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3 years ago

13

Tonya had a mixture of sand, salt, iron particles, and pebbles. What processes should be used to separate the substances in the

mixture from each other?

1 answer:

marusya05 [52]3 years ago

3 0

Explanation:

Put the pan into a water-containing dish and mix well. Now use a strainer to transfer the solution into another jar. The salt should disappear in it.

And using a tube with a filtrate, transfer the salts that has sand into another bottle with a filtrate. Therefore the sand is split. Eventually, when all the water vaporizes and the salt stays in the bottle, leave the extra solvent and heat it.

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When non metals react with other elements and compounds what happens to their valence electrons?

DiKsa [7]

Answer:

d

Explanation:

they either gain valence electrons or share them depending on what elements or compound they are reacting with

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2 years ago

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kenny6666 [7]

Answer:

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Explanation:

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2 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 :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

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

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

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

Explain in your own words the process using the following key words: carbon-di-oxide, water, sugar, oxygen, thylakoid, chloropla

AleksAgata [21]

Answer:

The Photosynthesis process

Explanation:

Plants, algae, and some other organisms can transform the sunlight energy into chemical energy. The photosynthesis process occur thanks to the chloroplasts. The chloroplast is an organelle found in all green plants. Inside of the chloroplast you can find the thylakoids which are arranged in stacks named grana, they have membranes with chloropyll a photosynthetic pigment, also you can find the photosystems, they are functional and structural units of protein complexes. The thylakoids capture the light and allow the reactions to transform CO2. The set of reactions that occurs in the chloroplasts are known as the Calvin cycle.

The general equation of photosynthesis is:

$6 CO_{2} + 6 H_{2} O + Energy -> C_{6} H_{12} 12O_{6} + 6 O_{2}$

6 CO2 + 6 H2O + Energy -> C6H12O6 + 6 O2

Carbon Dioxide + water + Light -> Glucose (sugar) + Oxygen

After, this glucose is transformed into pyruvate, and it allowed the release of denosine triphosphate (ATP) by cellular respiration. The ATP is an organic chemical that is requires for the cell to perform any process (any kind or work).

5 0

3 years ago

The heat of vaporization of a liquid is 84.0 J/g. How many joules of heat would it take to completely vaporize 172 g of this liq

Aliun [14]

Answer:

14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.

Explanation:

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to Q = m*L, where L is called the latent heat of the substance and depends on the type of phase change.

During the evaporation process, a substance goes from a liquid to a gaseous state and needs to absorb a certain amount of heat from its immediate surroundings, which results in its cooling. The heat absorbed is called the heat of vaporization.

So, it is called "heat of vaporization", the energy required to change 1 gram of substance from a liquid state to a gaseous state at the boiling point.

In this case, being:

Q=?
m= 172 g

L= 84 $\frac{J}{g}$

and replacing in the expression Q = m*L you get:

Q=172 g*84 $\frac{J}{g}$

Q=14,448 J

<u><em>14,448 J of heat would it take to completely vaporize 172 g of this liquid at its boiling point.</em></u>

5 0

3 years ago