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stepan [7]
2 years ago
14

Use the solubility rules from the Lab 4 introduction and your knowledge of qualitative separation schemes from the lab to answer

the following questions. The qualitative analysis experiment you did is actually an abbreviated version of a much larger analysis scheme in which many different cations are separated and identified. Suppose a mixture contains Ag , K , NH4 , Hg22 , Pb2 , Mg2 , Sr2 , Ba2 , Cu2 , Al3 and Fe3 .
(a) Which of the following ions could you separate, by causing them to precipitate, with the addition of HCl?
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
(b) After the addition of HCl, the above sample is centrifuged and decanted. Which of the following cations remaining in the supernatant could you separate, by causing them to precipitate, with the addition of H2SO4? (Hint: H2SO4 is a source of sulfate ions. Select all that apply.)
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
Chemistry
1 answer:
cluponka [151]2 years ago
4 0

Answer:

a13+a13

Explanation:

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2 years ago
The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.
Marina86 [1]

Answer:

Therefore it will take 7.66 hours for 80% of the lead decay.

Explanation:

The differential equation for decay is

\frac{dA}{dt}= kA

\Rightarrow \frac{dA}{A}=kdt

Integrating both sides

ln A= kt+c₁

\Rightarrow A= e^{kt+c_1}

\Rightarrow A=Ce^{kt}         [e^{c_1}=C]

The initial condition is A(0)= A₀,

\therefore A_0=Ce^{0.k}

\Rightarrow C=A_0

\therefore A=A_0e^{kt}.........(1)

Given that the Pb_{209}  has half life of 3.3 hours.

For half life A=\frac12 A_0 putting this in equation (1)

\frac12A_0=A_0e^{k\times3.3}

\Rightarrow ln(\frac12)= 3.3k     [taking ln both sides, ln \ e^a=a]

\Rightarrow k=\frac{ln \frac12}{3.3}

⇒k= - 0.21

Now A₀= 1 gram, 80%=0.8

and A= (1-0.8)A₀ = (0.2×1) gram = 0.2 gram

Now putting the value of k,A and A₀in the equation (1)

\therefore 0.2=1e^{(-0.21)\times t}

\Rightarrow e^{-0.21t}=0.2

\Rightarrow -0.21t= ln(0. 2)

\Rightarrow t= \frac{ln (0.2)}{-0.21}

⇒ t≈7.66

Therefore it will take 7.66 hours for 80% of the lead decay.

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3 years ago
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A chemist prepares a solution of sodium chloride(NaCl) by measuring out 25.4g of sodium chloride into a 100ml volumetric flask a
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Answer: 4 molL-1

Explanation:

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Answer:

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Explanation:

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