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Ber [7]
4 years ago
14

Find the potential difference required to accelerate protons from rest to 10% of the speed of light. (at this point, relativisti

c effects start to become significant.)
Chemistry
1 answer:
elixir [45]4 years ago
5 0
I don't know much about the potential difference but I certainly can calculate the energy of the proton <span>from rest to 10% of the speed of light.

Special relativity tells us that "</span><span>the energy needed to accelerate a particle (with mass) grow super-quadratically when the speed is close to c, and is ∞ when it is c."

Expressed the theory in equation is as follows:
</span>
ENERGY  = rest mass × speed of light squared / (1−(“percent of speed of light”)squared ) <span> = mc2/(1-("percent of speed of light")2)
</span>
Just plug-in the values and you can now have the answer what you are looking for!
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Which of the followings are true about D-glucose and L-glucose? A. They are furanose B. They are stereoisomers C. They are enant
zaharov [31]

Answer:

B. They are stereoisomers

C. They are enantiomers

Explanation:

Let us consider all the options

A. D and L-glucose are not necessarily furanose, they can also be in free form (open chain) or as a six-membered ring (pyranose)

B. These sugars are stereoisomers as they have the same molecular formula, same bonds but with the different spacial arrangement.

C. Two structures are called enantiomers, if they are stereoisomers and are mirror images of each other and are not-superimposable. The given pair of structures satisfy these conditions

D. Epimers are diastereoisomers (same molecular formula and connectivity having a different spacial arrangement but are not mirror images and non-superimposable) with only one different stereocenter (if there are more than one). This is not the case

E. All monosaccharides (any sugar that cannot be hydrolysed to a simpler sugar) are reducing sugars. So, this option is invalid

4 0
3 years ago
The carbon cycle involves an exchange of carbon between ___
Solnce55 [7]
<h2>Answer:</h2>

D) The geosphere, hydrosphere, and atmosphere

<h2>Explanation:</h2>

During the carbon cycle, various processes move carbon from the geosphere to the atmosphere such as volcanic eruptions release carbon dioxide from molten rock beneath the earth's surface to the atmosphere. Carbon can leave the soil through soil respiration which releases CO2, or by erosion  which can carry it into rivers or the ocean, where it then enters the hydrosphere. Carbon is found in the hydrosphere dissolved in ocean water and lakes and hence the cycle continues.

3 0
3 years ago
Read 2 more answers
Complete combustion of 2.0 metric tons of coal to gaseous carbon dioxide releases 6.6X10¹⁰ J of heat. Convert this energy to
Taya2010 [7]

First of all we should know that, 1 Joule = 0.000239 kilocalories.

So, 6.6×10^{10} J = 6.6×10^{10} × 0.000239 kilocalories

      6.6×10^{10} J = 15774000 kilocalories

                       = 1.58 × 10^{7} kilocalories

One joule is described as the quantity of electricity exerted when a pressure of 1 newton is implemented over a displacement of 1 meter. Within the SI machine, the unit of labor or electricity is the Joule.

The kilocalorie, or meals calorie, is the quantity of warmth required to elevate one kilogram of water 1 °C. warmness capability is the amount of heat required to raise one gram of material 1 °C beneath steady pressure. A kilocalorie is the amount of warmth required to raise the temperature of 1 kilogram of water one diploma Celsius.

Learn more about joule here:-

brainly.com/question/490326

#SPJ4

6 0
2 years ago
What happens when Barium (Ba), in group 2, reacts with oxygen (O) which is in group 6?
Digiron [165]

Answer: barium oxide

Explanation: The barium will give some of it's electrons up to the oxygen, and then they will both reach the stability of a noble gas. Then, they will both combinate to barium oxide, so the answer is barium oxide.

7 0
3 years ago
There are eight constitutional isomers with molecular formula C5H11NO2 that contain a nitro group (-NO2). Seven of these isomers
padilas [110]

Answer:

Explanation:

Seven of these isomers having similar pKa values are as follows

1)  CH₃ -CH(CH₃)-CH(NO₂)-CH₃

2 ) CH₃ -(CH₂)-CH(NO₂)-CH₂-CH₃

3 )  CH₃ -CH₂-CH₂-CH₂-CH₂-NO₂

4 ) CH₃ -CH₂-CH₂-CH(NO₂)-CH₃

5 )CH₃ -CH(CH₃)(NO₂)-CH₂-CH₃

6 ) CH₃ -CH(CH₃)-CH₂-CH₂-NO₂

7)CH₃ -CH₂-CH₂-CH(CH₃)-NO₂

CH₃-C(CH₃)₂-NO₂

The last isomer has different PKa value because of tertiary carbon attached to nitro group.

7 0
4 years ago
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