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Polyatomic ions. SO4 2- = sulfate OH - = hydroxide CO3 2- = carbonate NO3 - = nitrate NH4 + = ammonium For several common examples.
Well antimony carbonate if it exists would contain Sb3+ ions and CO32- ions-- to balance the charge the formula owuld be Sb2(CO3)3
A neutral solution of about 7 pH.
2 moles of Ca and 4 moles of OH
Please clarify the units of concentration of carbonate and bicarbonate. "mgl" is not a unit of concentration (it's not a unit of anything to my knowledge. To answer this question, you need the concentration of both ions. So either provide the amount of both ions AND the amount of water, or just specify the concentration (in unit of molarity, or moles per liter preferably).
A 22.5 gram sample of ammonium carbonate contains 4.5 moles of ammonium ions.
0,189 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.
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.
Ammonium carbonate = (NH4)2CO3. Its molar mass is 96g/mol. 6.965g/96 = 0.7255mol.
Number of moles = Weight (in g) / Molecular mass (in g/mol) = 8.718 / 96.09 = 0.0907 moles
Yes. An aqueous solution of ammonium carbonate would consist of dissociated ammonium ions and carbonate ions.
There are 2.8604 moles of ammonium ions in 6.955.
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.
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.
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.
0.00833 moles of CO3