesium nitrite Al(NO3)3 dichromic acid sodium cyanide Al(ClO4)3 P2 O5 calcium hydrogen carbonate ammonium acetate Fe So3 Al PO4 hydrosulfuric acid Ba Cr2 O7 aluminum permangante K O H H2 C2 O4 (aq) Iron (III) sulfate barium chlorate CCl4 magnesium nitrate KC2H3O2 Fe2 (CrO4)3 ammonium hydroxide Cr (C [cont.]
Asked by Longhorn_33 - Tue Dec 9 19:44:38 2008 - - 1 Answers - 0 Comments
A. Here are some of the answers: NO2-nitrite magnesium chloride-MgCl Al2 (SO4)3-Aluminum Sulfate aluminum oxalate- Al2 (C2O4) calcium hydroxide- Ca (OH) SrCo3-3- Strontium(III) Carbonate Be (ClO)2-Beryllium Hypochlorite sodium chlorite- Na2(CIO) Sn (HCO3)2-Tin(II) Hydrogen Carbonate Copper (II) phosphate-Cu3(Po4)2 Mg So4-Magnesium Sulfate
Answered by Sarah I - Tue Dec 9 20:13:57 2008
Q. i need some help with my chemistry, if you could help it will be greatly appreciated. some problems go from Name--> formula & formula --> Name. thanks very much. NO2 magnesium chloride Al2 (SO4)3 cabonic acid aluminum oxalate CoI3 BF3 NaNO3 lead (II) phosphite calcium hydroxide SrCo3 Be (ClO)2 sodium chlorite SO2 Sn (HCO3)2 Carbon Monoxide Copper (II) phosphate Mg So4 mercury (I) sulfide barium bromide Hg (HSO4)2 magnesium phosphate Ca C2 O4 magnesium nitrite Al(NO3)3 dichromic acid sodium cyanide Al(ClO4)3 P2 O5 calcium hydrogen carbonate ammonium acetate Fe So3 Al PO4 hydrosulfuric acid Ba Cr2 O7 aluminum permangante K O H H2 C2 O4 (aq) Iron (III) sulfate barium chlorate CCl4 magnesium nitrate KC2H3O2 Fe2 (CrO4)3 ammonium hydroxide Cr (C [cont.]
Asked by Longhorn_33 - Tue Dec 9 19:44:38 2008 - - 1 Answers - 0 Comments
A. Here are some of the answers: NO2-nitrite magnesium chloride-MgCl Al2 (SO4)3-Aluminum Sulfate aluminum oxalate- Al2 (C2O4) calcium hydroxide- Ca (OH) SrCo3-3- Strontium(III) Carbonate Be (ClO)2-Beryllium Hypochlorite sodium chlorite- Na2(CIO) Sn (HCO3)2-Tin(II) Hydrogen Carbonate Copper (II) phosphate-Cu3(Po4)2 Mg So4-Magnesium Sulfate
Answered by Sarah I - Tue Dec 9 20:13:57 2008
Reactions and Balancing Chemical Equations?
Q. From each question i need a) Type of reaction (ex: single displacement) b) the products c) balanced chemical equation Please help me with as such as you can and Thanks a lot guys 1. Alumin + Hydochloric Acid 2. Calcium hydroxide + nitric acid 3. Potassium chlorate heated 4. Magnesium + sulfur (S8) 5. Ammonium phosphate + alumin Chloride 6. Calcium oxide + water 7. Chromium + water 8. Tin + mercury(I) nitrate 9. Sodium Bromide + silver nitrate 10. hydrogen + oxygen 11. hydogen peroxide 12. fluorine + potasium bromide 13. Carbon dioxide + water 14. calcium chloride + ammonium hydroxide 15. Sodium + chlorine 16. bromine + sodium chloride 17. mercury (II) oxide heated 18. potassium + water 19. strontium carbonate + nitric… [cont.]
Asked by Ahmed E - Sun Oct 26 22:00:14 2008 - - 1 Answers - 0 Comments
A. 1. Alumin + Hydochloric Acid 2Al + 6HCl = 2AlCl3 + 3H2 [Single displacement] 2. Calcium hydroxide + nitric acid Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O [Double displacement] 3. Potassium chlorate heated 2KClO3 = 2KCl + 3O2 [decomposition] 4. Magnesium + sulfur (S8) 4Mg + S = Mg4S [synthesis] 5. Ammonium phosphate + alumin Chloride (NH4)3PO4 + AlCl3 = AlPO4 + 3NH4Cl [double replacement] 6. Calcium oxide + water CaO + H2O = Ca(OH)2 [Synthesis] 7. Chromium + water Cr + 6H2O = Cr(H2O)6 [synthesis] 8. Tin + mercury(I) nitrate Sn + 2HgNO3 = Sn(NO3)2 + 2Hg [Single displacement] 9. Sodium Bromide + silver nitrate NaBr + AgNO3 = NaNO3 + AgBr [double replacement] 10. hydrogen + oxygen 2H2 + O2 = 2H2O [Synthesis] 11. hydogen peroxide H2O2 = H2 +… [cont.]
Answered by kookiekim77 - Tue Oct 28 00:45:10 2008
Q. From each question i need a) Type of reaction (ex: single displacement) b) the products c) balanced chemical equation Please help me with as such as you can and Thanks a lot guys 1. Alumin + Hydochloric Acid 2. Calcium hydroxide + nitric acid 3. Potassium chlorate heated 4. Magnesium + sulfur (S8) 5. Ammonium phosphate + alumin Chloride 6. Calcium oxide + water 7. Chromium + water 8. Tin + mercury(I) nitrate 9. Sodium Bromide + silver nitrate 10. hydrogen + oxygen 11. hydogen peroxide 12. fluorine + potasium bromide 13. Carbon dioxide + water 14. calcium chloride + ammonium hydroxide 15. Sodium + chlorine 16. bromine + sodium chloride 17. mercury (II) oxide heated 18. potassium + water 19. strontium carbonate + nitric… [cont.]
Asked by Ahmed E - Sun Oct 26 22:00:14 2008 - - 1 Answers - 0 Comments
A. 1. Alumin + Hydochloric Acid 2Al + 6HCl = 2AlCl3 + 3H2 [Single displacement] 2. Calcium hydroxide + nitric acid Ca(OH)2 + 2HNO3 = Ca(NO3)2 + 2H2O [Double displacement] 3. Potassium chlorate heated 2KClO3 = 2KCl + 3O2 [decomposition] 4. Magnesium + sulfur (S8) 4Mg + S = Mg4S [synthesis] 5. Ammonium phosphate + alumin Chloride (NH4)3PO4 + AlCl3 = AlPO4 + 3NH4Cl [double replacement] 6. Calcium oxide + water CaO + H2O = Ca(OH)2 [Synthesis] 7. Chromium + water Cr + 6H2O = Cr(H2O)6 [synthesis] 8. Tin + mercury(I) nitrate Sn + 2HgNO3 = Sn(NO3)2 + 2Hg [Single displacement] 9. Sodium Bromide + silver nitrate NaBr + AgNO3 = NaNO3 + AgBr [double replacement] 10. hydrogen + oxygen 2H2 + O2 = 2H2O [Synthesis] 11. hydogen peroxide H2O2 = H2 +… [cont.]
Answered by kookiekim77 - Tue Oct 28 00:45:10 2008
From Yahoo Answer Search: 'mercury bromide'
Thu Nov 19 19:03:20 2009 [ refresh local cache ]
