The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
Hey there!
Na + FeBr₂ → NaBr + Fe
Firstly, balance Br.
Two on the left, one on the right. Add a coefficient of 2 in front of NaBr.
Na + FeBr₂ → 2NaBr + Fe
Next, balance Na.
One on the left, two on the right. Add a coefficient of 2 in front of Na.
2Na + FeBr₂ → 2NaBr + Fe
Lastly, balance Fe.
One on the left, one on the right. Already balanced.
Our final balanced equation:
2Na + FeBr₂ → 2NaBr + Fe
Hope this helps!
Im pretty sure the answer is distance and mass
The percentage of the sulfur (S) in the compound CuSO₄ is 20.1 %.
<h3>What is the mass percentage?</h3>
The percentage of an element in a compound can be determined as the number of parts by mass of that element present in 100 parts by mass of the given compound.
First, calculate the molecular mass of the given compound by the addition of the atomic masses of all the present elements in the molecular formula. Then, the percentage of the elements can be determined by dividing the total mass of the element by the molar mass of the compound multiplied by 100.
Given, the atomic mass of copper, sulfur, and oxygen is 63.55 g, 32.07 g, and 16.0g respectively.
The molecular mass of CuSO₄ = 63.55 + 32.07 + 4(16.0) = 159.62 g
The mass percentage of the sulphur = (32.07/159.62) × 100 = 20.1 %
Therefore, the mass percentage of the sulfur is equal to 20.1 %.
Learn more about the mass percentage, here:
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