The formula mass of sodium carbonate, Na2CO3 is 2(23.0)+12+3(16) = 106.0
Amount of sodium carbonate = 2.12/106.0 = 0.0200mol
the answer is 2
234 grams :)
234 grams
The formula mass of sodium carbonate, Na2CO3 is 2(23.0)+12+3(16) = 106.0Amount of sodium carbonate = mass of sample / formula mass = 4/106.0 = 0.0377mol
3 x 2 x 23 = 138g
the answer is 2
234 grams :)
424 divided by the atomic mass of sodium carbonate 105.99 g/mol
Dividing by the molar mass of sodium carbonate, we deduce that there are 4.25 x 10-5 moles in 4.5 x 10-3 grams of sodium carbonate.
6into23equals 138 6into23equals 138
0,028 moles carbonic are obtained.
234 grams
The formula mass of sodium carbonate, Na2CO3 is 2(23.0)+12+3(16) = 106.0Amount of sodium carbonate = mass of sample / formula mass = 4/106.0 = 0.0377mol
The molar mass of sodium hydrogen carbonate is 84 grams per mole, therefore 0.5 moles of it weighs 42 grams.
The carbonate ion is the conjugate base of a diprotic acid. If you react an equal number of moles of hydrochloric acid and sodium carbonate, the carbonate will only be partially neutralized you will get a mixture of sodium chloride and sodium bicarbonate. HCl + Na2CO3 --> NaHCO3 + NaCl Only by adding twice as many moles of HCl will you completely neutralize the sodium carbonate. 2HCl + Na2CO3 --> 2NaCl + H2O + CO2
My Shaft
3 x 2 x 23 = 138g