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saw5 [17]
2 years ago
13

Fe(NO3)3 Percent composition

Chemistry
1 answer:
Scilla [17]2 years ago
5 0

Answer:

Fe: 23.1%

N: 17.4%

O: 59.5%

Explanation:

You take the grams that each weigh on the periodic table and add it all up. Afterward you create fractions of each over the whole. (Example: Fe/Fe+N+O) then you divide and turn that decimal into a percentage.

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A Stimulus is the answer
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Complete this neutralization equation:<br> H2SO4 + Al(OH)3
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Answer:

H2SO4 + Al(OH)3 = Al2(SO4)3 + H2O

Explanation:

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Please answer as (a;3) &lt; example&gt;
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A;4, b;6, c;7, d;5, e;8, f;3
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When 500.0 g of water is decomposed by electrolysis and the yield of hydrogen is only 75.3%, how much hydrogen chloride can be m
Evgen [1.6K]

The amount of hydrogen chloride that can be made is 1064 g

Why?

The two reactions are:

2H₂O → 2H₂ + O₂ 75.3 % yield

H₂ + Cl₂ → 2HCl 69.8% yield

We have to apply a big conversion factor to go from grams of water (The limiting reactant), to grams of HCl, the final product. We have to be very careful with the coefficients and percentage yields!

500.0gH_2O*\frac{1moleH_2O}{18.01 gH_2O}*\frac{2 moles H_2}{2 moles H_2O}*\frac{2.015g H_2}{1 mole H_2}*\frac{75.3 actual g}{100 theoretical g}=42.12 g H_2

42.12H_2*\frac{1 mole H_2}{2.015gH_2}*\frac{2 moles HCl}{1 mole H_2}*\frac{36.46g}{1 mole HCl}*\frac{69.8 actualg}{100 theoreticalg} =1064gHCl

Have a nice day!

#LearnwithBrainly

7 0
3 years ago
Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests a
fiasKO [112]

Answer:

                    SEPARATION SCHEME FOR  CATIONS

GIVEN  CATIONS : Ag^{+} \ ,  Fe^{3+} , Cu^{2+}, Ni^{2+}

     

    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

     Reaction :  Cu^{2+}_{(aq)} + S^{2-}_{(aq)} ------> Cu_2S(ppt)

 

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

                 NH_3 (NH_3 \cdot H_2 O) , process the solution in a centrifuge,when the  

                 process  is done then sort out the  precipitate from the  solution

                 Now this precipitate is   Fe(OH)_3 and the remaining solution

                contains  (Ni (NH_3)_6)

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

                to it   this will dissolve it, then add a drop of NH_4SCN this will

                form  a precipitate  Fe(SCN)_{6}^{3-} which will have the color of

                 blood  indicating the presence of Fe^{3+}

             

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

                        Fe (OH)_{(s)} _3  + 3H^{+}_{aq} -------> Fe^{3+}_{aq} + 3H_2O_{(l)}

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

                      Now the remaining mixture contains Ni^{2+}

     

       

Explanation:

6 0
3 years ago
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