ApplId/ProductId | Drug name | Active ingredient | Form | Strenght |
---|---|---|---|---|

022119/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203321/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203543/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

203700/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203779/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203783/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203812/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203933/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203938/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204352/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 48.75mCi-487.5mCi per 13ML (3.75-37.5mCi per ML) |

204356/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

204366/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

204455/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204457/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204465/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204496/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204506/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204514/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204515/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per mL |

204535/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204539/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

204547/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

204667/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 18.8mCi-188mCi per 5ML (3.75-37.5mCi per ML) |

205687/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

207025/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204510/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi (3.75-37.5mCi per ML) |

ApplId/ProductId | Drug name | Active ingredient | Form | Strenght |
---|---|---|---|---|

022119/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203321/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203543/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

203700/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203779/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203783/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203812/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203933/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

203938/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204352/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 48.75mCi-487.5mCi per 13ML (3.75-37.5mCi per ML) |

204356/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

204366/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

204455/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204457/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204465/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204496/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204506/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204514/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204515/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per mL |

204535/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204539/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

204547/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

204667/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 18.8mCi-188mCi per 5ML (3.75-37.5mCi per ML) |

205687/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 3.75-260mCi per ML |

207025/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi per 8ML (3.75-37.5mCi per ML) |

204510/001 | AMMONIA N 13 | AMMONIA N-13 | INJECTABLE/INTRAVENOUS | 30mCi-300mCi (3.75-37.5mCi per ML) |

A licensed doctor will try to answer your question as quickly as possible.

There are a few. It's important to remember that N-13 has a very short half-life (10 minutes), so it disappears very quickly. This is both good and bad. In terms of safety and convenience for the patient, it's great. It decays very quickly, so after the patient has had the scan, it won't be long before all the radioactivity is gone and the patient can safely go home without being a risk to themselves or anyone else. It also means that it's safe to dispose of in the hospital- it doesn't form dangerous radioactive waste. Also, if there was an accident and it was released, it won't hang around and contaminate the hospital. The downside of all this is that it is also impossible to store for any period of time, as it simply decays. It needs to be made at the same place doing the scan, so they have to have the necessary equipment to do that. Answered by Donovan Mensch 1 month ago.

N-13 Ammonia Answered by Autumn Goldy 1 month ago.

Ammonia is a gas with a characteristic pungent odor. it is sold as a water solution for use in household cleaning. The gas is a compound of nitrogen and hydrogen in the atomic ratio 1:3. A sample of ammonia contains 7.933 g N and 1.712 g H . What is the atomic mass of N relative to H? The book answer is 13.90 13.90 what?! How? Help. Thank you. Answered by Ivette Fortado 1 month ago.

Ammonia is a gas with a characteristic pungent odor. it is sold as a water solution for use in household cleaning. The gas is a compound of nitrogen and hydrogen in the atomic ratio 1:3. A sample of ammonia contains 7.933 g N and 1.712 g H . What is the atomic mass of N relative to H? The book answer is 13.90 1.712 g is the mass of 3 H’s Mass of 1 H = 1.712 ÷ 3 = 0.570667 Relative mass = 7.933 ÷ 0.570667 = 13.90 When you see 2 sets of numbers, 1 : 3 and 7.933 , 1.712; 7.933 represents 1 and 1.712 represents 3 Sometimes I just divide and multiply until I get the answer. This provides a clue to the correct process to get the answer! This question is not very clear! Answered by Shavon Pinkert 1 month ago.

Pungent Odor Answered by Yelena Recht 1 month ago.

Need chemistry help ! Determine the amount of heat (in kJ) given off when 2.13x10^4 g of ammonia are produced according to the following equation. Assume that the reaction takes place under standard-state conditions at 25.0°C. N2(g) + 3 H2(g) 2 NH3(g) ΔH°rxn = -92.6 kJ and A 43 kg person drinks 280. g of milk, which has a "caloric" value of approximately 3.0 kJ/g. If only 19% of the energy in milk is converted to mechanical work, how high (in meters) can the person climb based on this energy intake? [Hint: The work done in ascending is given by mgh, where m is the mass (in kilograms), g the gravitational acceleration (9.8 m/s2), and h the height (in meters.)] Answered by Dell Parody 1 month ago.

1) N2(g) + 3 H2(g) = 2 NH3(g) = -92.6 kJ mol-1 how may moles of ammonia are present in 2.13x10^4 g n = m / MM n = 2.13x10^4 g / 17.03 g/mol = 1251 moles Therefore if we produced 1251 moles of NH3(g) we must of used (1251/2) 626 moles of N2 because they are in a 1:2 ratio as you can see from the balanced equation. For ever mole of ammonia made we used 1/2 a mole worth of N2. ΔH°rxn = -92.6 kJ mol-1 of N2 Energy produced = -92.6 kJ mol-1 x 626 mol = -57968 kJ Therefore we produced approx 58000 kJ of energy. 2) How much energy in total in the milk? E = 280 g x 3 kJ/g = 840 kJ We only absorb 19% of this so (840 kJ/100)x19 =160 kJ Work done = mgh We can do a maxium of 160k kJ of work so we need to rearrange the equation W / mg = h 1 joule = 1 kg m^2 s^-2 160 kJ = 160000 kg m^2 s^-2 h = 160000 kg m^2 s^-2 / 43 kg x 9.8 m s^-2 = 3721 m So you could climb 3721 m Answered by Carmelita Hine 1 month ago.

N2 + 3H2 ----------------------> 2NH3 so first find how many moles of ammonia you want if it's equals to N , N = 13.05/17 = 0.7676mol so now for 2 ammonia moles we need 3 H2 moles so lets find how many H2 moles needed to produce 0.7676 moles of ammonia N ( H2 ) = 3/2 * 0.7676 = 1.1514 moles Hope This Helps Answered by Vanda Hamner 1 month ago.

When three moles of a metal oxide, MO2, react with ammonia gas, the metal (M), water, and nitrogen gas are formed. A) write a balanced equation to represent this reaction. I got 3MO2+4NH --> 3M+6H20+2N B) When 13.8 g of ammonia react with an excess of metal oxide, 126g of M are formed. What is the molar mass of M? What is the identity of M? Answered by Sherley Wetmore 1 month ago.

Your equation is correct. 3MO2+4NH3 -- 3M+6H20+2N2 B) We need to convert 13.8 g of NH3 from a mass into moles n = m/MM n = 13.8 g / 17.03 g/mol = 0.81 mol If we look at the equation 3MO2+4NH3 -- 3M+6H20+2N2 4 moles of ammonia NH3 form 3 moles of the metal M. So they are in a ratio of 4:3 So for every mole of ammonia used NH3 we produce 0.75 moles of the metal M So we used 0.81 mol of ammonia means we produced 0.81 mol x 0.75 = 0.6075 mol We have 126 g of M and we now know that that amount equals 0.6075 mol So using n = m/MM MM = m / n MM = 126 g / 0.6075 mol = 207.41 g/mol What metal has the molar mass (MM) of 207.42 g/mol ? Well any search on google or on a periodic table will tell you it is Lead Pb The identity of M is therfore Lead, Pb. Answered by Buffy Hamalainen 1 month ago.

ammonia is NH3; nitrogen gas is N2. work it from there. Answered by Jade Demory 1 month ago.

You have the correct equation. First solve for n (the number of moles). Then, find the molar mass of ammonia, and then find the number of grams through the following conversion: n mol ammonia * (x grams ammonia / mol ammonia) = ? grams ammonia Answered by Lashunda Hufnagel 1 month ago.

You know the other quantities, so you can find n, the number of moles. If n = PV/RT, put in known values and obtain: n = (1.85)(0.125)/(.0821)(260) Then n = 0.01083 mole of ammonia. To convert to grams, you need to know the molecular weight of ammonia, 17. So 1 mole of ammonia has a mass of 17 grams and you have 0.01083 mole. Then you have 0.01083 times 17, or 0.184 gram of ammonia. Answered by Kristi Nibbs 1 month ago.

Yes! You've got it right. Calculate for n first and then multiply it by the molar mass. Use the formula PV = nRT. I presume you already used that. Convert the temperature to Kelvin. Just add 273. Derive the equation for n. After that, you can convert the moles to grams. You're on the right track. Don't worry. :) Answered by Refugia Millonzi 1 month ago.

You should use the ideal gas law : PV=nRT P- pressure: must always be in atm V- Volume: must always be in liters n- moles R- 0.08206 L atm ----------- K mol T- temperature: must always be in Kelvin 90+273= 363K You should convert temperature to kelvin by adding 273 to the celcius temperature. Then, input the values into the equation and solve. (2.13 atm) (6.43L) = (n)( 0.08206 L atm / K mol ) ( 363K) And solve Then convert moles to grams. Answered by Graciela Mamula 1 month ago.

mass of N in NH3 = 7.933 and this represents 1 atom of N mass of H in ammonia = 1.712 g which represents 3 atoms so mass of 1 atom of H in NH3 = 1 .712 / 3 = 0.570666667g mass ratio of N : H = 7.933 / 0.570666667 = 13 .90 Answered by Arlene Tonzi 1 month ago.

moles of N = 7.933/am(N) moles of H = 1.712/am(H) (moles of N)/(moles of H) = (7.933/1.712)*(amH)/am(N) 1/3 = (7.933/1.712)*(amH)/am(N) am(N)/am(H) = 3*(7.933/1.712) = 13.90 Answered by Jen Amstein 1 month ago.

I know this question has appeared already, but I've looked through all the results on Google for this question and I still don't understand it. If you haven't seen the question, here it is: Ammonia is a compound of nitrogen and hydrogen in the atomic ratio 1:3. A sample of ammonia contains 7.933gN and 1.712 gH. What is the atomic mass of N relative to H? Here is the method people used to find the solution: [7.933 g N / 1 mol] / [1.712 g H / 3 mol] = 13.90 g/mol N Here are my questions: What exactly is atomic mass, and is it usually recorded in g/mol format? Why did we divide 7.933 g (the mass of the nitrogen in the sample) by mol? And did that 1 mol come from the ratio 1:3? Is 1:3 the mole-mole ratio? is 7.933 g and 1.712 atomic weight or atomic mass or neither? Is it just mass of the elements in the sample? Why are we dividing g by mol? Why do you divide 7.933 / (1.712 / 3)? Thank you for your patience with my stupidity... Answered by Julissa Jastrzebski 1 month ago.

Atomic mass is the mass of an element that contains 22.06x10^23(1 mole) of atoms in grams(mass/mol). Yes, it is always recorded in g/mol format. Atomic mass of N = Mass of N/mol N => mol of N = Mass of N/Atomic mass of N = 7.933g/Atomic mass of N Like N, we work with H mol of H = Mass of H/Atomic mass of H = 1.712g/Atomic mass of H The ratio tell us that we have 3 moles of H for each mole of N... 3mol H = 1mol N =>mol of H/mol of N=3 =>(1.712g/Atomic mass of H)/(7.933g/Atomic mass of N)=3 =>Atomic mass of N/Atomic mass of H x 1.712g / 7.933g =3 =>Atomic mass of N/Atomic mass of H=7.933g x 3 / 1.712g = 13.90 g/mol Dear Yuna, You are not stupid...you just did n`t have enough knowledge. I hope you get what you want from this answer... Good luck! Answered by Linette Willner 1 month ago.

H Atomic Mass Answered by Jeanetta Labrador 1 month ago.

Atomic Weight Of N Answered by Vernon Abramowski 1 month ago.

Ammonia is NH_3 @ 17.0306 g/mol So im guessing access of hydrogen using law of conservation of mass 5.36 g N @ 14.006 g/mol and H @ 1 g/mol 5.36/14 = 0.38 ish mol ans * 17.0306 = 6.52 answer E. Answered by Tona Cappas 1 month ago.

google it werido Answered by Genevieve Pathak 1 month ago.