The empirical formula : MnO₂.
<h3>Further explanation</h3>
Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
- Find mol(mass : atomic mass)
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
- Divide by the smallest mol(Mn=0.0115)
Mn : O =

The empirical formula : MnO₂
Because of making the equestion to be more stable and form the best compound according to it's reactant.
Answer:
It evaporates and moves into the air.
Explanation:
When water is left out for a while, it evaporates into the air! :)
Sure. Whith what do you need help?
Answer:
3) NaCl.
Explanation:
<em>∵ ΔTf = iKf.m</em>
where, <em>i</em> is the van 't Hoff factor.
<em>Kf </em>is the molal depression freezing constant.
<em>m</em> is the molality of the solute.
<em>The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. </em>
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- For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.
<em>So, for sugar: i = 1.</em>
<em>∴ ΔTf for sugar = iKf.m = (1)(Kf)(2.0 m) = 2 Kf.</em>
<em></em>
- For most ionic compounds dissolved in water, the van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.
For NaCl, it is electrolyte compound which dissociates to Na⁺ and Cl⁻.
<em>So, i for NaCl = 2.</em>
<em>∴ ΔTf for NaCl = iKf.m = (2)(Kf)(1.0 m) = 2 Kf.</em>
<em></em>
<em>So, the right choice is: 3) NaCl.</em>
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