In order to obtain solid NaCl, the student should do a few steps.
First, he/she should do filtration. Pass the mixture through a filter paper, where all the sand should be filtered out already because they're not dissolved in the solution plus they're too small to pass through the filter paper.
Next, the filtrate should be left with NaCl (aqueous state). To seperate NaCl with the liquid, the student can either do evaporation or crystallization, depending on how pure or fast he/she wants the results to be. Evaporation involves heating the beaker or whatever apparatus under the bunsen burner until all the liquid has evaporated. Then, some white powder should be left, they're NaCl solid. For crystallization, the student should just put the beaker on a room condition environment, and wait. They might have to wait a month or so for the liquid to completely evaporate itself and left with clear and pure NaCl crystals.
From the balanced equation 2KClO3 → 2KCl + 3O2, the coefficients are the following:
coefficient 2 in front of potassium chlorate KClO3
coefficient 2 in front of potassium chloride KCl
coefficient 3 in front of oxygen molecule O2
We got this balanced equation by identifying the number of atoms of each element that we have in the given equation KClO3 → KCl + O2.
Looking at the subscripts of each atom on the reactant side and on the product side, we have
KClO3 → KCl + O2
K=1 K=1
Cl=1 Cl=1
O=3 O=2
We can see that the oxygens are not balanced. We add a coefficient 2 to the 3 oxygen atoms on the left side and another coefficient 3 to the 2 oxygen
atoms on the right side to balance the oxygens:
2KClO3 → KCl + 3O2
The coefficient 2 in front of potassium chlorate KClO3 multiplied by the subscript 3 of the oxygen atoms on the left side indicates 6 oxygen atoms just as the coefficient 3 multiplied by the subscript 2 on the right side indicates 6 oxygen atoms.
The number of potassium K atoms and chloride Cl atoms have changed as well:
2KClO3 → KCl + 3O2
K=2 K=1
Cl=2 Cl=1
O=6 O=6
We now have two potassium K atoms and two chloride Cl atoms on the reactant side, so we add a coefficient 2 to the potassium chloride KCl on the product side:
2KClO3 → 2KCl + 3O2, which is our final balanced equation.
K=2 K=2
Cl=2 Cl=2
O=6 O=6
The potassium, chlorine, and oxygen atoms are now balanced.
Answer:
Explanation:
Hello!
In this case for the solution you are given, we first use the mass to compute the moles of CuNO3:
Next, knowing that the molarity has units of moles over liters, we can solve for volume as follows:
By plugging in the moles and molarity, we obtain:
Which in mL is:
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