A metal M forms the oxide M2O. When 0.890 grams of M reacts with pure oxygen, 0.956 grams of M2O form. Write the balanced equati
2 answers:
The molar mass of M is 0.225g/mol and the element M is Hydrogen
If a metal M combines with an oxygen element to form the oxide, then the chemical reaction will be expressed as:
This shows that 4 moles of an unknown element M react with the oxygen element to produce the oxide
Given the following parameters
Mass of M = 0.890 grams
Mass of = 0.956 grams
Get the molar mass of M:
Molar mass = Mass/number of moles
Molar mass = 0.890/4
Molar mass = 0.225g/mol
Hence the molar mass of M is 0.225g/mol and the element M is Hydrogen
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Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
<u>What do we know about the unknown alkene? </u>
We know the product of the ozonolysis reaction (see figure 1). This reaction is an <u>oxidative rupture reaction</u>. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
<u>What is the product with the peroxyacid?</u>
This compound in the presence of alkenes will produce <u>peroxides.</u> Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product <u>2,2,3,3-tetrapropyloxirane.</u> (see figure 2)
