The water molecules are not completely removed so additional heating is required.
Explanation:
We have the copper (II) sulfate pentehydrate with the chemical formula CuSO₄ · 5H₂O.
molar mass of CuSO₄ · 5H₂O = 159.6 + 5 × 18 = 249.6 g/mole
Knowing this, we devise the following reasoning:
if in 249.6 g of CuSO₄ · 5H₂O there are 90 g of H₂O
then in 8 g of CuSO₄ · 5H₂O there are Y g of H₂O
Y = (8 × 90) / 249.6 = 2.88 g of water
mass of dried CuSO₄ = mass of CuSO₄ · 5H₂O - mass of H₂O
mass of dried CuSO₄ = 8 - 2.88 = 5.12 g
5.12 g is less that the weighted mass of 6.50 g. We deduce from this that the sample needs additional heating in order to remove all the water (H₂O) molecules.
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Greater the Ka value greater is the acid strength. Among these three acids HClO3 is the strongest acid, with greatest Ka and lowest pKa value (-1), Then comes HBrO3, its the second most strongest acid among the three, its Pka value is 0.7, higher than HClO3 but smaller than HIO3 (i.e. 0.77) which the weakest acid among the three.
<span>The answer is A. In chemistry, Gibbs Free Energy is defined as the enthalpy of the system minus the product of the temperature times the entropy of the system. To write it as a formula, G = H - TS, delta G=delta H-T*delta S, where delta denotes the change.</span>
Find the [OH-] in the solution. The pH is 9.5, so the pOH is 14 - 9.5 = 4.5.
[OH-] = 10^-4.5 M
Now use the dilution equation to find the new [OH-] after the volume is reduced from 150 mL to 50 mL:
M1V1 = M2V2
M1 = 10^-4.5 M
V1 = 150 mL
M2 = ?
V2 = 50 mL
(10^-4.5)(150) = M2(50)
M2 = 9.5 x 10^-5 M ≈ 1 • 10^-4 (We can only use one sig fig, because the pH was given to one decimal place.)
Now use this [OH-] to find pOH:
pOH = -log(1 x 10^-4) = 4.0
14 - pOH = pH, so the expected pH for the new solution is 10.
Speakers converts electrical energy into sound energy. So I think it would be radio. Hope this helps
