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1 year ago

Reaction between iodine solution and polysaccharides

Chemistry

2 answers:

Radda [10]1 year ago

8 0

Explanation:

7 4th CNN to CNN u 4th f7 4th d7 hmm

ElenaW [278]1 year ago

3 0

In the iodine test in this lodine and potassium iodide (KI) in water, the solution turns light orange brown colour.

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Explain the difference between a base unit and a derived unit

Scrat [10]

<span>A SI base unit is the single and direct measurement unit for a physical entity. E.g. For mass it is kg, for time it is sec, A derived unit is determined by a physical equation with the base units, like for velocity equals distance divided by time, in SI units: unit of v (velocity)= m (meter) divided by sec (seconds).

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2 years ago

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OCAS GAME

makkiz [27]

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3 years ago

5 Facts about cell division

Kay [80]

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2 years ago

Purification of copper can be achieved by electrorefining copper from an impure copper anode onto a pure copper cathode in an el

Dovator [93]

Answer: 406 hours

Explanation:

$Q=I\times t$

where Q= quantity of electricity in coloumbs

I = current in amperes = 39.5 A

t= time in seconds = ?

The deposition of copper at cathode is represented by:

$Cu^{2+}+2e^-\rightarrow Cu$

$96500\times 2=193000$ Coloumb of electricity deposits 1 mole of copper

i.e. 63.5 g of copper is deposited by = 193000 Coloumb

Thus 19.0 kg or 19000 g of copper is deposited by = $\frac{193000}{63.5}\times 19000=57748032$ Coloumb

$57748032=39.5\times t$

$t=1461975sec=406hours$ (1hour=3600s)

Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A

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2 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$