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.
Explanation:
its b cz the gain electrons i think
4,410 kJ
Explanation:
Gravitational Potential Enegry (GPE) is calculated as;
GPE = <em>m*g*h</em> where;
m = mass (kg)
g = gravity (m/s²)
h = height (meters)
= 90 * 9.8 * 5000
= 4,410,000 joules
= 4,410 kJ
The empirical formula for a compound composed of 0.0683 mol of carbon ( C ), 0.0341 mol of hydrogen ( H ), and 0.1024 mol of nitrogen ( N ) is
.
<h3>What is the empirical formula?</h3>
An empirical formula tells us the relative ratios of different atoms in a compound.
Given data:
Moles of carbon = 0.0683 mol
Moles of hydrogen = 0.0341 mol
Moles of nitrogen = 0.1024 mol
Dividing each mole using the smallest number that is divided by 0.0341 moles.
We get:
Carbon= 2
Hydrogen=1
Nitrogen=3
The empirical formula for a compound is
.
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Answer:
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Explanation: