In fact, this problem belongs to the chemistry section. Recall that many other sciences require mathematical calculations. The problem will belong to Mathematics only if no knowledge of other sciences are required to solve the problem.
Solubility for the given substances is measured in grams per 100 g of water at a particular temperature (20 deg.C).
This means that the mass (assumed to be the solute) will not change the solubility, just the minimum quantity of solvent (water) will.
Thus the solubility of sodium chloride will remain L=36 g/100g H2O for any quantity of solute. Similarly, the solubility of lead nitrate will remain as K=54 g/100 g H2O.
The reason that they remain constant is because the quantity of solvent (water) is fixed at 100 g. Varying amount of solute will affect the quantity of solvent required, but not the solubility.
I'll leave it to you to calculate the difference between K & L.
I think this should be the answer!!
The way you wrote them was right
Answer:
153.86 un²
Step-by-step explanation:
Area circle = πr²
Radius is the distance of the center to the edge(perimeter)
Radius = 7
Area circle = 3.14(7)²
Area circle = 3.14 * 49 = 153.86
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-Chetan K
Answer:
y=5
Step-by-step explanation:
First, distribute the 2 to the 3y and -7
6y - 14 = 16
Second, add 14 to both sides.
6y = 30
Third, divide both sides by 6
y = 5
As you can see y equals 5!
I hope the explanation helped and have a great day!