20.06 g of Hg and 1.6 g of O₂
<u>Explanation:</u>
To Find:
Number of Mercury and oxygen that can be obtained from 21.7 g of HgO
First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,
2 HgO (s) → 2Hg(l) + O₂ (g)
21.7 g of HgO = 
= 0.1 mol of HgO.
As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .
So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)
= 20.06 g of Hg
Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂
Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO
Li2O + H2O → 2LiOH
This is the answer
3rd blank write 2
Answer:
See the image 1
Explanation:
If you look carefully at the progress of the SN2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the back side relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way. (see image 2)
The result of this backside attack is that the stereochemical configuration at the central carbon inverts as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane. (see image 3)
What this means is that SN2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product.
Answer:
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