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prisoha [69]
3 years ago
9

The entropy change for a real, irreversible process is equal to:______.

Chemistry
1 answer:
taurus [48]3 years ago
8 0

Answer:

The entropy change for a real, irreversible process is equal to <u>zero.</u>

The correct option is<u> 'c'.</u>

Explanation:

<u>Lets look around all the given options -:</u>

(a)  the entropy change for a theoretical reversible process with the same initial and final states , since the entropy change is equal and opposite in reversible process , thus this option in not correct.

(b) equal to the entropy change for the same process performed reversibly ONLY if the process can be reversed at all. Since , the change is same as well as opposite too . Therefore , this statement is also not true .

(c) zero. This option is true because We generate more entropy in an irreversible process. Because no heat moves into or out of the surroundings during the procedure, the entropy change of the surroundings is zero.

(d) impossible to tell. This option is invalid , thus incorrect .

<u>Hence , the correct option is 'c' that is zero.</u>

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Your calculated density of aluminum is d = 2.69 g/cm3. Aluminum’s accepted density is 2.70 g/cm3. Without writing the "%" sign,
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Answer:

0.37 %

Explanation:

Given that:

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Accepted density of aluminum = 2.70 g/cm³

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Thus, applying values as:

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dmitriy555 [2]
It would be in the transition metals
5 0
3 years ago
How many moles of oxygen are needed to react with 87 grams of aluminum
labwork [276]

Answer:

2.4 moles of oxygen are needed to react with 87 g of aluminium.

Explanation:

Chemical equation:

4Al(s)  + 3O₂(l)   → 2AlO₃(s)

Given data:

Mass of aluminium = 87 g

Moles of oxygen needed = ?

Solution:

Moles of aluminium:

Number of moles of aluminium= Mass/ molar mass

Number of moles of aluminium= 87 g/ 27 g/mol

Number of moles of aluminium= 3.2 mol

Now we will compare the moles of aluminium with oxygen.

                              Al         :         O₂

                               4          :         3

                               3.2       :         3/4×3.2 = 2.4 mol

2.4 moles of oxygen are needed to react with 87 g of aluminium.

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2 years ago
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