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kirza4 [7]
3 years ago
12

Which is the product of hydrolysis of an animal fat by a strong base?

Chemistry
2 answers:
Ira Lisetskai [31]3 years ago
6 0
Soap this process is called saponification
VladimirAG [237]3 years ago
5 0
///The answer is soap///
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Part of molten rock at mid-ocean ridges
erastovalidia [21]

Answer:

Magma

Explanation:

Magma is part of molten rock at mid-ocean ridges.

Hope this helps!

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NEED HELP ASAP!
xz_007 [3.2K]
0.074Hz is the answer to that
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PLEASE HELP ME I NEED TO SUBMIT RIGHT NOW
sasho [114]

Answer:  number of protons?

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Please help it’s a test due in five minutes! I’ll give 10 points
barxatty [35]

Answer: negative acellaration or mass.

Explanation:

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3 0
3 years ago
Be sure to answer all parts. for each reaction, find the value of δso. report the value with the appropriate sign. (a) 3 no2(g)
Maurinko [17]

Answer:

Change in entropy for the reaction is

ΔS° = -268.13 J/K.mol

Explanation:

To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.

3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)

From Literature.

S°(NO₂) = 240.06 J/K.mol

S°(H₂O) = 69.91 J/K.mol

S°(HNO₃) = 155.60 J/K.mol

S°(NO) = 210.76 J/K.mol

These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.

Note that

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]

= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol

Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]

= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol

ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)

ΔS° = 521.96 - 790.09 = -268.13 J/K.mol

Hope this Helps!!

4 0
3 years ago
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