The masses of the components are obtained as;
- Sodium hydrogen carbonate = 3.51 g
- Sodium carbonate = 8.708 g
<h3>What is decomposition?</h3>
The term decomposition has to do with the breakdown of the given substance into its components. The components of sodium hydrogen carbonate could be identified as water vapor, carbon dioxide gas and sodium carbonate. Among these products that have been listed here, we can see that it is only the sodium carbonate that remains as a solid. The others are gases that move away from the system that is under study.
Now putting down the equation of the reaction, we have;

Now, the loss in mass must be due to the carbon dioxide and the water. Hence we obtain the loss in mass to be 10.000 g - 8.708 g = 1.292 g
Mass of sodium hydrogen carbonate = 2 * 88 g/mol * 1.292 g/62 g/mol
= 3.51 g
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A pH scale runs from 1 to 14 with 7 being neutral.
1-6 has base like properties
8-14 has avid line properties
since this solution has a pH scale of 4.... the solution is basic
I can't actually answer this one if the empirical formula is not given. Luckily, I've found a similar problem from another website. The problem is shown in the picture attached. It shows that the empirical formula is CH₂O. Let's calculate the molar mass of the empirical formula.
Molar mass of E.F = 12 + 2(1) + 16 = 30 g/mol
Then, let's divide this to the molar mass of the molecular formula.
Molar mass of M.F/Molar mass of E.F = 180/30 = 6
Therefore, let's multiply 6 to each subscript in the empirical formula to determine the actual molecular formula.
<em>Actual molecular formula = C₆H₁₂O₆</em>
Answer:
Answer D => E°(Mg°/Cu⁺²) = 0.34 + 2.37 = 2.71v
Explanation:
(Oxidation) => Mg°(s) => Mg⁺²(aq) + 2e⁻ E°(Mg°/Mg⁺²) = -2.37 v
(Reduction) => Cu⁺²(aq) + 2e⁻ => Cu°(s) E°(Cu⁺²/Cu°) = +0.34 v
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Net Rxn => Mg°(s) + Cu⁺²(aq) => Mg⁺²(aq) + Cu°(s)
Std Cell Potential (25°C/1Atm) = E°(Redn) = E°(Oxidn) = +0.34v - (-2.37v)
= 0.34v + 2.37v = 2.72v