(need all boxes answered)
bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. its formula is nahco3. find the mass percentages (mass %) of na, h, c, and o in sodium hydrogen carbonate. you may use whole numbers for the masses.
the mass of one mole of nahco3 is:

Answer & Explanation:

∴ The mass of 1.0 mol of NaHCO₃ = (no. of moles of NaHCO₃)(the molecular mass of NaHCO₃)

∴ The mass of 1.0 mol of NaHCO₃ = (1)(the molecular mass of NaHCO₃) = (the molecular mass of NaHCO₃).

∴ The mass of 1.0 mol of NaHCO₃ = (atomic weight of Na) + (atomic weight of H) + (atomic weight of C) + 3(atomic weight of O) = (23.0 g of Na) + (1.0 g of H) + (12.0 g of C) + 3(16.0 g of O) = (23.0 g of Na) + (1.0 g of H) + (12.0 g of C) + (48.0 g of O) = 84.0 g.

For Na:

The mass% of Na = (the mass of Na)/(the mass of  NaHCO₃) = (23.0 g)/(84.0 g) x 100 = 27.381%.

For H:

The mass% of H = (the mass of H)/(the mass of  NaHCO₃) = (1.0 g)/(84.0 g) x 100 = 1.19%.

For C:

The mass% of C = (the mass of C)/(the mass of  NaHCO₃) = (12.0 g)/(84.0 g) x 100 = 14.285%.

For O:

The mass% of O = (the mass of O)/(the mass of  NaHCO₃) = (48.0 g)/(84.0 g) x 100 = 57.143%.

The sum of % of elements = mass% of Na + mass% of H + mass% of C + mass% of O = 27.381% + 1.19% + 14.285% + 57.143% = 99.999% ≅ 100.0%.

Here's what I get.

Explanation:

b. examine and describe cells