What is the ph level of ammonia?
113751
Ammonia is a gas at room temp, dissolved in water it it very basic and would have a high pH of around 10 depending on the concentration
149153
This is a limiting reagent problem. You must calculate the moles of NH3 produces from the given amount of grams of N2 and H2.\n\n3.00g N2 x 1mol/28.02g x 2 mol NH3/1 mol N2 = .214 mol NH3\n\n.54g H2 x 1 mol/2.02g H2 x 2 mol NH3/3 mol H2 = .178 mol NH3\n\nFrom this calculation you now know that the H2 is the limiting reagent. The next part to calculate the grams of NH3 goes like this:\n\n.178 mol NH3 x 17.04g/1 mol = 3.03 g NH3
419911
That's easy. [Co(NH3)6]Cl3. This is coordination chemistry, dude. There can be other products depending on the ammonia concentration. But this is your best bet. I've been doing these reaction in the lab way way back.
634185
uric acid, ammonia, and urine
373054
HNO3 and NH4NO3. A buffer is a solution of a weak acid with the corresponding weak base. It is set up to maintain a specific range os Ph, determined by the proportions of acidvs base and by the POh of the acid/base. The amoutnt of H+ that a buffer syatem can absorb before the buffer breaks is determined by the streangth, ie molarity, of the solution. a
282642
actually, pure water does not have a pH of 7.0, it is slightly acidic, around 5.6-5.8 from what i have seen at work.
452634
N2 +3H2 ---> 2NH3\n\nso, H2 Mwt is 2g/mol and NH3 Mwt is 34g/mol\nas per above eqn, 6g of H2 will give u 34g of NH3\nSo, 9.988g of H2 will give 56.6g of NH3\n\nYou got it?
761517
ammonia, NH3, dissolves in water by forming the ammonium ion.\n\nNH3+H2O--> NH4+ + OH-\n\nthe ammonia gains a proton from the water\na proton is hydrogen(atomic number 1=1 proton)
592234
3g of hidrogen (plus 14 g of nitrogen) make up 17g of ammonia (H+H+H+N=1+1+1+14), so:\n\n3g  H____17g NH3\n50g H____=283.33g NH3\n\nI assumed a 100% efficient reaction (that means every bit of reactant will actually react).
754704
You should write NH4(+) otherwise you will not get the oxidation number right.  H is 1 so H4 is 4.\n\nIn order to be left with 1(+) you need 4 - 3 = 1.  So N will be -3.\n\nCheck -3 + 4*1 = +1 OK.
243948
Yes, using the David Hasselhoff equation is much easier. The HH equation is:\n\npH = pKa + log(base/acid)\n\nFill in the information you know (either from the problem or a table).\n\n9.00 = 9.25 + log(0.10/x)\n\nSolve for the concentration of NH4.\n\n0.178 = [NH4+]\n\nSince there is 2.0L of solution, use that to solve for the number of moles.\n\n0.178M * 2.0L = 0.356 moles NH4Cl.\n\n\n\nIn case you were wondering, you have to look up the Kb of ammonia (NH3) and convert it to the Ka of ammonium.\n\nKw/Kb = Ka\n\n1e-14/1.78e-5 = Ka = 5.62e-10\n\nTake the -log of that to find the pKa of ammonium, which is needed in the equation.\n\n\nHope this helped.
482772
Ph is A measure of the acidity or alkalinity of a solution, numerically equal to 7 for neutral solutions, increasing with increasing alkalinity and decreasing with increasing acidity. The pH scale commonly in use ranges from 0 to 14.
122833
pH level, buffering capacity (alkalinity), general hardness and salinity
813106
Just an initial note: NH3 is ammonia\n\nA. \nInitial math:\nSince you have 8.5 grams of NH3, you have to convert grams mols, then into molecules. \nSo you do: 8.5 grams of NH3 / 17 grams in 1 mole of NH3 and X 6.022 x 10^23 molecules in one mole\nYou get 3.011 x 10 23 molecules, and from that you subtract 1.5 x 10^23 molecules, and you get: \n<b>1.511 x 10^23</b>\n\nB. Then, you convert 1.5 x 10^23 to moles by dividing by 6.022 x 10^23 moles, which equals .249 moles of NH3. \nThen you convert 8.5 grams of NH3 to moles too by dividing 17 grams of NH3 per mole again, and get: .5 moles! \nso: .5 mol - .249 mol = <b>.241 mol</b>\n\nC. \nFrom before, you know 1.5 x 10^23 molecules of NH3 is .249 moles. You convert this to grams by multiplying .249 by 17 grams per mol and get: 4.233 grams. \nSo: 8.5 grams - 4.233 grams = <b>4.267 grams</b>\n\nD. From before, you know you have .241 moles of ammonia left. So, you multiply this by avocado's number (6.022 x 10^23) and get: 1.45 x 10^23 molecules of ammonia. Then, you can see that there are 3 atoms of hydrogen of per molecule of ammonia, and multiply the number of molecules by 3 and get: \n<b> 4.354 x 10^23</b> \n\nI think those should be right. Sorry about the sig figs though!
233434
This is just a simple stoichiometry problem.  You have 200g of NH3, your first job is to calculate how many moles of NH3 this is.  From the periodic table, find the atomic weights of N and H to calculate the molar mass of NH3:  \n\nN=14 and H=1, so NH3 = 17.  \n\n200g NH3 X (1 mol NH3 / 17g NH3) = 11.76 mol NH3\n\nOkay, now that we know we're working with 11.76 moles of NH3, we can use the mole ratio from the given chemical equation to figure out how many moles of oxygen are needed.  Look at the coefficients in the equation -- we need 5 moles of O2 for every 4 moles of NH3:\n\n11.76 mol NH3 X (5 mol O2)/(4 mol NH3) = 14.7 mol O2\n\nNow we know we need 14.7 mol O2.  All we have to do now is convert that to grams, again using the atomic mass from the periodic table.  The atomic mass of O is 16, so O2=32  \n\n14.7 mol O2 X 32g/mol = 470.4 g O2\n\nSUMMARY:\nConvert grams ammonia to moles ammonia using atomic weights.\nUse the mole ratio to determine needed moles of O2.\nConvert moles O2 to grams O2 using atomic weights.
750432
Amount of N2 which reacted = 21/ relative molecular mass of N2\n                                               = 21/ 28\n                                               = 0.75mol.\nMass of H = (0.75 x 6/2) x 2 = 4.5g of H2\nMass of NH3 = (0.75 x 2) x 17 =25.5g of NH3.\n\nThe value in the parentheses are your mol values of H and NH3 respectively.
1066669
we know that pH is ranged from totally basic to totally acidic, with neutrality (water) being at pH=7; so just display a range from 0 to 14 and color code it (red for acid and blue for basic) and place your findings there (ie before and after situations of NaCl towards the pH of the solution). Also place analogies (ie. the results of the pH due to NaCl concentration of 10% is equivalent to household ammonia)\n\nWhen explaining something to a non-science person, visuals and analogies always helps.