0.1875 mol of Na2CO3 needed to react with the 750 mL of 0.250 M H2SO4.
The given reaction between Na2CO3 and H2SO4 are already balanced.
Na2CO3 + H2SO4 → Na2SO4 + CO2.H2O
First, let us determine the number of moles of H2SO4.
Molarity (mol/L) = number of moles of H2SO4 / volume of H2SO4
0.250 M = number of moles of H2SO4 / [(750 mL)(1 L/1000 mL)]
number of moles of H2SO4 = 0.1875 mol
Then, let us determine the number of moles of Na2CO3 that would react with 0.1875 mol H2SO4.
number of moles of Na2CO3 = moles of H2SO4 * stoichiometric ratio
number of moles of Na2CO3 = 0.1875 mol H2SO4 * (1 mol Na2CO3/ mol H2SO4)
number of moles of Na2CO3 = 0.1875 mol Na2CO3
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Physical damage.
If you burned the pencil, it would be considered chemical.
Done with an experiement and if he has enough data he can compare it and make his results more accurate please make me brainliest
Just took the test and it was 32grams. On my test it was Block B: 32 Grams
