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Pie
2 years ago
10

The solubility of CaSO4 in pure water at 0oC is 1.09 gram(s) per liter. The value of the solubility product is g

Chemistry
1 answer:
malfutka [58]2 years ago
7 0

Answer: See image attached!

Good luck with your future exams!

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According to the graph, what is the instantaneous speed of the object at 5 seconds? *
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Is carbon dioxide a iconic or covalent compound
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3 0
2 years ago
You make a solution by putting 45.6g of iron lll carbonate into 167ml of water. What is it's molarity?
ludmilkaskok [199]
1. The molar mass of Fe2(CO3)3 is 291.72 g/mol. This means that 45.6 g is equivalent to 0.156 mol. Dividing by the 0.167 L of water gives a solution of 0.936 M.
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4 0
3 years ago
Match the solutions to the descriptions of the freezing points.a. One mole of the ionic compound Na3PO4 dissolved in 1000 g H2O
rosijanka [135]

Explanation:

Depression in Freezing point

= Kf × i × m

where m is molality , i is Van't Hoff factor, m = molality

Since molality and Kf remain the same

depression in freezing point is proportional to i

i= 2 for CuSO4 ( CuSO4----------> Cu+2 + SO4-2

i=1 for C2h6O

i= 3 for MgCl2 ( MgCl2--------> Mg+2+ 2Cl-)

So the freezing point depression is highest for MgCl2 and lowest for C2H6O

so freezing point of the solution = freezing point of pure solvent- freezing point depression

since MgCl2 has got highest freezing point depression it will have loweest freezing point and C2H6O will have highest freezing point

5 0
2 years ago
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