Answer:
because water wets stuff expect for fire.
Answer:
(S)-3-methoxy-3-methylbutan-2-ol
Explanation:
In this case, we have an <u>epoxide opening in acid medium</u>. The first step then is the <u>protonation of the oxygen</u>. Then the epoxide is broken to generate the most <u>stable carbocation</u>. The nucleophile (
) will attack the carbocation generating a new bond. Finally, the oxygen is <u>deprotonated</u> to obtain an ether functional group and we will obtain the molecule <u>(S)-3-methoxy-3-methylbutan-2-ol</u>.
See figure 1
I hope it helps!
The major species in solution when solid ammonium bromate is dissolved in water is shown below
Answer:
5.00 mol Mg
10.0 mol Cl
40.0 mol O
Explanation:
Step 1: Given data
Moles of Mg(ClO₄)₂: 5.00 mol
Step 2: Calculate the number of moles of Mg
The molar ratio of Mg(ClO₄)₂ to Mg is 1:1.
5.00 mol Mg(ClO₄)₂ × 1 mol Mg/1 mol Mg(ClO₄)₂ = 5.00 mol Mg
Step 3: Calculate the number of moles of Cl
The molar ratio of Mg(ClO₄)₂ to Cl is 1:2.
5.00 mol Mg(ClO₄)₂ × 2 mol Cl/1 mol Mg(ClO₄)₂ = 10.0 mol Cl
Step 4: Calculate the number of moles of O
The molar ratio of Mg(ClO₄)₂ to Cl is 1:8.
5.00 mol Mg(ClO₄)₂ × 8 mol O/1 mol Mg(ClO₄)₂ = 40.0 mol O