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KiRa [710]
2 years ago
9

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

8 0
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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