NaoH HAVE 6.022(10)23 IN HALF DROP.
You have to realise that a drop from the burette for instance is insignificant, if you are dealing with at least 10ml solution which you usually deal with on a titration. If you don't want to regard it as insignificant, then if NaOH is in the burette, then the solution doesn't become more concentrated with NaOH because that drop escaped.
No, they are not. "0.5 mole of NaOH" means that you have half a mole of sodium hydroxide. "0.5M of NaOH" means that you have half a mole of sodium hydroxide for every liter of solution. "0.5M" is also commonly written as "0.5 mol/L" or "mol L-1".
volts.
Molarity = moles of solute/Liters of solution 3.42 M NaOH = 1.3 moles NaOH/Liters NaOH Liters NaOH = 1.3 moles NaOH/3.42 M NaOH = 0.38 Liters
I assume you mean 32.0 grams of NaOH and 450 milliliters of NaOH. Molarity = moles of solute/Liters of solution ( 450 ml = 0.450 liters ) get moles of NaOH 32.0 grams NaOH (1 mole NaOH/39.998 grams) = 0.800 moles NaOH Molarity = 0.800 moles NaOH/0.450 liters = 1.78 Molar NaOH
You have to realise that a drop from the burette for instance is insignificant, if you are dealing with at least 10ml solution which you usually deal with on a titration. If you don't want to regard it as insignificant, then if NaOH is in the burette, then the solution doesn't become more concentrated with NaOH because that drop escaped.
There is one hydrogen atom in one formula unit of NaOH.
No, they are not. "0.5 mole of NaOH" means that you have half a mole of sodium hydroxide. "0.5M of NaOH" means that you have half a mole of sodium hydroxide for every liter of solution. "0.5M" is also commonly written as "0.5 mol/L" or "mol L-1".
Take half volume of 1.0 M NaOH and add another half volume of water. Or Take 20.0 gram NaOH , carefully dissolve this completely in ca. 0.9 L water and then fill up to 1.0 L end volume.
Half a unit.Half a unit.Half a unit.Half a unit.
For sodium oxide, the empirical formula is the same as the formula unit, Na2O. (If any formula unit or molecular formula contains an atomic symbol with no following subscript, the empirical and actual formulas will be the same.)
3.42 moles NaOH (39.998 grams/1 mole NaOH) = 137 grams NaOH
volts.
Yes you have to drop the fuel tank to access the fuel gauge sending unit on top of the tank.
Molarity = moles of solute/Liters of solution 3.42 M NaOH = 1.3 moles NaOH/Liters NaOH Liters NaOH = 1.3 moles NaOH/3.42 M NaOH = 0.38 Liters
8 grams NaOH (1 mole NaOH/39.998 grams) = 0.2 moles NaOH
It is 1.5 times the unit price.