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.
There are 2.8604 moles of ammonium ions in 6.955.
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.
A 22.5 gram sample of ammonium carbonate contains 4.5 moles of ammonium ions.
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.
There are 2.8604 moles of ammonium ions in 6.955.
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.
0,189 moles
A 22.5 gram sample of ammonium carbonate contains 4.5 moles of ammonium ions.
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.
assuming that 8.778 is in grams then there are 0.1069 moles in 8.778 grams of ammonium carbonate here is the math:(NH4)2CO3 N2=14.01g H8=8.08g C=12.01g O3=48.00 14.01+8.08+12.01+48.00=82.10g/mole 8.778g X 1 mole/82.10g=0.1069moles
Number of moles = Weight (in g) / Molecular mass (in g/mol) = 8.718 / 96.09 = 0.0907 moles
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.
Chemical formula for Ammonium Carbonate: (NH4)2CO3 So there are two Nitrogen atoms in that molecule.
Ammonium carbonate is (NH4)2CO3. From this formula you can see that it contains 4 elements: nitrogen, hydrogen, carbon, and oxygen.