can someone please tell me what formula's to use for the 2 following chemistry problems.

1. Calculate the solubility of potassium bromide at 23 degrees Celsius. hint. Assume that the solubility increases by an equal amount for each degrees between 20 degrees C and 30 degrees C. at 20 degrees Celsius it is 65.2.
at 30 degrees Celsius it is 70.6.

2. a saturated solution of barium chloride at 30 degrees Celsius contains 150 g of water. How much additional barium chloride can be dissolved by heating this solution to 60 degrees Celsius.
F.Y.I. these are the entire questions I think I could figure it out if I knew what formula to use.

thanking you in advance

#1.

Take the solubility at 30 and subtract solubility at 20 and divide by 10. That gives you an average solubility per degree C. You have it at 20, then add 3*solubility/degree and you will have it at 23 c.

#2. Surely you have a solubility curve or a graph that will tell you the solubility at 60 C. There is nothing in the problem that tells you what the solubility is in the saturated solution.

I will make some assumptions for #2.

That the 85.5 g solubility you quote later is 85.5g BaCl2/100 g water @ 60 C. If that's so then you will be able to dissolve 85.5g x (150gH2O/100 gH2O) = approx 128 g BaCl2 that can be dissolved at 60 C. The only piece of information needed from the 30 C solution is the 150 g H2O. Basically, this is the question you are being asked although it is masquerading with that other information. .
If the solubility of BaCl2 at 60 C is 85.5 g BaCl2/100 g H2O, how much BaCl2 can be dissolved in 150 g H2O at 60 C? Stating it that way makes it simpler I think.

Tero aama urgent randi

For the first problem, you can use the linear interpolation formula to calculate the solubility of potassium bromide at 23 degrees Celsius.

Linear Interpolation Formula:
y = y1 + [(x - x1)/(x2 - x1)] * (y2 - y1)

In this formula:
- y represents the desired value (solubility at 23 degrees Celsius)
- y1 and y2 are the known values (solubility at 20 and 30 degrees Celsius, respectively)
- x represents the variable (temperature)
- x1 and x2 are the known values of the variable (20 and 30 degrees Celsius, respectively)

Given data:
x1 = 20 degrees Celsius, y1 = 65.2 (solubility at 20 degrees Celsius)
x2 = 30 degrees Celsius, y2 = 70.6 (solubility at 30 degrees Celsius)
x = 23 degrees Celsius (what we need to find)

Using the formula, you can substitute the values into the formula and solve for y:

solubility at 23 degrees Celsius = y = 65.2 + [(23 - 20)/(30 - 20)] * (70.6 - 65.2)

Now, for the second problem, you need to calculate the additional amount of barium chloride that can be dissolved by heating the solution.

To do that, you need to use the solubility curve. The solubility of a compound usually increases with increasing temperature. The additional amount of barium chloride that can be dissolved is the difference between the solubility at 60 degrees Celsius and the initial amount of barium chloride in the saturated solution at 30 degrees Celsius.

You will need to find the solubility values of barium chloride at 30 and 60 degrees Celsius. Once you have those values, you can calculate the difference and determine the additional amount that can be dissolved.

You may need to refer to a solubility chart or lookup table specific to barium chloride at different temperatures to find the solubility values.

Unfortunately, without specific values for solubilities, it is not possible to provide the exact calculations for these problems. However, the explanations above outline the general approach and formulas required to solve both problems.