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Ivan
4 years ago
11

93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0 C.what would be the volume of this dry gas at STP

Chemistry
1 answer:
ExtremeBDS [4]4 years ago
5 0
Let us assume that the oxygen behaves as an ideal gas such that we can use the ideal gas equation to solve for the number of moles of O2.
                       PV = nRT   ; n = PV/RT
Substituting the known values,
               n = (0.930 atm)(93/1000 L) / (0.0821 L.atm/mol.K)(10 + 273.15K)
                             n = 3.72 x 10^-3 mols
At STP, the volume of each mol of gas is equal to 22.4 L.
                          volume = (3.73 x 10^-3 mols) x (22.4 L/1 mol)
                           volume = 0.0833 L or 83.34 mL
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The Question is incomplete here is the complete question " Suppose an iron atom in the oxidation state +3 formed a complex with three hydroxide anions and three water molecules. Write the chemical formula of this complex.

Answer:

{Fe(OH)3(H20)3}

Explanation:

Oxidation state is the electron gained or lost by and atom, So if iron is in +3 state in formula it must have lost three electron.

We know that OH posses the oxidation state of -1 and water have zero oxidation state. SO, Let's take iron equal to y and find its oxidation state in the formula

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Using the appropriate ksp value from appendix d in the textbook, calculate the ph of a saturated solution of ca(oh)2 .
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The ph of a saturated solution of Ca(OH)2 is 12.35

CALCULATION:
For the reaction 
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we will use the Ksp expression to solve for the concentration [OH-] and then use the acid base concepts to get the pH:
     Ksp = [Ca2+][OH-]^2

The listed Ksp value is 5.5 x 10^-6. Substituting this to the Ksp expression, we have
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Taking the cube root, we now have
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We know that the value of [OH-] is actually equal to 2s:
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We can now calculate for pOH:          
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Therefore, the pH is
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4 years ago
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