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2 years ago

Finally, it is time to return to Earth and review how scientists here classify organisms:

Chemistry

1 answer:

lilavasa [31]2 years ago

6 0

5. Eubacteria

6. Plantae

7. Animalia

8. Protist (technically not a kingdom)

9. Archaebacteria

10. Fungi

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How long would it take a current of 10A to deposit 6.36g of copper during the electrolysis of copper (ll) tetraoxosulphate VI so

zhuklara [117]

It take 0.54 hours to deposit 6.36g of copper

<h3>Further explanation</h3>

Faraday's Law I

"The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis

W = e.i.t / 96500

$\tt \large {\boxed {\bold {W \: = \: \dfrac {e \times i \: x \: t} {96500}}}}$

e = equivalent = Ar / valence

i = current, A

t = time, s

W=6.36 g

e = 63.5 : 2 =31.75

i = 10 A

$\tt t=\dfrac{W\times 96500}{e.i}\\\\t=\dfrac{6.36\times 96500}{31.75\times 10 }\\\\t=1933.04~s\approx 0.54~hours$

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3 years ago

Suppose that some FeSCN2+ is added to the above solution to shift the equilibrium. When equilibrium is re-established, the follo

kiruha [24]

Explanation:

The given reaction equation will be as follows.

$[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]$

Let is assume that at equilibrium the concentrations of given species are as follows.

$[Fe^{3+}] = 8.17 \times 10^{-3}$ M

$[SCN^{-}] = 8.60 \times 10^{-3}$ M

$[FeSCN^{2+}] = 6.25 \times 10^{-2}$ M

Now, first calculate the value of $K_{eq}$ as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

= $\frac{8.17 \times 10^{-3} \times 8.60 \times 10^{-3}}{6.25 \times 10^{-2}}$

= $11.24 \times 10^{-4}$

Now, according to the concentration values at the re-established equilibrium the value for $[FeSCN^{2+}]$ will be calculated as follows.

$K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}$

$11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}$

$[FeSCN^{2+}] = 5.66 \times 10^{-2}$ M

Thus, we can conclude that the concentration of $[FeSCN^{2+}]$ in the new equilibrium mixture is $5.66 \times 10^{-2}$ M.

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