Answer:
The reactive nucleophile is Ketone.
Explanation:
In organic chemistry, The process of acid - catalyzed aldol condensation starts from when ketone (or any aldehyde) is converted to an -enol, after which it attacks another ketone/aldehyde that has already been activated by parbonyl oxygen protonation.
The process of this is that first of all the ketone undergoes tautomerization to form -enol. Thereafter, the other carbonyl will undergo protonation which makes the carbon activated towards attack. Now, the nucleophilic enol will be added to the carbonyl in a [1,2]-addition reaction and we will now use deprotonation to obtain the neutral Aldol product.
Now, since only the ketone can produce an -enol, thus it is the nucleophile as aldehydes are better electrophiles
The equation for the nuclear fusion reaction is,
4 ¹₁H → ₂⁴He + 2 ₁⁰e
Calculation of mass defect,
Δm = [mass of products - mass of reactants]
= 4(1.00782) - [4.00260 + 2(0.00054858)]
= 0.0275828 g/mole
Given that,
Mass of Hydrogen-1 = 2.58 g
The no. of moles of ₁¹H = 2.58 g / 1.00782 = 2.56 moles
Therefore, the mass defect for 2.58 g of ₁¹H is,
= 2.56 moles * (0.0275828 g / 4) = 0.01765 x 10⁻³ kg
Energy for (0.01765 x 10⁻³ kg) is,
= (0.01765 x 10⁻³ kg) (3.0 x 10⁸)² = 1.59 x 10¹² J
Answer:
The suitable equation for this reaction is
2CO + O₂ -----> 2CO₂
Here, we are given that we have 2 grams of O₂
From the equation, we can see that 2 * Moles of O₂ = Moles of CO₂
Moles of O₂:
2/32 = 1/16 moles
Therefore, the number of moles of CO₂ is twice the moles of O₂
Moles of CO₂ = 2 * 1/16
Moles of CO₂ formed = 1/8 moles
Mass of CO₂ formed = Molar mass of CO₂ * Moles of CO₂
Mass of CO₂ formed = 44 * 1/8
Mass of CO₂ formed = 5.5 grams
Hence, option B is correct
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What’s the question Bc I don’t understand what u are trying to ask
Answer:
6,000kg/m3
Explanation:
6.00g/1cm3 x 1kg/1000g x 1cm3/0.000001m3
= 6.00kg/0.001m3
= 6,000kg/m3
