Answer:
But since the solubility product constant for each compound is provided, their relative solubility can be ranked from highest to lowest. Depending on the ranking above, it is evident that aluminum hydroxide Al(OH)3 A l ( O H ) 3 has the lowest solubility at 25 Celsius degreesAs temperature increases, its solubility increases as well. Notice, however, that it does not increase significantly. In fact, you can expect to be able to dissolve no more than 40 g of sodium chloride per 100 g of water at 80∘C
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.
The answer to this problem is quite simple, it’s 9
Answer: 306 grams are there in 2.50 moles of potassium chlorate
Explanation
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the number of moles, we use the equation:
given mass of potassium chlorate = ?
Molar mass of potassium chlorate = 122.55 g/mol
Putting in the values we get:
306 grams are there in 2.50 moles of potassium chlorate
