2.49x10-1mol NH3
Source: e2020
2.49x10-1mol NH3
Source: e2020
It is about .415 moles
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45 grams H20 x (1 mole H20/18 grams H2O) x (6.02E23 molecules H20/1 mole H2O) the grams H2O and moles H2O cancel out. When you punch it into your calculator, the answer comes out to: =1.505E24 molecules H2O
150/132 equals 1.136moles
.150 M is the molarity of the solution, which is the number of moles per liter. So all you need to do is multiply the molarity by the number of liters. So .150 moles/liter x .550 L = .0825 moles
The approximate molar mass of Al(OH)3 = 27 + 48 + 3 = 78 g/mol150 g x 1 mol/78 g = 1.92 moles
The reaction equation is 2 NaHCO3 + H2SO4 = 2 CO2 + 2H2O + Na2SO4. This means that for every mole of sulfuric acid, two moles of NaHCO3 are needed. 150 grams of H2SO4 is 1.53 moles, so 3.06 moles of NaHCO3 are required.
150 M NaOH = 150 mole / liter 20.0 mL = 0.0200 L 150 * 0.0200 = 3 moles
150 g x 1 mol/68.154 g = 2.2 moles (2 sig figs)
45 grams H20 x (1 mole H20/18 grams H2O) x (6.02E23 molecules H20/1 mole H2O) the grams H2O and moles H2O cancel out. When you punch it into your calculator, the answer comes out to: =1.505E24 molecules H2O
150/132 equals 1.136moles
150
.150 M is the molarity of the solution, which is the number of moles per liter. So all you need to do is multiply the molarity by the number of liters. So .150 moles/liter x .550 L = .0825 moles
The approximate molar mass of Al(OH)3 = 27 + 48 + 3 = 78 g/mol150 g x 1 mol/78 g = 1.92 moles
150
This depends on many things including temperature, pressure, number of moles and molecular weight or the density
The reaction equation is 2 NaHCO3 + H2SO4 = 2 CO2 + 2H2O + Na2SO4. This means that for every mole of sulfuric acid, two moles of NaHCO3 are needed. 150 grams of H2SO4 is 1.53 moles, so 3.06 moles of NaHCO3 are required.
Divide by molar mass of NaCl (not NaCI ! ) which is 58.44 g/mol NaCl.
No you need more copper