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guapka [62]
3 years ago
5

A compound is 44.7% P and the rest O. Determine its emperical formula.

Chemistry
1 answer:
svet-max [94.6K]3 years ago
6 0

Answer:  

P₅O₁₂

<em>Explanation:  </em>

Assume that you have 100 g of the compound.  

Then you have 44.7 g P and 55.3 g O.  

1. Calculate the <em>moles</em> of each atom  

Moles of P  = 44.7 × 1/30.97 = 1.443 mol Al  

Moles of O = 55.3 × 1/16.00 = 3.456 mol O  

2. Calculate the <em>molar ratios</em>.  

P: 1.443/1.443 =   1  

O: 3.456/1.443 = 2.395

3. Multiply by a number to make the ratio close to an integer

P:  5 × 1         =  5

O: 5 × 2.395 = 11.97

3. Determine the <em>empirical formula </em>

Round off all numbers to the closest integer.  

P:   5

O: 12

The empirical formula is <em>P₅O₁₂</em>.  

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Explanation:

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3 0
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Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests a
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Answer:

                    SEPARATION SCHEME FOR  CATIONS

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    Step 1:   Add 6mol/dm^3 of HCl to the mixture solution

    Result : This would cause a precipitate of AgCl to be formed

    Reaction :  Ag^{+} _{(aq)} + Cl^{-} _{(aq)}  ---------> AgCl(ppt)

    Step 2 : Next is to remove the precipitate and add H_2S to the remaining          

                 solution in the presence of 0.2 \ mol/dm^3 of HCl

     Result : This would cause a precipitate of CuS to be formed

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     Step 3: Next remove the precipitate then add 6 \ mol/dm^3 of aqueous      

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                 process  is done then sort out the  precipitate from the  solution

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                 Next take out the precipitate to a different beaker and add HCl

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                 blood  indicating the presence of Fe^{3+}

             

   Reaction :   F^{3+}_{(aq)} + 30H^-_{(aq)} --------->Fe(OH)_3_{(aq)}

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                         Fe^{3+} + 6SCN^{-} -----> Fe(SCN)_6 ^{3-}

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Explanation:

6 0
3 years ago
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Answer:

9842kJ of energy are released

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Based on the reaction:

2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂(g) + 10 H₂O (g) + 5720 kJ

<em>When 2 moles of C₄H₁₀ react with 13 moles of O₂ there are released 5720 kJ of energy</em>

As molar mass of C₄H₁₀ is 58.12g/mol, moles in 200.0g of the hydrocarbon are:

200.0g C₄H₁₀ ₓ (1mol / 58.12g) =<em> 3.441 moles of C₄H₁₀</em>

<em> </em>

As 2 moles of C₄H₁₀ release 5720kJ of energy, 3.441 moles of C₄H₁₀ release:

3.441 moles C₄H₁₀ ₓ (5720kJ / 2 moles C₄H₁₀) = <em>9842kJ of energy are released</em>

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