How many molecules are there in 21.4 grams of Ca(OH)2?
1 answer:
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Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).
Answer : The value of of the weak acid is, 4.72
Explanation :
First we have to calculate the moles of KOH.
Now we have to calculate the value of of the weak acid.
The equilibrium chemical reaction is:
Initial moles 0.25 0.03 0
At eqm. (0.25-0.03) 0.03 0.03
= 0.22
Using Henderson Hesselbach equation :
Now put all the given values in this expression, we get:
Therefore, the value of of the weak acid is, 4.72
<u>Answer:</u>
<em>That oxidation number of chlorine is -1 while that of oxygen is -2</em>
<u>Explanation:</u>
When it comes to manganese chloride to stabilize the chlorine Ion. Two manganese ions shall be required. Hence the formula for the compound becomes while the formula for manganese (iv) chloride will be
similarly.
When oxygen and magnesium oxide is considered the Two ions of Manganese will satisfy to ions of oxygen. Which cancels and becomes while for manganese (iv) oxide the formula becomes
.