Chemistry, 01.07.2019 00:10 lazerlemon500

(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:

(need all boxes answered) bicarbonate of soda (sodium hydrogen carbonate) is used in many comm

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Answer from: yeet1273

Answer & Explanation:

If we have 1.0 mol of NaHCO₃:

∴ 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.

To find the mass% of each element:

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%.

Answer from: angelinad02

Here's what I get.

Explanation:

$\text{The mass of one mole of NaHCO$_{3}$ is:}\\\\\text{$\boxed{23}$ g Na +$\boxed{1}$ g H + $\boxed{12}$ g C + $\boxed{48}$ g O = $\boxed{84}$ g}$

$\text{mass $\%$ Na = $\boxed{23}$ g $/$ $\boxed{84}$ g $\times$ 100 = $\boxed{27.38}$ $\%$}\\\\\text{mass $\%$ H = $ \quad \boxed{1}$ g $/$ $\boxed{84}$ g $\times$ 100 = $\boxed{1.19}$ $\%$}\\\\\text{mass $\%$ C = $\:\: \boxed{12}$ g $/$ $\boxed{84}$ g $\times$ 100 = $\boxed{14.29}$ $\%$}\\\\\text{mass $\%$ O = $\:\: \boxed{48}$ g $/$ $\boxed{84}$ g $\times$ 100 = $\boxed{57.14}$ $\%$}$

$\boxed{27.38} + \boxed{1.19} + \boxed{14.29} + \boxed{ 57.14} = \boxed{100.00}$

Answer from: Quest

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.

chemical decomposition is opposite of chemical synthesis.

Answer from: Quest

That is a lot of moles that you have