Ammonia is NH3 (not NH2) so its molar mass is 17 g/mol. In 1.2*10^3 g there are
1.2*10^3(g) / 17 (g/mol) = (70.6 =) 71 mol NH3.
Ammonia is NH3.
1200 g of ammonia is equivalent to 70,46 moles.
How many moles are in 1.2 x 10^3 grams of ammonia, NH3
286 (mol BaCl2) * 208.23 (g/mol BaCl2) = 59,553.78 = 59.6*103 g BaCl2 = 59.6 kg BaCl2
The formula unit for the usual form of solid ammonium carbonate is (NH4)2CO3.H2O. This formula shows that each formula unit contains two atoms of nitrogen. Because nitrogen forms diatomic molecules at standard temperature and pressure, the number of moles of nitrogen is therefore the same as the number of formula units of ammonium carbonate, stated to be 650. The gram formula unit mass of this solid ammonium carbonate is 114.10. Therefore, 114.10(650) or 7.42 X 103 grams of the solid, to the justified number of significant digits, will be required.
One step at a time.1/103 = 0.001 M HCl, so.....Molarity = moles of solute/Liters of solution ( 25 ml = 0.025 Liters )0.001 M HCl = X moles HCl/0.025 Liters= 2.5 X 10 - 5 moles HCl========================now, balanced eqiationNaOH + HCl --> NaCl + H2O ( all one to one )( now drive reaction towards mass NaOH )2.5 X 10 - 5 moles HCl (1 moles NaOH/1 mole HCl)(39.998 grams/1 mole NaOH)= 10 -4 grams caustic soda needed==========================
How many atoms of gallium are in 2.85 x 103 g of gallium?
[ 34.3(gC3H8) / 44(g/mol C3H8)] * [ 3mol CO2 / 1molC3H8 ] * 44(g/mol CO2) = 103 gram CO2
A lot! 9.51 X 103 grams Pb (1 mole Pb/207.2 grams) = 45.9 moles of lead ===============
One mole=1000 Milli moles One mole=6.022 x 1023 molecules of the substance Therefore, one Milli mole of ammonia has 6.022 x 1023 /103 = 6.022 x 1020 molecules of ammonia.
103 pounds = about 46,700 grams.
103 grams = 0.22707613 pounds103 grams = 0.23 pounds
The gram molecular mass of C2H4 is 28.05. Therefore, 1.26 X 103 grams constitutes (1.26/28.05) X 103 or about 44.9 moles. Each mole contains Avogadro's number of molecules; therefore, 1.26 X 103 grams contains 2.71 X 1025 molecules, to the justified number of significant digits.
286 (mol BaCl2) * 208.23 (g/mol BaCl2) = 59,553.78 = 59.6*103 g BaCl2 = 59.6 kg BaCl2
The balanced equation for the reaction is 2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O. This shows that 13 moles of diatomic oxygen are required to burn 2 moles of butane. By proportionality, (4.8/2)13 or 31.2 moles of oxygen are required to burn 4.8 moles of butane. This corresponds to 31.2(32) or 1.0 X 103 grams of oxygen.
Each mole of boron atoms has a mass of 10.811 grams, as indicated by the gram atomic mass or weight of boron. Therefore, 585 moles has a mass of about 6.32 X 103 grams, to the same number of significant digits as 585.
28 grams in an oz Source: Drug dealer
20.4