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3 years ago

What are fossil fuels made from ?

2 answers:

6 0

fossil fuel is a general term for buried combustible geologic deposits of organic materials, formed from decayed plants and animals that have been converted to crude oil, coal, natural gas, or heavy oils by exposure to heat and pressure in the earth's crust over hundreds of millions of years.

Andreas93 [3]3 years ago

5 0

Fossil fuels are buried geologic deposits of organic materials. They are made of decaying plants and animals that have been turned into to natural oil. A few examples of fossil fuels are coal and oil.

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sodium-25 was to be used in an experiment but it took 3.0 minutes to get the sodium from the reactor to the laboratory if 5.0 mg

Hoochie [10]

3 half lives later... 5/2 = 2.5mg, then 2.5/2 = 1.25 mg, then 1.25/2 = .625 mg left

7 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 :

<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

What are diatomic molecules and are they compounds? PLEASE ANSWER ASAP!!

Fantom [35]

Answer:

Diatomic molecules consist of two atoms that are chemically bonded. The two atoms can be the same or different chemical elements. As for whether or not they are compounds, there is not technically an answer. This is because all compounds are molecules, but not all molecules are compounds. For example diatomic molecules that comprise the chemical compounds nitric acid, carbon monoxide, and hydrogen chloride are made up of two different elements. As you can see, most diatomic molecules are not made up of the same kind of elements and not every diatomic molecule comprises a chemical compound.

hope this helps :)

Explanation:

4 0

3 years ago

890J of heat are applied to a piece of aluminum, causing a 4.6°C increase in its temperature. The specific heat of aluminum is 0

Semmy [17]

Answer:

The answer would be C 214g

Explanation:

890j of heat causes 4.6°c increase in temperature

specific heat of aluminium after is o.9022 j /g°c

now by using the formula .The mass of aluminium would be c that is 214 g

7 0

3 years ago

Which two minerals make up the complex hydroxyapatite?

matrenka [14]

Ca₁₀(PO₄)₆(OH)₂ - hydroxyapatite

b. phosphorus (P) and calcium (Ca)

6 0

3 years ago

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