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Ammonium sulphate consists of two ions: ammonium (NH4+) and sulphate (SO4^2-).

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How many ammonium ions are present in 10.7g of ammonium phosphate?

The formula for the most common form of ammonium phosphate is (NH4)3PO4.3 H2O, and its gram formula mass is 203.13. The formula shows that there are 3 ammonium ions in each formula unit. 10.7g/203.13 is 5.27 X 10-2 formula units. Therefore, the number of ammonium ions present in 10.7g of this ammonium phosphate is 3 X 5.27 X 10-2 X Avogadro's Number or 9.52 X 1019 ammonium ions, to the justified number of significant digits.


How many moles of ammonium ions are in 8.738 g of ammonium carbonate?

To determine the number of moles of ammonium ions in 8.738 g of ammonium carbonate, first calculate the molar mass of ammonium carbonate (NH4)2CO3. Then, divide the given mass by the molar mass to find the number of moles. Since there are two ammonium ions in one formula unit of ammonium carbonate, multiply the number of moles by 2 to get the moles of ammonium ions.


How do you know how many ammoniums will combine with how many nitrate ions?

Ammonium ions have a 1+ charge, and nitrate ions have a 1- charge. One ammonium ion combined with one nitrate ion will produce an ionic compound with no overall charge. NH4+ + NO3- ---> NH4NO3


How many moles of ammonium ions are in 8.903 g of ammonium carbonate?

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.


How many ions in ammonium phosphate?

Ammonium phosphate is a salt produced from the reaction between ammonia and phosphoric acid. It consists of one ammonium ion (NH4+) and one phosphate ion (PO4^3-), so overall it contains two ions - one ammonium ion and one phosphate ion.

Related Questions

How many ammonium ions are present in 10.7g of ammonium phosphate?

The formula for the most common form of ammonium phosphate is (NH4)3PO4.3 H2O, and its gram formula mass is 203.13. The formula shows that there are 3 ammonium ions in each formula unit. 10.7g/203.13 is 5.27 X 10-2 formula units. Therefore, the number of ammonium ions present in 10.7g of this ammonium phosphate is 3 X 5.27 X 10-2 X Avogadro's Number or 9.52 X 1019 ammonium ions, to the justified number of significant digits.


How many ions in ammonium sulfate?

Ammonium sulfate contains two types of ions: ammonium ions (NH4+) and sulfate ions (SO4^2-). Therefore, there are 5 ions in total in ammonium sulfate, which is composed of (NH4)2SO4.


How many hydrogen atoms are contained in the formula for ammonium sulfate?

Ammonium sulphate (NH4)2SO4 contains two ammonium ions per unit so there are total 8 hydrogen atoms.


How many atoms of hydrogen are present in NH4 2S?

There are 8 atoms of hydrogen present in NH4 2S. This is because there are 4 hydrogen atoms in each ammonium ion (NH4+) and there are 2 ammonium ions in NH4 2S.


How many moles of ammonium ions does a 22.5 gram sample of ammonium carbonate?

To calculate the number of moles of ammonium ions in a 22.5 gram sample of ammonium carbonate, you need to first determine the molar mass of ammonium carbonate. Then, divide the given mass by the molar mass to find the number of moles. After that, since there are 2 ammonium ions in one molecule of ammonium carbonate, you will need to multiply the result by 2 to determine the number of moles of ammonium ions.


How many moles of ammonium ions are in 8.738 g of ammonium carbonate?

To determine the number of moles of ammonium ions in 8.738 g of ammonium carbonate, first calculate the molar mass of ammonium carbonate (NH4)2CO3. Then, divide the given mass by the molar mass to find the number of moles. Since there are two ammonium ions in one formula unit of ammonium carbonate, multiply the number of moles by 2 to get the moles of ammonium ions.


How many micromoles of Ammonia present in 25ml of 6mM Ammonium Sulphate?

Assuming that "mM" means "millimolar", the solution specified contains 6 millimoles of ammonium sulphate per liter. Therefore, 25 ml of the solution contains 6(25/1000) = 0.15 millimoles. By definition, there are 1000 micromoles per millimole. Therefore, 0.15 millimoles = 150 micromoles.


How do you know how many ammoniums will combine with how many nitrate ions?

Ammonium ions have a 1+ charge, and nitrate ions have a 1- charge. One ammonium ion combined with one nitrate ion will produce an ionic compound with no overall charge. NH4+ + NO3- ---> NH4NO3


How many moles of ammonium ions are in 6.965 g of ammonium carbonate?

To find the number of moles of ammonium ions in 6.965 g of ammonium carbonate, you first need to calculate the molar mass of ammonium ions, which is 18.0399 g/mol. Then, divide the given mass by the molar mass to find the number of moles. So, 6.965 g / 18.0399 g/mol = 0.386 moles of ammonium ions.


How many moles of ammonium ions are in 6.965 of ammonium carbonate?

To find the number of moles of ammonium ions, we first calculate the molar mass of ammonium carbonate (NH4)2CO3. It is 96.086 g/mol. Then we can find the moles of (NH4)2 cations in 6.965 g by dividing the mass by the molar mass. This gives us 0.0726 moles of ammonium ions.


How many moles of ammonium ions are in 8.903 g of ammonium carbonate?

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.


How many atoms are present in ammonium nitrate?

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