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1 year ago

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atmospheric stability and wind conditions control group of answer choices the vertical and horizontal dispersion of pollutants t

he quantity of pollution emitted into the atmosphere from industrial and mobile sources. the rate of photochemical reactions taking place. the increased use of fuel for heating or cooling

1 answer:

Semenov [28]1 year ago

3 0

The correct answer is

atmospheric stability and wind conditions control the vertical and horizontal dispersion of pollutants.

What is meant by atmospheric stability?

Atmospheric stability is a measure of atmospheric status which determines whether or not air will rise, sink, or be neutral. In general, stability refers to air tendency to rise or to resist vertical motion.

Three Types of Stability

1. An unstable atmosphere will enhance or encourage the vertical movement of air.

2. A stable atmosphere will suppress or resist vertical motion.

3. A neutral atmosphere will neither suppress nor enhance vertical motion.

Learn more about atmospheric stability here :

brainly.com/question/4553502?referrer=searchResults

#SPJ4

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When does the dipole arrow point to Atom A? (Use electronegativity in your explanation)

Marrrta [24]

Answer:

Dipole, less electronegativity, higher electronegativity

8 0

3 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

<u>Answer:</u> The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

<u>Explanation:</u>

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

<u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

7 0

3 years ago

Read that From picture and tell me the answer please

Dafna11 [192]

Answer:

1g Hydrogen

Explanation:

<h3><u>Getting</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>equation</u><u>:</u></h3>

Calcium in water reacts vigorously to give a cloudy white <em>Precipitate</em><em> </em>(compound) called Calcium hydroxide alongwith the evolution of Hydrogen gas.

$\boxed{ \mathsf{Ca + H_2O \rightarrow Ca(OH)_2 + H_2}}$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3><u>Balancing</u><u> </u><u>the</u><u> </u><u>equation</u><u>:</u><u> </u></h3>

This reaction is not in it's balanced form! The number of atoms of Hydrogen on the left is 2 while that on the right is 4,I.e.,they're not equal.

Adding a 2 in front of H2O solves the problem by making the number of atoms of each element on both the sides equal.

$\mathsf{Ca +2 H_2O \rightarrow Ca(OH)_2 + H_2}$

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3><u>Observations</u><u>:</u></h3>

Looking into the equation more carefully, we see:

<u><em>1</em><em> </em></u><em>atom</em><em> </em><em>of</em><em> </em><em>Calcium</em><em> </em><em>reacts</em><em> </em><em>with</em><em> </em><em><u>2</u></em><em> </em><em>molecules</em><em> </em><em>of</em><em> </em><em>water</em><em> </em><em>to</em><em> </em><em>give</em><em> </em><u><em>1</em><em> </em></u><em>molecule</em><em> </em><em>of</em><em> </em><em>Calcium</em><em> </em><em>Hydroxide</em><em> </em><em>alongwith</em><em> </em><em><u>1</u></em><em> </em><em>molecule</em><em> </em><em>of</em><em> </em><em>Hydrogen</em><em> </em><em>gas</em><em>.</em>

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3><u>Gram</u><u> </u><u>atomic</u><u> </u><u>and</u><u> </u><u>molecular</u><u> </u><u>masses</u><u> </u></h3>

<u>Mass</u><u> </u><u>of</u><u> </u><u>one</u><u> </u><u>atom</u><u> </u><u>of</u><u> </u><u>Calcium</u><u> </u>= it's gram atomic mass

= 40 g

<u>Mass of one "molecule" of Hydrogen</u>

= it's Gram molecular mass

= gram mass of one atom × number of atoms in one molecule

= 1 × 2

= 2 g

So,

according to our observation:

One atoms of Calcium gives one molecule of Hydrogen <em>(during the particular reaction)</em>

=><u> 40g of Calcium gives = 2g of Hydrogen</u>

•°• 1 g of Calcium gives = $\frac{2}{40}$

= $\frac{1}{20}$ g Hydrogen

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

<h3><u>Answer</u><u>:</u></h3>

We're provided with 20g of Calcium,

=> 20g of Calcium gives = 20 × $\frac{1}{20}$ g H2

<u>= 1 g H2</u>

_______________

Hope this helps!

6 0

3 years ago

A green laser pointer emits light with a wavelength of 536 nm. What is the frequency of this light?

anastassius [24]

Answer:

Your answer is 5.63 × 10^14 Hz

We know the value of the speed of light and we are given the wavelength. All we have to do is rearrange the equation to solve for the frequency:

frequency: v = c/ λ

Now just plug in the given values:

v = 3.00 × 10 ^ 8 / 530 × 10^-9

v = 5.66 × 10 ^ 14 Hz

I hope this helps you.

3 0

3 years ago

How many atoms does 5.06 grams of oxogen have

blagie [28]

Answer: 1.90x10²³ atoms of O

Explanation: First convert the mass of O into moles:

5.06 g O x 1 mole O / 16 g O

= 0.32 moles O

Next use Avogadro's Number to find the number of atoms of O

0.32 moles O x 6.022x10²³ atoms O / 1 mole O

= 1.90x10²³ atoms O

7 0

4 years ago