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2 years ago

How is impure copper refined by electrolysis

Chemistry

1 answer:

Gelneren [198K]2 years ago

6 0

Answer:

By giving electricity to copper compound solution.

Explanation:

Electrolysis is one of the major way of refined copper. The copper containing solution has two electrodes.i) positive electrodes called anode. ii) negative electrodes called cathode. When electricity is pass into the copper containing solution electrolysis process is starts and impure copper is formed in anode and pure copper is formed in cathode.

So, We can get pure copper in cathode through electrolysis.

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Find density of nitrogen dioxide at 75*C and 0.805 atm.

Eva8 [605]

Answer:

1 (348) (D2) = 273 (2.05) (0.805) D2= 1.29 g/L

Explanation:

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3 years ago

How many bonds are broken in 2O2

nirvana33 [79]

Answer:

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2 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0

2 years ago

Read 2 more answers

It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?

Mnenie [13.5K]

From the calculations, the half life of the material is 6.5 days.

<h3>What is radioactivity?</h3>

The term radioactivity has to do with the spontaneous disintegration of a specie.

Uisng the formula;

N=Noe^-kt

N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq

No = amount initially present = 1.75 x 10^12 Bq

k = rate constant = ?

t = time taken = 55 days

Hence;

4.995 ×10^9 = 1.75 x 10^12e^-55k

4.995 ×10^9/1.75 x 10^12 = e^-55k

2.85 * 10^-3 = e^-55k

ln2.85 * 10^-3 = -55k

k = ln2.85 * 10^-3/-55

k = 0.1066 day-1

Half life = 0.693/ 0.1066 day-1

= 6.5 days

Learn more about radioactivity:brainly.com/question/1770619

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4 0

1 year ago

6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)

deff fn [24]

Answer:

1120 gm

Explanation:

6. Consider the reaction: CzHo (g) + 02 (8) - 4 CO2(g) + 6H2O (1)

(a) Balance the equation.

(b) How many grams of oxygen are required to react with 10 moles of ethane for a complete

combustion reaction?

FIRST, CORRECT THE EQUATION THEN BALANCE

2C2H6(G) + 7O2------------> 4CO2 + 6H2O

so for 10 moles of ethane, we need

7 X 5 = 35 MOLES O2

=35 MOLES O2

O2 HAS A MOLAR MASS OF 2X16 = 32 gm

35 MOLES OF O2 HAS A MASS OF 35 X 32 =1120 gm

6 0

2 years ago