Answer:
There are 3, 64 moles of NaCl.
Explanation:
First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate themoles in 213 grams of NaCl, making a simple rule of three:
Weight NaCl= Weight Na + Weight Cl = 23 g + 35, 5 g= 58, 5 g/ mol
58,5 g ------1 mol NaCl
213 g---------x= (213 g x 1 mol NaCl)/ 58, 5 g= <em>3, 64 mol NaCl</em>
5mg in liter is 5 ppm
Explanation: 1 ppm is one part per million.
1 ppm is 1 mg is one part of million from 1 kg = 1000 000 mg
1 litre water is 1. Kg.
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A gas with a vapor density greater than that of air, would be most effectively displaced out off a vessel by ventilation.
The two following principles determine the type of ventilation: Considering the impact of the contaminant's vapour density and either positive or negative pressure is applied.
Consider a vertical tank that is filled with methane gas. Methane would leak out if we opened the top hatch since its vapour density is far lower than that of air. A second opening could be built at the bottom to greatly increase the process' efficiency.
A faster atmospheric turnover would follow from air being pulled in via the bottom while the methane was vented out the top. The rate of natural ventilation will increase with the difference in vapour density. Numerous gases that require ventilation are either present in fairly low concentrations or have vapor densities close to one.
Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
