1.54 moles H2O (18.016 grams/1 mol H2O ) =27.7 grams
Molarity = moles of solute/Liters of solution Find moles glucose, which is molecular formula------C6H12O6 154 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.8548 moles C6H12O6 Molarity = 0.8548 moles C6H12O6/1 Liter = 0.855 M glucose ---------------------------
2,74 moles of CCL4 is equivalent to 421,44 g.
7.4x10^23 x ( 1 mol AgNO3/ 6.02x10^23) x (169.88g AgNO3/ 1 mol AgNO3)=2.1x10^2g AgNO3no of molecules=7.4x10^23Mr of AgNO3=169.88we can find the no of moles, thereforein 1mol there are 6.02x10^23 molecules7.4x10^23 molecules represent (7.4x10^23/6.02x10^23)molesin theory when we multiply molar mass (Mr) with the no of moles, we obtain the mass of the substance in gram:Mr x no of Mol= Mass in grams(7.4x10^23/6.02x10^23)x169.88= 2.1x10^2
154 psi
One liter of ozone gas at 1 atmosphere and 0°C has the mass of is 2.141 gm. At 20°C (STP) that would be 1.995 gm. 3.36 L would be 6.70 gm (to three sig figs).
Molarity = moles of solute/Liters of solution Find moles glucose, which is molecular formula------C6H12O6 154 grams C6H12O6 (1 mole C6H12O6/180.156 grams) = 0.8548 moles C6H12O6 Molarity = 0.8548 moles C6H12O6/1 Liter = 0.855 M glucose ---------------------------
Divide by molar mass and check the units(italicalized):0.140 (g HCl) / 36.45 (g.mol-1HCl) = 3.84*10-3 mol HCl
2,74 moles of CCL4 is equivalent to 421,44 g.
7.4x10^23 x ( 1 mol AgNO3/ 6.02x10^23) x (169.88g AgNO3/ 1 mol AgNO3)=2.1x10^2g AgNO3no of molecules=7.4x10^23Mr of AgNO3=169.88we can find the no of moles, thereforein 1mol there are 6.02x10^23 molecules7.4x10^23 molecules represent (7.4x10^23/6.02x10^23)molesin theory when we multiply molar mass (Mr) with the no of moles, we obtain the mass of the substance in gram:Mr x no of Mol= Mass in grams(7.4x10^23/6.02x10^23)x169.88= 2.1x10^2
The equivalent is 154 moles.
154 grams of water is approximately 11 tablespoons. Other substances will vary.
154 grams
70,000 grams is 154 pounds and 5.18 ounces.
154 psi
154
154g = 5.432oz
There is 0.77 cups in 154 grams of granulated sugar. This means that there is 1 gram in 0.005 cups of granulated sugar.