2 1. Biology is the study of

A student has a mixture of solid sand and nacl(aq)in a flask to obtain solid nacl from this mixture the student should

maria [59]

In order to obtain solid NaCl, the student should do a few steps.

First, he/she should do filtration. Pass the mixture through a filter paper, where all the sand should be filtered out already because they're not dissolved in the solution plus they're too small to pass through the filter paper.

Next, the filtrate should be left with NaCl (aqueous state). To seperate NaCl with the liquid, the student can either do evaporation or crystallization, depending on how pure or fast he/she wants the results to be. Evaporation involves heating the beaker or whatever apparatus under the bunsen burner until all the liquid has evaporated. Then, some white powder should be left, they're NaCl solid. For crystallization, the student should just put the beaker on a room condition environment, and wait. They might have to wait a month or so for the liquid to completely evaporate itself and left with clear and pure NaCl crystals.

5 0

3 years ago

Which one of the following represents the number of units of each substance

netineya [11]

Answer:B

Explanation:Coefficient represents the number of units of each substance.

4 0

3 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

(a) As per question, lead is oxidized and copper is reduced.

The oxidation-reduction half cell reaction will be,

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

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

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

(b) As per question, lead is reduced and copper is oxidized.

The oxidation-reduction half cell reaction will be,

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

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

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

In substitution reactions, (CH3)3C-I reacts at the same rate with Br- and Cl- even though Br- is a more reactive nucleophile tha

Kamila [148]

Answer:

A. (CH3)3C-I reacts by SN1 mechanism whose rate is independent of nucleophile reactivity.

Explanation:

We must recall that (CH3)3C-I is a tertiary alkyl halide. Tertiary alkyl halides preferentially undergo substitution reaction via SN1 mechanism.

In SN1 mechanism, the rate of reaction depends solely on the concentration of the alkyl halide (unimolecular mechanism) and is independent of the concentration of the nucleophile. As a result of this, both Br^- and Cl^- react at the same rate.

5 0

3 years ago

Is this equation balanced or unbalanced? C+02=CO2

LenKa [72]

This equation is balanced

4 0

3 years ago