if you have 2.60x10^23 molecules of (NH4)3PO4 how many
grams to you have


if you have 2.60x10^23 molecules of (NH4)3PO4 how many grams to you have
1. How many GRAMS of ammonium phosphate, (NH4)3PO4, are present in 2.81×1022 formula units of this compound?
How many moles of nitrogen atoms and hydrogen atoms are present in 1.4 moles of (NH4)3PO4?
23. How many molecules of carbon monoxide, CO, comprise 4.82 grams?
If you have 6.25 x 1024 molecules of sulfur tetrafluoride, how many grams do you have?
Ammonium phosphate ((NH4)3PO4) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H3PO4) with ammonia (NH3). What mass of ammonium phosphate is produced by the reaction of 4.81 g of ammonia?
Ammonium phosphate ((NH4)3PO4) is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid (H3PO4) with liquid ammonia. Calculate the moles of ammonia needed to produce 1.90mol of ammonium phosphate.
how many grams of HNO3 are in 4.055 x10^22 molecules of HNO3? This is how they have the equation set up. 4.055 x 10^22 molecules HNO3 x 1 mol HNO3 / 6.022 x 10^23 molecules HNO3 x 63.018 g HNO3/ 1 mol HNO3= I cant get the correct answer using their equation.
2(NH4)3PO4 + 3 Pb(NO3)2 ------> 6NH4NO3 + Pb3(PO4)2 how may moles of (NH4)3NO3 would be required to react with 9 moles of Pb(NO3)2?
Ammonium phosphate [(NH4)3PO4) is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid H3PO4 with liquid ammonia. Calculate the moles of phosphoric acid needed to produce 1.30 mol of ammonium phosphate. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
How many grams of Al are in 1.38 * 1023 molecules of Al2O3 if the atomic mass of Al is 27.0 g/mole, O is 16.0 g/mole, and Al2O3 is 102.0 g/mole (the final answer should have the correct number of significant figures)? Here is my work: (1.38*10^23 molecules) (1mol / 6.022*10^23 molecules) (27.0 gram/ mol)= 6.1873 I counted 3 Sig Fig so I rounded my answer to 6.19.... Another thought was (1.38*10^23 molecules) (54grams /1 mol) (1mol/6.022*10^23) but that answer...