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insens350 [35]
3 years ago
13

An organism had 1,000 grams of carbon-14 (a radioactive form of carbon) in it when it died. How much remains after five half-liv

es?
Chemistry
1 answer:
Katarina [22]3 years ago
5 0

Answer:

After 5th half life the remaining mass is 31.25 g.

Explanation:

Given data:

Total mass of carbon-14 = 1000 g

Mass remain after 5 half lives = ?

Solution:

At time zero = 1000 g

At first half life = 1000 g/2 = 500 g

At second half life =  500 g/ 2= 250 g

At third half life = 250 g/ 2 = 125 g

At 4th half life = 125 g/2 = 62.5 g

At 5th half life = 62.5 g/2 = 31.25 g

Thus after 5th half life the remaining mass is 31.25 g.

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A generic weak acid with formula HA has a Ka = 2.76 x 10-8. Calculate the Kb for the conjugate base of the acid.
Reika [66]

Answer:

3.62x10⁻⁷ = Kb

Explanation:

The acid equilibrium of a weak acid, HX, is:

HX + H₂O ⇄ X⁻ + H₃O⁺

Where Ka = [X⁻] [H₃O⁺] / [HX]

And basic equilibrium of the conjugate base, is:

X⁻ + H₂O ⇄ OH⁻ + HX

Where Kb = [OH⁻] [HX] / [X⁻]

To convert Ka to Kb we must use water equilibrium:

2H₂O ⇄ H₃O⁺ + OH⁻

Where Kw = 1x10⁻¹⁴ = [OH⁻] [H₃O⁺]

Thus, we can obtain:

Kw = Ka*Kb

Solving for Kb:

Kw / Ka = Kb

1x10⁻¹⁴ /  2.76x10⁻⁸ =

3.62x10⁻⁷ = Kb

4 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:

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