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3 years ago

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Use the graph to answer the following questions. Which state has the highest level of air pollution particles for the year 2014?

Which state shows the least change in air pollution particles from 2006 to 2015? What action is most likely to have helped reduce the rate of air pollution particles?

2 answers:

Mice21 [21]3 years ago

7 0

Answer:

Maryland, Wyoming, reduced burning of fossil fuels

Explanation:

edgeunity

Vikki [24]3 years ago

6 0

Answer: Maryland, Wyoming, reduced burning of fossil fuels

Explanation: I took the test and got it right.

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Suppose that X represents an arbitrary cation and that Y represents an anionic species. Using the charges indicated in the super

dmitriy555 [2]

Answer:

See explanation.

Explanation:

Hello!

In this case, when having the cationic and anionic species with the specified charges, in order to abide by the net charge rule, we need to exchange the charges in the form of subscripts and without the sign, just as shown below: $X^{m+}Y^{n-}\rightarrow X_nY_m$

Thus, for all the given combinations, we obtain:

- Y⁻

$X^+Y^-\rightarrow XY\\\\X^{2+}Y^-\rightarrow XY_2\\\\X^{3+}Y^-\rightarrow XY_3$

- Y²⁻

$X^+Y^{2-}\rightarrow X_2Y\\\\X^{2+}Y^{2-}\rightarrow X_2Y_2\rightarrow XY\\\\X^{3+}Y^{2-}\rightarrow X_2Y_3$

- Y³⁻

$X^+Y^{3-}\rightarrow X_3Y\\\\X^{2+}Y^{3-}\rightarrow X_3Y_2 \\\\X^{3+}Y^{3-}\rightarrow X_3Y_3\rightarrow XY$

Best regards!

8 0

2 years ago

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

5 0

3 years ago

1. What is the MAIN reason the Phoenicians are important for marine science?

Vikki [24]

Answer:

D

Explanation:

The Phoenicians contributed to ocean exploration by establishing the first trade routes throughout the Mediterranean, even as far north as Great Britain.

3 0

2 years ago

State three evidences that the particles of substances are in continuous motion

Solnce55 [7]

The three evidences that show that the particles of substances are in continuous motion are given below:

1. DIFFUSION: Diffusion is the movement of particles of a substance through a medium, from a region of higher concentration to a region of lower concentration. Particles of substances (solid, liquid or gas) tend to move from an area where more of them are present to an area where only few are present until they are evenly distributed. For example, a can of gasoline left in an open air can easily catch fire because the particles of gasoline can be easily carried by air to the nearest source of ignition.

2. OSMOSIS: Osmosis is the process by which solvent molecules move from a less concentrated solution to a more concentrated solution through a semi permeable membrane; the movement typically continues until the concentrations of solutes on both sides of the membrane are equal. Osmosis is a very crucial process in the biological systems of living organisms, where the plasma membrane, which is semi permeable in nature only allow specific small particles to pass through. Water molecules pass through the plasma membrane by diffusion via the phospholipid bilayers; osmosis thus provide the major mean via which water is moved in and out of living cells

3. BROWNIAN MOVEMENT: When a pollen grain, which is suspended in a drop of water is observed under a microscope, it will be seen that the pollen is moving about in a zig-zag manner. The zig zag movement of the pollen is due to the continuous movement of the particles of water on which the pollen was suspended. This phenomenon was first observed by a botanist, a British scientist called Robert Brown in 1827 and that was why the phenomenon was named after him.

3 0

3 years ago

Read 2 more answers

Titration of a 25.0 mL sample of barium hydroxide requires 34.45 mL of 0.100 M nitric acid. What is the molarity of the barium h

Mariana [72]

The titration reaction is written as follows:

Ba(OH)2 + 2HNO3 = Ba(NO3)2 + 2H2O

The calculations are as done as follows:

0.100 mol HNO3 / L (.03445 L solution ) = 0.003445 mol HNO3

0.003445 mol HNO3 ( 1 mol Ba(OH)2 / 2 mol HNO3 ) = 0.001723 mol Ba(OH)2

0.001723 mol / 0.025 L = 0.07 M Ba(OH)2

8 0

3 years ago