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Sveta_85 [38]
3 years ago
7

Elaborate on the difference in natural occurrences between fission and fusion reactions. A) Neither fission nor fusion reactions

naturally occur on earth or outer space. B) Fission reactions do not naturally occur while fusion reactions occur in the stars. C) Both fission and fusion reactions naturally occur in the pressure of the earth's core. D) Fusion reactions do not naturally occur while fission reactions occur in the volcanoes.
Chemistry
2 answers:
vazorg [7]3 years ago
8 0
<span>Okay then I would go with choice B since fusion takes place in the sun which is a giant star.</span>
11111nata11111 [884]3 years ago
7 0

Answer: Option 'B' is correct.

Explanation:

Nuclear fission means the splitting of larger atoms into two or more smaller atoms whereas Nuclear fusion means the splitting of two or more atoms into larger atoms.

Nuclear fission does not occur naturally but nuclear fusion occurs in stars etc.

So, only statement is correct regarding the difference in natural occurrences between fission and fusion.

Hence, Option 'B' is correct.

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At a certain temperature the vapor pressure of pure methanol is measured to be . Suppose a solution is prepared by mixing of met
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Answer:

Partial pressure is 0.13 atm

Explanation:

CHECK THE COMPLETE QUESTION BELOW :

At a certain temperature, the vapor pressure of pure methanol is measured to be 0.43atm. Suppose a solution is prepared by mixing 88.2 g of methanol and 116.g of water. Calculate the partial pressure of methanol vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal.

Using Raoult´s law for ideal soultions we have

P(A) = X(A) *Pº(A)

where P(A) is the partial vapor pressure pressure of methanol,

X(A) is the mole fraction of solute (methanol) in solution,

Pº(A) is the vapor pressure of pure solute

Raoult's law states that the vapor pressure of a solution is dependent on the mole fraction of a solute added to the solution.

Raoult's law can be expressed below

Psolution = ΧsolventP0solvent.

Expressing it interns of the constituents given in the question we have

P(CH₃OH) = X(CH₃OH) x Pº(CH₃OH)

To calculate the mole fraction of CH₃OH, we make use of the formula below :

X(A) = mol (A) / ntotal

Ntotal = (sum of number of moles of A )+( moles solvent)

mol (CH₃3OH) can be calculated as :: 88.2 g/ 32 g/mol = 2.76 mol of CH₃OH

mol (H₂O) can be calculated as ::116 g/ 18 g/mol = 6.44 mol

total n = (6.44 + 2.76) mol = 9.20 mol

To calculate the partial pressure the we say;

P(CH₃OH) = (2.76 mol CH + 9.20 mol) x( 0.43 atm) = 0.13 atm

Hence, the partial pressure rounded to two significant figures is 0.13 atm

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