A student determines that a 1.5 g mixture of CaCO3(s) and NaHCO3(s) contains 0.010 mol of NaHCO3(s). Based on the student's
measurement, what is the mass percent of Na in the mixture?

We are given:

Mass of the mixture = 1.5 grams

Number of Moles of NaHCO₃ = 0.01 mole

Mass of 0.01 moles of NaHCO₃:

Mass = number of moles * Molar mass

Mass = 0.01 * 84

Mass = 0.84 grams

Mass% of Na in NaHCO₃:

Molar mass of NaHCO₃ = 84 grams/mol

Molar mass of Na = 23 grams / mol

Mass% = (Molar mass of Na / Molar mass of NaHCO₃) * 100

Mass% = (23/84) * 100

Mass% = 27%

Mass of Na in 0.01 moles:

Since the Mass% of Na is 27% and the total mass is 0.84 grams

So, we can say that 27% of 0.84 is the mass of Na

Mass of Na = 0.84 * 0.27

Mass of Na = 0.23 grams

Mass% of Na in the mixture:

Total mass of the mixture = 1.5 grams

Mass of Na in the mixture = 0.23 grams

Mass% of of Na in the Mixture = (Mass of Na / Mass of Mixture) * 100

Mass% of Na = (0.23 / 1.5) * 100

Mass% of Na = 15.3%

Therefore, we have 15.3% Na in the given mixture by mass

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.

answer: i dont know the answer but this might

explanation:

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