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3 years ago

The energy exchanges between space, the atmosphere, and Earth’s surface produce what

Chemistry

2 answers:

nekit [7.7K]3 years ago

8 0

Answer: It produces climate or weather.

Explanation: The energy exchanges between space, the atmosphere, and the Earth's surface produces climate or weather which is responsible for the various life forms that exists on the earth. can be defined as the day to day changes in the temperature and precipitation. Whereas climate refers to the the average changes in atmospheric conditions for a longer period of time.

Sauron [17]3 years ago

7 0

The energy exchanges between space,the atmosphere and the Earth surface produce Weather and Climate

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Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

<u>Answer:</u> The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u> $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

<u>Reduction half reaction:</u> $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

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

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account

hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

3 0

3 years ago

How many molecules are there in 24 grams of FeF(3)?

Burka [1]

The SAME number of molecules are in ANY “mole” of a compound or element. So, you only need to ... 24 g116 g/mol=0.207 moles of FeF3.

5 0

2 years ago

How many moles of carbon are in 25 grams of carbon

Brut [27]

Answer:

One mole of carbon would look like 25/12.01

Explanation:

Firstly, you will divide 25 by 12.01 and get 2.081598

We know 1 mole equals the gram per atomic mass, so one mole of carbon is 12.01 grams. In conclusion, it would look like 25/12.01.

4 0

3 years ago

A liquid has an empirical formula CCl2, and a boiling point of 1 21 oC. When vapourised, the gaseous compound has a density of 4

Darya [45]

Based on the data given, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

<h3>How can molar mass of a gas be obtained from density, temperature and pressure?</h3>

The molar mass of a gas can be obtained from density, temperature and pressure using the formula below:

molar mass = density × molar gas constant × temperature/pressure

Molar gas constant, R = R = 0.082 L.atm/mol/K.

Temperature = 150 °C = 423 K

Pressure = 785 torr = 1.033 atm

density = 4.93 g/L

molar mass of gas = 4.93 × 0.082 × 423/1.033

molar mass of gas = 165.5 g/mol

Then, molecular weight of the gas = 165.5 amu

Therefore, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu

Learn more about molar mass of a gas at: brainly.com/question/26215522

6 0

1 year ago