It's found inside of a cell attached to the cytoplasm and attached to the endoplasmic reticulum. hope this helps
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: 10.9 mol.
Explanation:
- To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.
Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
- Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.
Using cross multiplication:
1.0 mole of PbO₂ → 2.0 moles of H₂O
5.43 moles of PbO₂ → ??? moles of water
The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.
Answer:
Hydrogen and Cobalt
Explanation:
Break up each individual element
- Hope that helps! Please let me know if you need further explanation.
