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zloy xaker [14]
3 years ago
8

Difference between desirable changes and undesirable changes​

Chemistry
1 answer:
kkurt [141]3 years ago
3 0

Answer:

...n..m..mm .......mmnjjjjjjj ygttgyujjunjhh

Explanation:

ghyhhhhjn

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What is the radius of a hydrogen atom whose electron is bound by 0.544 ev? express your answer with the appropriate units?
insens350 [35]
First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5

The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm
6 0
3 years ago
Each A atom is adjacent to 3 B atoms. What is the A-C-B bond angle?
Nutka1998 [239]
Using the simulation to build a system with 5 bonds. The resulting structure is called the Trigonal bipyramidal structure. The two different sites in a trigonal bipyramid are labeled as A and B in the drawing to the right. 
In this case, therefore, the bond angle A-C-B is 90 Degrees (right angle)
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3 years ago
Answer choices are A. Constant speed<br> B. Stopped<br> C.Moving sideways<br> D.Accelerating
Maurinko [17]
B.stopped
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7 0
2 years ago
I need help answering this question
ohaa [14]
The correct answer is c because enzymes increase the rate of all biochemical processes
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Read 2 more answers
At a certain temperature, the vapor pressure of pure benzene ‍ is atm. A solution was prepared by dissolving g of a nondissociat
erastova [34]

Answer:

Molar mass of solute: 300g/mol

Explanation:

<em>Vapor pressure of pure benzene: 0.930 atm</em>

<em>Assuming you dissolve 10.0 g of the non-volatile solute in 78.11g of benzene and vapour pressure of solution was found to be 0.900atm</em>

<em />

It is possible to answer this question based on Raoult's law that states vapor pressure of an ideal solution is equal to mole fraction of the solvent multiplied to pressure of pure solvent:

P_{sln} = X_{solvent}P_{solvent}^0

Moles in 78.11g of benzene are:

78.11g benzene × (1mol / 78.11g) = <em>1 mol benzene</em>

Now, mole fraction replacing in Raoult's law is:

0.900atm / 0.930atm = <em>0.9677 = moles solvent / total moles</em>.

As mole of solvent is 1:

0.9677× total moles = 1 mole benzene.

Total moles:

1.033 total moles. Moles of solute are:

1.033 moles - 1.000 moles = <em>0.0333 moles</em>.

As molar mass is the mass of a substance in 1 mole. Molar mass of the solute is:

10.0g / 0.033moles = <em>300g/mol</em>

8 0
3 years ago
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