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The molar mass of anhydrous sodium carbonate is 106 g.The molar mass of decahydrated sodium carbonate is 286 g.
106 grams
The molar mass of ammonium dichromate is 252,07 g.
molar mass of unknown/molar mass of empirial = # of empirical units in the molecular formula. Example: empirical formula is CH2O with a molar mass of 30. If the molar mass of the unknown is 180, then 180/30 = 6 and molecular formula will be C6H12O6
The density or some other information must be given that allow you to find the molar mass. Calculate the empirical formula mass. Divide molar mass by empirical formula mass. This answer is multiplied by all subscripts of the empirical formula to get the molecular formula.
Ammonium carbonate = (NH4)2CO3. Its molar mass is 96g/mol. 6.965g/96 = 0.7255mol.
The molar mass of anhydrous sodium carbonate is 106 g.The molar mass of decahydrated sodium carbonate is 286 g.
The molar mass of sodium carbonate to the nearest gram is 105,99 g.
Question : How many moles of ammonium ions are in 8.754g of ammonium carbonate? Let's use the criss-cross method: (NH4+, CO32-)→(NH4)2CO3 That menas in each mole of ammonium carbonate((NH4)2CO3 )there are 2moles of ammonium (NH4). Let's find the quantity of mole(N) in 8.75g of ammonium carbonate((NH4)2CO3 ). N=mass/Molar mass (#) Molar mass((NH4)2CO3 )=2(14.0067+41.0079)+12.011+315.999)g/mol=96.09g/mol. (#) Gives us : N=(8.754g)/(96.09g/mol)=0.0911mole Let's K be the quantity of ammonium mole included in 8.75g of ammonium carbonate. As I stated before each mole of ammonium carbonate contains 2moles of ammonium, therefore : K=2*0.0911mole=1.1822mole Best regards, BILL JESY FOREVER 7171.
Ammonium carbonate is (NH4 )2 CO3 and the molar mass is 96.0878 so you just divide 8.790 g/96.0878 and you get 0.91478835 and since there are two ammonium (NH4 )2 ions you multiply 0.91478835 * 2 and get 0.1830 mol.
106
Formula : (NH4)2C2O4 Molar mass : 124
106 grams
The molar mass of ammonium dichromate is 252,07 g.
the empirical formula and the molar mass
ammonium carbonate: (NH4)2CO3 = 96.09 g/mol
molar mass of unknown/molar mass of empirial = # of empirical units in the molecular formula. Example: empirical formula is CH2O with a molar mass of 30. If the molar mass of the unknown is 180, then 180/30 = 6 and molecular formula will be C6H12O6