each mol of h3po4 has 6.022*10^23 molecules
you have 1.5 mol, so 1.5 * 6.022*10^23 molecules = 9.033*10^23 molecules
each molecule of h3po4 has 1 atom of p,
so you have 9.033*10^23 atoms p.
150 g x 1 mol/68.154 g = 2.2 moles (2 sig figs)
Divide by molar mass of NaCl (not NaCI ! ) which is 58.44 g/mol NaCl.
3.5M / 1000, x 150 = how many moles you need. = .525 moles. moles = mass / molec mass. molec mass if cabr2 is 199.86 g/mol. .525 x 199.86 is 104.93g CaBr2
The atomic mass of Copper is 63.5 grams One mole of any element has a mass equal to the atomic mass. 0.75 grams of Cu = x moles of Cu 63.5 grams of Cu = 1 mole of Cu Set up a proportion and solve for x Divide 0.75 / 63.5 = x /1 0.75 ÷ 63.5 = x
2NaCl + H2SO4 ----> 2HCl + Na2SO4 *molar mass of H2SO4 is 98.09g *molar mass of NaCl is 58.44g *molar mass HCl is is 36.46g Moles NaCl = 150 g / 58.44 = 2.56 The ratio between NaCl and H2SO4 is 2 : 1 so NaCl is the limiting reactant We would get 2.56 mol HCl => 2.56 mol x 36.46 g/mol = 93.3 g
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)
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.
150/132 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
1 mol = 6,022 x 10^23 molecules of HI. So: 6,022E23 *0,3 = Your answer !
The approximate molar mass of Al(OH)3 = 27 + 48 + 3 = 78 g/mol150 g x 1 mol/78 g = 1.92 moles
This depends on many things including temperature, pressure, number of moles and molecular weight or the density
Divide by molar mass of NaCl (not NaCI ! ) which is 58.44 g/mol NaCl.
No you need more copper
Several part problem. Get molarity of NaHCO3. (150 ml)( M NaHCO3) = (150 ml)(0.44 M HCl) = 0.44 M NaHCO3 --------------------------- get moles NaHCO3 ( 150 ml = 0.150 Liters ) 0.44 M NaHCO3 = moles NaHCO3/0.150 Liters = 0.066 moles NaHCO3 ---------------------------------------get grams 0.066 moles NaHCO3 (84.008 grams/1 mole NaHCO3) = 5.54 grams NaHCO3 needed ---------------------------------------------answer