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We can see that the solubility of salt increases with increasing temperature. This happens with most substances.
To find out the maximum mass of copper sulfate that can be dissolved in water at these temperatures, just interpret the graph.
Considering Y-axis as g copper sulfate/100 g water and the X-axis as the temperature in °C:-
<u>1)</u>
a: <u>0 °C - 14 g of copper sulfate/100 g of water</u>
b: <u>50 °C - 34 g of copper sulfate/100 g of water</u>
c: <u>90 °C - 66 g of copper sulfate/100 g of </u><u>water</u>
<u>2)</u> From the graph, we can infer that temperature affects the solubility of the salt.
<em>Answered</em><em> </em><em>by</em><em> </em><em>Benjemin360</em><em> </em>:)
Answer:
A variable shape that adapts to fit its container.
The procedure, which can be used to determine more accurately the concentration of the unknown acid is TO BACK-TITRATE WITH ADDITIONAL HYDROCHLORIC ACID TO NEUTRALIZE THE ADDITIONAL SODIUM HYDROXIDE THAT WAS ADDED.
Monoprotic acids are acids that can donate only one proton per each molecule and they have only one equivalence point. Examples of monoprotic acids are HCI, HNO3 and CH3COOH.
The back titration method is typically used when one needs to determine the concentration of an analyte provided there is a known molar concentration of excess reactants.
From the information given in the question above, we are told that excess NaOH was added. To correct this mistake, the right thing to do is to use additional HCl to carry out back titration, taking note of the quantity of acid that will be needed to neutralize the excess NaOH.
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Answer : The concentration of HI (g) at equilibrium is, 0.643 M
Explanation :
The given chemical reaction is:
Initial conc. 0.10 0.10 0.50
At eqm. (0.10-x) (0.10-x) (0.50+2x)
As we are given:
The expression for equilibrium constant is:
![K_c=\frac{[HI]^2}{[H_2][I_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BHI%5D%5E2%7D%7B%5BH_2%5D%5BI_2%5D%7D)
Now put all the given values in this expression, we get:
x = 0.0713 and x = 0.134
We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.
Thus, we are taking value of x = 0.0713
The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M
Thus, the concentration of HI (g) at equilibrium is, 0.643 M
