Which 2 characteristics must and object have in order to be considered matter?

The molar mass of carbon dioxide (CO2) is 44.01 g/mol. The molar mass of water (H2O) is 18.01 g/mol. A reaction uses 528 g of CO

Black_prince [1.1K]

CO₂ + H₂O -------> H₂CO₃

moles of CO₂ = $\frac{MASS of Carbon}{ MOLAR MASS}$

= $\frac{528 g}{44.01 g / mol}$

= 11.997 mol

mole ratio of CO₂ : H₂O =   1 : 1

  ∴ moles of H₂O = (`11.997 mol ) * 1
= 11.997 mol
   ≈ 12 mol

6 0

3 years ago

The surface temperatures on land vary more over the course of a year than the surface temperatures on the ocean because land has

bogdanovich [222]

Answer:

Lower heat capacity

Explanation:

The heat or thermal capacity is a physical property defined as the amount of heat a material need in order to elevate a unit in its temperature, this means that water increases its temperature more easily than land.

I hope you find this information useful and interesting! Good luck!

7 0

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.

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

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

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