When HCl is added to a saturated solution of PbCl2, the solubility of PbCl2 would decrease so precipitation would occur. The decrease in the solubility is due to the common ion effect or the presence of Cl- ions in both compounds.
Answer:
Final mass = 159.5 g
Final temperature = 10 C
Final density = 1.00 g/ml
Explanation:
<u>Given:</u>
Beaker 1:
Mass of water = 44.3 g
Temperature = 10 C
Beaker 2:
Mass of water = 115.2 g
Temperature = 10 C
Density of water at 10C = 1.00 g/ml
<u>To determine:</u>
The final mass, temperature and density of water
<u>Calculation:</u>
Since there is no change in temperature, the final temperature will be 10 C
Density of a substance is an intensive property i.e. it is independent of the mass. Hence the density of water will remain constant i.e. 1.00 g/ml
Answer:
The correct answer is 40.7 grams.
Explanation:
Based on the given information, the volume of copper sulfate added to the solution is 255 ml or 0.255 L. The molarity of copper sulfate is not given, so let us consider it to be 1 M, which can also be written as 1 moles per liter.
The moles of copper sulfate can be determined by using the formula,
Moles = Molarity * Volume in Liters
Moles of CuSO4 = 1 moles/Liter * 0.255 L
Moles of CuSO4 = 0.255 moles
The mass of CuSO4 added in the solution will be,
Mass of CuSO4 = Moles * Molecular mass
= 0.255 moles * 159.609 grams per mole
= 40.7 grams.
So it's good to map out what you know you have and work from there:
We have two liter measurements and one mole measurement, and we need to find the moles.
For this problem, think of it this way: 46 liters of gas = 1.4 moles.
If one side changes, the other has to as well (if the liters decrease, the moles decrease. if the liters increase, so do the moles.) What you can do is put this into a fraction:
<span><u>1.4 moles</u></span>
46 L <span> </span>
if we know that each liter of gas is equal to x amount of moles, we know that 11.5 liters equals some amount of moles. You can put this into a fraction too, and make it equal to the other fraction:
<span><u>1.4 moles</u></span> = <u>x moles</u>
46 L 11.5 L
Then get your calculator out and do some algebra.
11.5 * (1.4/46) = x
The answer should come out to be: 0.35 moles