The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
Sodium borohydride is a relatively selective reducing agent Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.
The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone
The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .
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Answer:
The answer to your question is 1.36 x 10²³ atoms
Explanation:
Data
number of atoms = ?
mass of the sample = 34.2 g
Molecule = Cl₂O₅
Process
1.- Calculate the molar mass of Cl₂O₅
Cl₂O₅ = (35.5 x 2) + (16 x 5) = 71 + 80 = 151 g
2.- Calculate the atoms of Cl₂O₅
151 g of Cl₂O₅ ---------------- 6 .023 x 10²³ atoms
34.2 g of Cl₂O₅ ------------ x
x = (34.2 x 6.023 x 10²³) / 151
x = 1.36 x 10²³ atoms
Answer:
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Explanation:
Answer:
You should follow these steps:
Count each type of atom in reactants and products.
Place coefficients, as needed, in front of the symbols or formulas to increase the number of atoms or molecules of the substances.
Repeat steps 1 and 2 until the equation is balanced.
Explanation:
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