When NH3 gas is introduced at one end of a long tube while HCl is introduced simultaneously at the other end..
Q. a ring of white ammonium chloride is observed to form in the tube after a few minutes. The ring is closer to the HCl end of the tube than the NH3 end. Why is this?
Asked by Shelby M - Tue Jan 1 10:50:41 2008 - - 3 Answers - 0 Comments
A. The NH3 diffuses more quickly than the HCl. This is because the size of a molecule of NH3 is smaller than that of a molecule of HCl. The mollecular weight of ammonia is 17, while that of Hydrogen Chloride is about 38. According to graham's law of effusion, a gas effuses (diffuses) at a greater rate when the molecular mass is smaller, which is another way of saying the particles can move more quickly through the air in the tube. Therefore, NH3 molecules are moving more quickly down the tube than the HCl molecules and they meet more towards the HCl end. When they meet, the reaction occurs between the two and the ring forms where the reation occurs. Therefore, the ring is close to the HCl side.
Answered by e1b23456 - Tue Jan 1 10:53:32 2008
Q. a ring of white ammonium chloride is observed to form in the tube after a few minutes. The ring is closer to the HCl end of the tube than the NH3 end. Why is this?
Asked by Shelby M - Tue Jan 1 10:50:41 2008 - - 3 Answers - 0 Comments
A. The NH3 diffuses more quickly than the HCl. This is because the size of a molecule of NH3 is smaller than that of a molecule of HCl. The mollecular weight of ammonia is 17, while that of Hydrogen Chloride is about 38. According to graham's law of effusion, a gas effuses (diffuses) at a greater rate when the molecular mass is smaller, which is another way of saying the particles can move more quickly through the air in the tube. Therefore, NH3 molecules are moving more quickly down the tube than the HCl molecules and they meet more towards the HCl end. When they meet, the reaction occurs between the two and the ring forms where the reation occurs. Therefore, the ring is close to the HCl side.
Answered by e1b23456 - Tue Jan 1 10:53:32 2008
What is the equivalent in moles of 135 L of ammonia (NH3) gas at STP?
Q. What is the equivalent in moles of 135 L of ammonia (NH3) gas at STP?
Asked by m C - Thu Feb 8 02:15:53 2007 - - 2 Answers - 0 Comments
A. The formula that you need to know is 22.4 L = 1 mole so 135 L / 22.4 L * 1 mole = 6.03 moles (135 L divided by 22.4 L times 1 mole equals 6.03 moles) Thanks
Answered by Josh - Thu Feb 8 02:28:00 2007
Q. What is the equivalent in moles of 135 L of ammonia (NH3) gas at STP?
Asked by m C - Thu Feb 8 02:15:53 2007 - - 2 Answers - 0 Comments
A. The formula that you need to know is 22.4 L = 1 mole so 135 L / 22.4 L * 1 mole = 6.03 moles (135 L divided by 22.4 L times 1 mole equals 6.03 moles) Thanks
Answered by Josh - Thu Feb 8 02:28:00 2007
In the gas phase one mole of N2 and three moles of H2 react to form two moles of NH3?
Q. In the gas phase one mole of N2 and three moles of H2 react to form two moles of NH3 At T = 773 Kelvin , Kp = 0.00067 atm-2 Calculate Kp, T = 773 Kelvin , for the reaction where two moles of N2 and six moles of H2 react to form four moles of NH3 .
Asked by student - Wed Feb 27 23:13:58 2008 - - 1 Answers - 0 Comments
A. 0.00067^2 = 4.489e-9
Answered by hehe - Thu Feb 28 16:33:34 2008
Q. In the gas phase one mole of N2 and three moles of H2 react to form two moles of NH3 At T = 773 Kelvin , Kp = 0.00067 atm-2 Calculate Kp, T = 773 Kelvin , for the reaction where two moles of N2 and six moles of H2 react to form four moles of NH3 .
Asked by student - Wed Feb 27 23:13:58 2008 - - 1 Answers - 0 Comments
A. 0.00067^2 = 4.489e-9
Answered by hehe - Thu Feb 28 16:33:34 2008
Two vessels are labeled A and B. Vessel A contains NH3 gas, and vessel B contains He gas. Both A and B are at?
Q. Two vessels are labeled A and B. Vessel A contains NH3 gas, and vessel B contains He gas. Both A and B are at the same temperature. If the average kinetic energy of NH3 is 7.1 10-21 J/molecule, calculate the mean-square speed of He atoms in m2/s2...?
Asked by Nick W - Tue Mar 24 13:04:59 2009 - - 1 Answers - 0 Comments
Q. Two vessels are labeled A and B. Vessel A contains NH3 gas, and vessel B contains He gas. Both A and B are at the same temperature. If the average kinetic energy of NH3 is 7.1 10-21 J/molecule, calculate the mean-square speed of He atoms in m2/s2...?
Asked by Nick W - Tue Mar 24 13:04:59 2009 - - 1 Answers - 0 Comments
The Rate of diffusion of an unknown gas is exactly half that of ammonia gas, NH3.?
Q. Calculate the molar mass of the unknown gas. Thanks!
Asked by BOB - Wed Jan 23 21:21:40 2008 - - 1 Answers - 0 Comments
A. Gramham's Law Kinetic energy x molar mass = Kinetic energy x molar mass Kinetic energy = thermal energy (aka heat) If the temp is the same, then molar mass ratios are the same. If the unknown is 2x as slow as ammonia - then its twice as heavy.
Answered by drbillmacmo - Wed Jan 23 21:35:56 2008
Q. Calculate the molar mass of the unknown gas. Thanks!
Asked by BOB - Wed Jan 23 21:21:40 2008 - - 1 Answers - 0 Comments
A. Gramham's Law Kinetic energy x molar mass = Kinetic energy x molar mass Kinetic energy = thermal energy (aka heat) If the temp is the same, then molar mass ratios are the same. If the unknown is 2x as slow as ammonia - then its twice as heavy.
Answered by drbillmacmo - Wed Jan 23 21:35:56 2008
Why is NH3 a gas at room temperature, but H2O is a liquid?
Q. Why is NH3 a gas at room temperature, but H2O is a liquid?
Asked by Shelby M - Tue Jan 1 11:13:47 2008 - - 4 Answers - 0 Comments
A. Both NH3 and H2O form hydrogen bonds. Each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of + hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia. In ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there aren't enough lone pairs to go around to satisfy all the hydrogens. Therefore each ammonia molecule can only bond to 2 other molecules. Sciman's point about electronegativity does play a part in determining the strength of a hydrogen bond but in this case is is more… [cont.]
Answered by Brandon M - Tue Jan 1 11:45:03 2008
Q. Why is NH3 a gas at room temperature, but H2O is a liquid?
Asked by Shelby M - Tue Jan 1 11:13:47 2008 - - 4 Answers - 0 Comments
A. Both NH3 and H2O form hydrogen bonds. Each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of + hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia. In ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there aren't enough lone pairs to go around to satisfy all the hydrogens. Therefore each ammonia molecule can only bond to 2 other molecules. Sciman's point about electronegativity does play a part in determining the strength of a hydrogen bond but in this case is is more… [cont.]
Answered by Brandon M - Tue Jan 1 11:45:03 2008
Chemistry question - NH3 gas?
Q. Chemistry question - NH3 gas? Consider the following equilibrium: N2 (g) + 3 H2 (g) 2 NH3 (g) H=-92.94 kJ Which of the following changes will shift the equilibrium to the right? Tick all correct answers. A. decreasing the pressure. B. decreasing the volume. C. increasing the temperature. D. increasing the volume. E. removing some N2. F. removing some NH3. G. decreasing the temperature. H. adding aome NH3. I. adding some N2. J. increasing the pressure. Thanks for any helper!
Asked by WWW - Mon Sep 21 02:37:08 2009 - - 2 Answers - 0 Comments
A. Ill write the letters which will shift it towards the right F. G. I. J The volume ones are ambigous, if it was increase/decrease the volume of the whole system (which i assume is), then it will have no effect on the equilbrium.
Answered by darkf1re - Mon Sep 21 03:59:00 2009
Q. Chemistry question - NH3 gas? Consider the following equilibrium: N2 (g) + 3 H2 (g) 2 NH3 (g) H=-92.94 kJ Which of the following changes will shift the equilibrium to the right? Tick all correct answers. A. decreasing the pressure. B. decreasing the volume. C. increasing the temperature. D. increasing the volume. E. removing some N2. F. removing some NH3. G. decreasing the temperature. H. adding aome NH3. I. adding some N2. J. increasing the pressure. Thanks for any helper!
Asked by WWW - Mon Sep 21 02:37:08 2009 - - 2 Answers - 0 Comments
A. Ill write the letters which will shift it towards the right F. G. I. J The volume ones are ambigous, if it was increase/decrease the volume of the whole system (which i assume is), then it will have no effect on the equilbrium.
Answered by darkf1re - Mon Sep 21 03:59:00 2009
How many liters of NH3 gas at 25 degrees celcius and 1.46 atm pressure are required to prepare 2.00L of a 3.50
Q. How many liters of NH3 gas at 25 degrees celcius and 1.46 atm pressure are required to prepare 2.00L of a 3.50
Asked by Emily R - Wed Apr 23 12:01:09 2008 - - 1 Answers - 0 Comments
Q. How many liters of NH3 gas at 25 degrees celcius and 1.46 atm pressure are required to prepare 2.00L of a 3.50
Asked by Emily R - Wed Apr 23 12:01:09 2008 - - 1 Answers - 0 Comments
How many molecules of NH3 gas are present in a 1.00-L flask of NH3 gas at STP:?
Q. How many molecules of NH3 gas are present in a 1.00-L flask of NH3 gas at STP:?
Asked by Antonio H - Wed Dec 5 20:07:11 2007 - - 1 Answers - 0 Comments
A. 1L 22.4mol/L 6.02 10^23molecules/mol = 1.348 10^25molecules of NH3
Answered by Igthomque - Fri Dec 7 19:41:02 2007
Q. How many molecules of NH3 gas are present in a 1.00-L flask of NH3 gas at STP:?
Asked by Antonio H - Wed Dec 5 20:07:11 2007 - - 1 Answers - 0 Comments
A. 1L 22.4mol/L 6.02 10^23molecules/mol = 1.348 10^25molecules of NH3
Answered by Igthomque - Fri Dec 7 19:41:02 2007
Write a balanced equation for the reaction of nitrogen gas and hydrogen gas to produce Ammonia gas (NH3)?
Q. B) If 212.5g of ammonia gas is produced, how many molecules of hydrogen gas must have reacted with nitrogen? I need full steps...I don't get it...
Asked by Darius P - Sun Feb 22 15:04:22 2009 - - 1 Answers - 0 Comments
Q. B) If 212.5g of ammonia gas is produced, how many molecules of hydrogen gas must have reacted with nitrogen? I need full steps...I don't get it...
Asked by Darius P - Sun Feb 22 15:04:22 2009 - - 1 Answers - 0 Comments
What is the % yield? If 1.70 Mg of ammonia, NH3, is reacted 2.80 Mg oxygen, O2, according to NH3(gas + 2 O2 (g?
Q. full question: What is the % yield? If 1.70 Mg of ammonia, NH3, is reacted 2.80 Mg oxygen, O2, according to NH3(gas + 2 O2 (g) ---> HNO3(aq) + H2O(l) , and 3.20 Mg of nitric acid, HNO3, actaully rercovered, What is the % yield? Identify the limiting agent. (1 Mg=10^6g = 1 metric ton)
Asked by MO - Thu Jul 16 14:21:59 2009 - - 1 Answers - 0 Comments
A. NH3(g) + 2 O2 (g) ---> HNO3(aq) + H2O(l) If all NH3 reacts: (1.70 Mg)(Mmol NH3 / 17 Mg)(1 Mmol HNO3 / 1 Mmol NH3) = 0.1 Mmol HNO3 If all O2 reacts: (2.80 Mg)(Mmol O2 / 32 Mg)(1 Mmol HNO3 / 2 Mmol O2) = 0.04375 Mmol HNO3 O2 will be the limiting reagent. (0.04375 Mmol HNO3)(63 Mg / Mmol) = 2.756 Mg HNO3 The actual yield is greater than the theoretically optimum yield -- the values in this problem are bogus. % yield = 3.20/2.756 * 100 = 116% (?)
Answered by not_publius - Sun Jul 19 15:15:23 2009
Q. full question: What is the % yield? If 1.70 Mg of ammonia, NH3, is reacted 2.80 Mg oxygen, O2, according to NH3(gas + 2 O2 (g) ---> HNO3(aq) + H2O(l) , and 3.20 Mg of nitric acid, HNO3, actaully rercovered, What is the % yield? Identify the limiting agent. (1 Mg=10^6g = 1 metric ton)
Asked by MO - Thu Jul 16 14:21:59 2009 - - 1 Answers - 0 Comments
A. NH3(g) + 2 O2 (g) ---> HNO3(aq) + H2O(l) If all NH3 reacts: (1.70 Mg)(Mmol NH3 / 17 Mg)(1 Mmol HNO3 / 1 Mmol NH3) = 0.1 Mmol HNO3 If all O2 reacts: (2.80 Mg)(Mmol O2 / 32 Mg)(1 Mmol HNO3 / 2 Mmol O2) = 0.04375 Mmol HNO3 O2 will be the limiting reagent. (0.04375 Mmol HNO3)(63 Mg / Mmol) = 2.756 Mg HNO3 The actual yield is greater than the theoretically optimum yield -- the values in this problem are bogus. % yield = 3.20/2.756 * 100 = 116% (?)
Answered by not_publius - Sun Jul 19 15:15:23 2009
At STP, which gas has properties most similar to those of an ideal gas?1.NH3 2.CO2 3.O2 4.H2?
Q. At STP, which gas has properties most similar to those of an ideal gas? NH3 CO2 O2 H2
Asked by Q+A 22 - Sun Nov 25 11:24:46 2007 - - 3 Answers - 0 Comments
A. The lightest gas with the simplest structure will have the least attraction for other molecules. And the molecule that has the least attractive forces with other molecules will have the most ideal behavior. The answer, of course, will be hydrogen. In fact the only gas that would be more ideal in behavior would be helium. Answer: H2
Answered by Dennis M - Sun Nov 25 11:34:24 2007
Q. At STP, which gas has properties most similar to those of an ideal gas? NH3 CO2 O2 H2
Asked by Q+A 22 - Sun Nov 25 11:24:46 2007 - - 3 Answers - 0 Comments
A. The lightest gas with the simplest structure will have the least attraction for other molecules. And the molecule that has the least attractive forces with other molecules will have the most ideal behavior. The answer, of course, will be hydrogen. In fact the only gas that would be more ideal in behavior would be helium. Answer: H2
Answered by Dennis M - Sun Nov 25 11:34:24 2007
At equilibrium a mixture of N2, H2, NH3 gas at 500 C is determined to consist of...?
Q. 0.601 mol/L of N2 0.392 mol/L of H2 0.200 mol L of NH3 What is the equalibrium constant for the reaction: N2(g) + 3 H2(g) <==> 2 NH3(g) at this temperature?
Asked by Lycan - Thu Apr 17 22:10:47 2008 - - 1 Answers - 0 Comments
Q. 0.601 mol/L of N2 0.392 mol/L of H2 0.200 mol L of NH3 What is the equalibrium constant for the reaction: N2(g) + 3 H2(g) <==> 2 NH3(g) at this temperature?
Asked by Lycan - Thu Apr 17 22:10:47 2008 - - 1 Answers - 0 Comments
what is the density of a sample of ammonia gas (NH3) if the pressure is 0.580 atm and the temperature is 91.5?
Q. degress celcius? answer in units of g/L?
Asked by jennyyy - Sun May 25 16:25:03 2008 - - 1 Answers - 0 Comments
Q. degress celcius? answer in units of g/L?
Asked by jennyyy - Sun May 25 16:25:03 2008 - - 1 Answers - 0 Comments
Hydrogen gas, H2, reacts with nitrogen gas, N2 to form ammonia gas- NH3 according to the equation...?
Q. Hydrogen gas, H2, reacts with nitrogen gas, N2 to form ammonia gas- NH3 according to the equation: 3H2(g)+N2(g)---> 2NH3(g) How many grams of NH3 can be produced from 2.54 mol of N2? How many grams of H2 are needed to produce 10.14g of NH3? and How many molecules (not moles) of NH3 are produced from 8.85x10^-4 g of H2? I am unsure how to work this problem- do I first convert to moles? Any help will be appreciated, thank you.
Asked by Ash 5 - Wed Oct 15 10:29:10 2008 - - 1 Answers - 0 Comments
A. If you are given the mass of product or reactant, convert it to moles, find the amount of moles of the chemical you are asked about, and then convert it into grams. Example: How many grams of NH3 can be produced from 2.54 mol of N2? From the equation we see 2 mol of NH3 are produced from each mol of N2. That means 2.54 mol of N2 will give us 5.08 mol of NH3. One mol of NH3 has the mass of 14+3*1=17. Multiply that by 5.08 and you have the answer. The second question is the same, just convert 10.14 g NH3 into moles first. In the third question, just find the number of moles of NH3 produced and multiply it by Avogadro's number to find the number of molecules. Hope that helps.
Answered by Caps Lock (TRS_BC) - Wed Oct 15 10:37:29 2008
Q. Hydrogen gas, H2, reacts with nitrogen gas, N2 to form ammonia gas- NH3 according to the equation: 3H2(g)+N2(g)---> 2NH3(g) How many grams of NH3 can be produced from 2.54 mol of N2? How many grams of H2 are needed to produce 10.14g of NH3? and How many molecules (not moles) of NH3 are produced from 8.85x10^-4 g of H2? I am unsure how to work this problem- do I first convert to moles? Any help will be appreciated, thank you.
Asked by Ash 5 - Wed Oct 15 10:29:10 2008 - - 1 Answers - 0 Comments
A. If you are given the mass of product or reactant, convert it to moles, find the amount of moles of the chemical you are asked about, and then convert it into grams. Example: How many grams of NH3 can be produced from 2.54 mol of N2? From the equation we see 2 mol of NH3 are produced from each mol of N2. That means 2.54 mol of N2 will give us 5.08 mol of NH3. One mol of NH3 has the mass of 14+3*1=17. Multiply that by 5.08 and you have the answer. The second question is the same, just convert 10.14 g NH3 into moles first. In the third question, just find the number of moles of NH3 produced and multiply it by Avogadro's number to find the number of molecules. Hope that helps.
Answered by Caps Lock (TRS_BC) - Wed Oct 15 10:37:29 2008
nitorgen gas reacts with hydrogen gas to form ammonia gas, NH3(g)?
Q. balance the equation if 4.43 g of ammonia gas are produced how many grams of nitrogen gas would be needed? if 21.7g of ammonia are produced, how many molecules of hydrogen would be used?
Asked by tennispanther33 - Mon Jan 21 15:26:46 2008 - - 1 Answers - 0 Comments
A. N2 (g) + 3H2 (g) --> 2NH3 (g)
Answered by Christine - Mon Jan 21 15:37:08 2008
Q. balance the equation if 4.43 g of ammonia gas are produced how many grams of nitrogen gas would be needed? if 21.7g of ammonia are produced, how many molecules of hydrogen would be used?
Asked by tennispanther33 - Mon Jan 21 15:26:46 2008 - - 1 Answers - 0 Comments
A. N2 (g) + 3H2 (g) --> 2NH3 (g)
Answered by Christine - Mon Jan 21 15:37:08 2008
Please solve: 105mL of pure water(at 4 celsius) is saturted with NH3 gas, producing a solution of density....
Q. 0.9g/mL.If this solution contains 30% of NH3 by weight, calculate its volume.
Asked by Sarang - Wed Jun 6 05:16:39 2007 - - 1 Answers - 0 Comments
A. the mass of 105mL of water at 4 celsius is 105g saturated the water represent (100-30)% =70%=0.7 so the mass of the solution is 105/0.7 = 150g The density is 0.9 meaning 1 mL weights0.9g the volume is 150/0.9 =166.7mL
Answered by maussy - Wed Jun 6 05:42:57 2007
Q. 0.9g/mL.If this solution contains 30% of NH3 by weight, calculate its volume.
Asked by Sarang - Wed Jun 6 05:16:39 2007 - - 1 Answers - 0 Comments
A. the mass of 105mL of water at 4 celsius is 105g saturated the water represent (100-30)% =70%=0.7 so the mass of the solution is 105/0.7 = 150g The density is 0.9 meaning 1 mL weights0.9g the volume is 150/0.9 =166.7mL
Answered by maussy - Wed Jun 6 05:42:57 2007
three 1 L flasks at STP. Flask A contains NH3 gas, Flask B contains NO2 gas, and Flask C contains N2 gas. A.?
Q. three 1 L flasks at STP. Flask A contains NH3 gas, Flask B contains NO2 gas, and Flask C contains N2 gas. A.?
Asked by thajoka23 - Mon Jun 2 22:13:30 2008 - - 1 Answers - 0 Comments
A. Please spell out your question, so others may help you.
Answered by Hahaha - Tue Jun 3 21:18:24 2008
Q. three 1 L flasks at STP. Flask A contains NH3 gas, Flask B contains NO2 gas, and Flask C contains N2 gas. A.?
Asked by thajoka23 - Mon Jun 2 22:13:30 2008 - - 1 Answers - 0 Comments
A. Please spell out your question, so others may help you.
Answered by Hahaha - Tue Jun 3 21:18:24 2008
Calculate the density of ammonia (NH3) gas in the units of g/l in a 4.32L container at 837mm Hg and 45.0 deg.C?
Q. Calculate the density of ammonia (NH3) gas in the units of g/l in a 4.32L container at 837mm Hg and 45.0 deg.C?
Asked by Keta - Mon Nov 17 14:22:54 2008 - - 2 Answers - 0 Comments
A. Use PV = nRT to find n, number of moles (T in degrees K; P in units that match the units you are using for R) Multiply by molar mass to find grams ammonia, and then divide by volume. Does the volume matter in this question?
Answered by Paul B - Mon Nov 17 14:31:24 2008
Q. Calculate the density of ammonia (NH3) gas in the units of g/l in a 4.32L container at 837mm Hg and 45.0 deg.C?
Asked by Keta - Mon Nov 17 14:22:54 2008 - - 2 Answers - 0 Comments
A. Use PV = nRT to find n, number of moles (T in degrees K; P in units that match the units you are using for R) Multiply by molar mass to find grams ammonia, and then divide by volume. Does the volume matter in this question?
Answered by Paul B - Mon Nov 17 14:31:24 2008
The density of NH3 gas at STP is?
Q. a) 0.760g/mL b) 0.760 g/L c) 1.32 g/mL d) 1.32g/L
Asked by Victoria - Sun Apr 26 16:42:12 2009 - - 1 Answers - 0 Comments
A. The density of NH3 is 0.6826 g/L at 25 degrees Celsius and 1 atm (STP). From your choices, the answer I would pick is b) 0.760 g/L. Hope this helped!
Answered by amrka5 - Sun Apr 26 17:55:59 2009
Q. a) 0.760g/mL b) 0.760 g/L c) 1.32 g/mL d) 1.32g/L
Asked by Victoria - Sun Apr 26 16:42:12 2009 - - 1 Answers - 0 Comments
A. The density of NH3 is 0.6826 g/L at 25 degrees Celsius and 1 atm (STP). From your choices, the answer I would pick is b) 0.760 g/L. Hope this helped!
Answered by amrka5 - Sun Apr 26 17:55:59 2009
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