All categories

3 years ago

What are items that are cycled through the biosphere in biogeochemical cycles

1 answer:

4 0

Some examples are the water cycle, the carbon cycle, the phosphorus cycle and the nitrogen cycle.

Hope this helps xox :)

You might be interested in

Any suggestions or answers

leva [86]

Explanation: energy levels are three(counting number of cyecles) outer electrons are 3 also

Answer:3,3

6 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

2 years ago

The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6

s344n2d4d5 [400]

Answer:

the conversion factor is f= 6 mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6 mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

3 0

3 years ago

A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

likoan [24]

Given :

A 3.82L balloon filled with gas is warmed from 204.9K to 304.8 K.

To Find :

The volume of the gas after it is heated.

Solution :

Since, their is no information about pressure in the question statement let us assume that pressure is constant.

Now, we know by ideal gas equation at constant pressure :

$\dfrac{V_1}{V_2} = \dfrac{T_1}{T_2}\\\\\dfrac{3.82}{V_2}= \dfrac{204.9}{304.8}\\\\V_2 = \dfrac{304.8}{204.9} \times 3.82\\\\V_2 = 5.68 \ L$

Hence, this is the required solution.

3 0

2 years ago

Which of the following ions would represent the ion of an element from Group 2 on the Periodic Table? A2+ Y3+ X2- Z+

Luba_88 [7]

A2+. Group 2 elements form cations with 2+ charge.

7 0

2 years ago