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Nataliya [291]
2 years ago
15

How quickly would groundwater flow through rock high porosity and high permeability?

Chemistry
1 answer:
lozanna [386]2 years ago
8 0
Do you mean an actual rate of fast or slow? It'd flow quickly.
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Determine the electron geometry (eg) and molecular geometry (mg) of CO2.
eimsori [14]

Answer:

eg=linear, mg=linear

Explanation:

First of all, it must be stated that most triatomic molecules are either linear or bent. This depends on the electron geometry of the molecule and the number of bonding groups, multiple bonds and lone pairs present.

CO2 contains four regions of electron density and two bonding groups. For a specie containing two bonding groups, a linear molecular geometry is expected with an angle of 180°.

For a specie having two bonding groups and no lone pairs with multiple bonds, the expected electron geometry is also linear.

6 0
3 years ago
An acidic fog in pasadena was found to have a ph of 2.50. which expression represents this ph measurement?
joja [24]

Supposing a temperature of 25 degrees and supposing that all activity coefficients are 1 

pH = -log[H+] 
pOH = -log[OH-] 
pH + pOH = 14 

Thus a pH of 2.50 would mean that the [H+], the concentration of the hydrogen ion, would be 10^(-2.50) 

pH + pOH = 14 
pOH = 14 - pH = 14 - 2.5 = 11.5 
MOH- levels would be coordinated with pOH 
pOH = -log[OH-] ==> [OH-] = [MOH-] = 10^-pOH = 10^-11.5 = 3.2 x 10^-12 

Therefore, MOH¯ = 3.2 × 10¯12 M 

5 0
3 years ago
The purpose of an incline plane is to reduce which of the following?
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The purpose of an inclined plane is to reduce force.
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2 years ago
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Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --
frosja888 [35]

Answer :

(a) The anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

Thus, the anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}

The overall cell reaction will be,

2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}

E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V

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

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}

E^o=1.54V-(0.34V)=1.20V

Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}

where,

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

E_{cell} = emf of the cell = ?

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

E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}

E_{cell}=1.022V

Therefore, the emf of cell potential is 1.022 V

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The best method to improve air quality is to
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Stop littering and polluting
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