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2 years ago

Which of the following is a cause of eutrophication of a lake?

Chemistry

2 answers:

cluponka [151]2 years ago

4 0

Answer:

storm runoff

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Anastaziya [24]2 years ago

4 0

Eutrophication is caused by an excess amount of nutrients within the water. These nutrients can be collected within a lake by runoff from a agricultural area filled with fertilizers, etc etc. The algae within the body of water would grow at a rapid rate, causing a depletion for oxygen within the water and the lack of sunlight underneath the water surface.

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The pictures are the answers

Mice21 [21]

Answer:

Explanation:

If the students want to know at what percent of CO2 in the air the plant will grow at the fastest, then the percent of CO2 should be a different value for each plant in the table.

There are 2 tables that have different values for the CO2 - the tables in answer choices C and D.

Since the students only want to know how the amount of CO2 affects the plant, every other variable should remain constant.

The only answer choice that has a changing value for the percent of CO2 and a constant value for every other variable is C.

4 0

3 years ago

A voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn

nlexa [21]

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_{[Ni^{2+}/Ni]}=-0.23V$

$E^0_{[Zn^{2+}/Zn]}=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^{2+}+2e^-$ $E^0_{[Zn^{2+}/Zn]}=-0.76V$

Reaction at cathode (reduction) : $Ni^{2+}+2e^-\rightarrow Ni$ $E^0_{[Ni^{2+}/Ni]}=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^{2+}(aq)\rightarrow Zn^{2+}(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Ni^{2+}/Ni]}-E^o_{[Zn^{2+}/Zn]}$

$E^o=(-0.23V)-(-0.76V)=0.53V$

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

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

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:

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(0.100)}{(1.50)}$

$E_{cell}=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^{2+}$ has fallen to 0.500 M.

New concentration of $Ni^{2+}$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Ni^{2+}]}$

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

$E_{cell}=0.53-\frac{0.0592}{2}\log \frac{(1.1)}{(0.500)}$

$E_{cell}=0.52V$

(c) Now we have to calculate the concentrations of $Ni^{2+}$ and $Zn^{2+}$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}+x]}{[Ni^{2+}-x]}$

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

$0.45=0.53-\frac{0.0592}{2}\log \frac{(0.100+x)}{(1.50-x)}$

$x=1.49M$

The concentration of $Ni^{2+}$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^{2+}$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

5 0

3 years ago

When 25.0 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced?

densk [106]

Answer: 77.7g

Explanation:Please see attachment for explanation

4 0

3 years ago

While out on a hike, you find an unknown mineral. How would you test the mineral to find out what it is?

notsponge [240]

Are there answer choices or is it just right it your self?

6 0

3 years ago

It is dangerous to have over 1.3 ppm of copper (Cu) in drinking water. A small town investigates their water supply and finds th

lianna [129]

Answer:

The town should find a new source of water.

Explanation:

Step 1: Convert the mass of copper to milligrams

We will use the conversion factor 1 g = 1000 mg.

0.002 g × (1000 mg/1 g) = 2 mg

Step 2: Convert the mass of solution to kilograms

We will use the conversion factor 1 kg = 1000 g.

1000 g × (1 kg/1000 g) = 1 kg

Step 3: Calculate the concentration of Cu in ppm (mg/kg)

2 mg/1 kg = 2 ppm

This is over 1.3 ppm, so the town should find a new source of water.

3 0

2 years ago