There are 2.8604 moles of ammonium ions in 6.955.
A 22.5 gram sample of ammonium carbonate contains 4.5 moles of ammonium ions.
There are 9 moles of NH4 (Ammonium ions) in Ammonium carbonate. There are 2 moles of NH4 per molecule and 4.5 molecules, so 2 moles times 4.5 is 9 moles.
As the formula for ammonium carbonate is (NH4)2CO3, there are two moles of ammonium for every mole of ammonium carbonate. Just double the given number.
Ammonium carbonate = (NH4)2CO3. Its molar mass is 96g/mol. 6.965g/96 = 0.7255mol.
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.
0,189 moles
A 22.5 gram sample of ammonium carbonate contains 4.5 moles of ammonium ions.
There are 9 moles of NH4 (Ammonium ions) in Ammonium carbonate. There are 2 moles of NH4 per molecule and 4.5 molecules, so 2 moles times 4.5 is 9 moles.
As the formula for ammonium carbonate is (NH4)2CO3, there are two moles of ammonium for every mole of ammonium carbonate. Just double the given number.
Number of moles = Weight (in g) / Molecular mass (in g/mol) = 8.718 / 96.09 = 0.0907 moles
Ammonium carbonate = (NH4)2CO3. Its molar mass is 96g/mol. 6.965g/96 = 0.7255mol.
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.
0.0999458293608 moles
The most common form of solid ammonium carbonate is a hydrate with formula (NH4)2CO3.H2O and a gram formula unit mass of 114.10. The formula shows that each formula unit contains 2 ammonium ions. The number of formula units of ammonium carbonate is 8.903/114.10 or 0.078028. The number of formula units of ammonium ions is twice this, or 0.1561, to the justified number of significant digits.
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.
I suppose that the answers are: - 0,9 moles aluminium ions - 2,7 moles chloride ions
0.00833 moles of CO3