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yanalaym [24]
2 years ago
13

An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 6.50 g of this compound produced 9.53 g o

f carbon dioxide and 3.90 g of water. If 6.50 g of the unknown compound contained 0.217 mol of C and 0.433 mol of H, how many moles of oxygen, O, were in the sample?
Chemistry
1 answer:
OverLord2011 [107]2 years ago
4 0

Answer:

There were 0.216 moles of oxygen in the sample

Explanation:

Step 1: Data given

Mass of compound = 6.50 grams

Mass of CO2 = 9.53 grams

Molar mass of CO2 = 44.01 g/mol

Mass of H2O = 3.90 grams

Molar mass H2O = 18.02 g/mol

Step 2: Calculate moles CO2

Moles CO2 = 9.53 grams / 44.01 g/mol

Moles CO2 = 0.217 moles

Step 3: Calculate moles C

For 1 mol CO2 we have 1 mol C

For 0.217 moles CO2 we have 0.217 moles C

Step 4: Calculate mass C

Mass C = 0.217 moles C *12.01 g/mol

Mass C = 2.61 grams

Step 5: calculate moles H2O

Moles H2O = 3.90 grams / 18.02 g/mol

Moles H2O = 0.216 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.216 moles H2O we have 0.433 moles H

Step 7: Calculate mass H

Mass H = 0.433 moles * 1.01 g/mol

Mass H = 0.437 grams

Step 8: Calculate mass O

Mass O = 6.50 grams - 2.61 grams - 0.437 grams

Mass O = 3.453 grams

Step 9: Calculate moles O

Moles O = 3.453 grams / 16.0 g/mol

Moles O = 0.216 moles

There were 0.216 moles of oxygen in the sample

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The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
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The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
     NaA → Na+ + A- 
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
     Kb = [HA][OH-] / [A-] 
           = x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
     Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
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