<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
carbon atoms form 2 bonds with sharing valence electrons
The base units of length and volume are linked in the metric system. By definition, a liter<span> is equal to the volume of a cube exactly 10 </span>cm<span> tall, 10 </span>cm<span> long, and 10 </span>cm<span> wide. Because the volume of this cube is 1000 cubic </span>centimeters<span> and a </span>liter<span> contains 1000 milliliters, 1 milliliter is equivalent to 1 cubic centimeter.</span>
The correct option is B. To get the number of atom for each compound, each element in the compound will be counted as an atom. For instance, for Fe[ClO4]2, there are 1 atom of Fe, 2 atoms of Cl, and 8 atoms of O, making a total of 11 atoms [1 + 2 + 8= 11]. The other options have less than 11 atoms.
Answer:
Supersaturated solution.
Explanation:
Hello!
In this case, according to the types of solution in terms of the relative amounts of solute and solvent, we can define a point called solubility at which the amount of solute is no longer dissolved in the solvent; thus, a value of solute/solvent less than the solubility is related to unsaturated solutions, equal to the solubility is related to the saturated solutions and more than the solubility to supersaturated solutions.
Thus, since solubility is temperature-dependent, at 30 °C the solubility of sodium chloride is 36.09 g per 100 mL of water; which means that, since the solution has 50 g of sodium chloride, more than 36.09 g, we infer this is a supersaturated solution.
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