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3 years ago

For your 13th answer in #4, what is the other name for this particle?

Chemistry

2 answers:

solong [7]3 years ago

8 0

Answer:

The First layer is the troposphere. The troposphere is what we live in. It creates weather for us. Second layer is the Stratosphere. You will find the ozone layer in it. The ozone layer helps protect us from ultraviolet radiation (UV) from the sun. In fact, the ozone layer absorbs most of the UV radiation the sun sends to us. Life as we know it wouldn't be possible without this layer of protection. Third layer is Mesosphere. The Mesosphere is very important for earth's protection. The mesosphere burns up most meteors and asteroids before they are able to reach the earth's surface. Fourth layer is Thermosphere. The Thermosphere absorbs the sun radiation making it very hot.

Explanation: I don't know if you include this one but the fifth layer is the Exosphere. Although the exosphere is the most distance layer of earth's atmosphere it is the layer that is the planet's first line of defense against the sun's rays. It is also the first layer to come into contact and protect the earth from meteors, asteroids, and cosmic rays.

Musya8 [376]3 years ago

5 0

Answer:

Explanation:

The really interesting characteristic of the atmosphere is that each layer has a different temperature range. That is truly incredible when you list them.

The first layer is the troposphere. It is the layer closest to the earth's surface. It is 12 km high and goes from 15oC to - 56oC

The next layer is the Stratosphere. It is 38 km above the Troposphere and the temperature goes from -56 to -2.5

The third layer is Mesosphere. It is 40 km above the Stratosphere and the temperature goes from -2.5 to -86.5

The next layer is the Thermosphere. It is 710 km above the Mesosphere. It goes from - 86.5 to 1200o C So this goes from - 100 roughly to + 1200. That is a massive distance. It speaks wonders of how balanced this planet really is.

The final layer is the Exosphere. It goes up to 3000 km. The temperature is consistently at 1200oC

What I don't understand is how you get the layers from a 13 answer of something. What is the first name of the particle? That would seem to have nothing to do with this answer. Please explain because this was a good deal of work that might be for nothing. Mind you the answer is very interesting.

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When a variable is determined by a factor outside of the function or model being evaluated, it is said to be A. statistically i

inna [77]

Answer:

The correct answer is B. exogenous

Explanation:

Let us try to describe exogenous and endogenous variables an exogenous variable value is influenced only by factors outside a model or system and is forced onto the model, while a change in an exogenous variable is known as an exogenous change. Also an endogenous variable is one whose value is influenced only by the system or model under study.

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4 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

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3 years ago

All organic compound soluble in WATEe and conduct electrically

Delvig [45]

Explanation:

Not all organic substances are soluble in water and will not conduct electricity. Only polar organic compounds are soluble in water and will conduct electricity.

The general rule of solubility is that like dissolves likes.
Polar compounds will only dissolve other polar compounds too. Water is a polar solvent.
Non-polar compounds will only dissolve in non-polar compounds too.
Since water is a polar organic compound, only polar organic compounds will dissolve in it.
Not all organic compounds are polar.
When a compound dissolves in water, it produces ions that makes it able to conduct electricity.
Fats and oil are insoluble in water

learn more:

Inorganic compounds brainly.com/question/5047702

#learnwithBrainly

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3 years ago

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The patient will have less hemoglobin in the blood

The patient will have a reduced oxygen supply to the cells

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