Hi!
The radical bromination reaction of C₆H₅CH₂CH₃ is performed through a mechanism in which radical reactions are involved. This compound is an alkylbenzene compound, and the carbon that is more reactive towards radical bromination is the carbon bonded to the aromatic ring because in the reaction mechanism the intermediaries are stabilized by resonance in the aromatic ring.
A terminal substitution will not occur because substitution there will not be stabilized by resonance. The compound that will be formed in this reaction would be:
C₆H₅CH₂CH₃ + Br₂ → C₆H₅CH₂(Br)CH₃ + HBr
<span>Na + Cl = NaCl
answer : </span><span>synthesis reaction .
hope this helps!
</span>
Answer:
C4H8O4
Explanation:
To determine the molecular formula, first, let us obtain the empirical formula. This is illustrated below:
From the question given, we obtained the following information:
C = 45.45%
H = 6.12%
O = 48.44%
Divide the above by their molar mass
C = 45.45/12 = 3.7875
H = 6.12/1 = 6.12
O = 48.44/16 = 3.0275
Divide by the smallest
C = 3.7875/3.0275 = 1
H = 6.12/3.0275 = 2
O = 3.0275/3.0275 = 1
The empirical formula is CH2O
The molecular formula is given by [CH2O]n
[CH2O]n = 132.12
[12 + (2x1) + 16]n = 132.12
30n = 132.12
Divide both side by the coefficient of n i.e 30
n = 132.12/30 = 4
The molecular formula is [CH2O]n = [CH2O]4 = C4H8O4
Answer:
The mass of xenon in the compound is 2.950 grams
Explanation:
Step 1: Data given
Mass of XeF4 = 4.658 grams
Molar mass of XeF4 = 207.28 g/mol
Step 2: Calculate moles of XeF4
Moles XeF4 = mass XeF4 / molar mass XeF4
Moles XeF4 = 4.658 grams / 207.28 g/mol
Moles XeF4 = 0.02247 moles
Step 3: Calculate moles of xenon
XeF4 → Xe + 4F-
For 1 mol xenon tetrafluoride, we have 1 mol of xenon
For 0.02247 moles XeF4 we have 0.02247 moles Xe
Step 4: Calculate mass of xenon
Mass xenon = moles xenon * molar mass xenon
Mass xenon = 0.02247 moles * 131.29 g/mol
Mass xenon = 2.950 grams
The mass of xenon in the compound is 2.950 grams
Answer:
Mg₃N₂
Explanation:
The empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound. Using molecular mass of Mg (24,305g/mol) and mass of nitrogen (14,006g/mol), moles of each element are:
0,73g × (1mol / 24,305g) = 0,03 moles of Mg
0,28g × (1mol / 14,006g) = 0,02 moles of N
Dividing each value in 0,01 to obtain natural numbers:
0,03 moles of Mg / 0,01 = 3
0,02 moles of N / 0,01 = 2.
Thus, empirical formula is: <em>Mg₃N₂</em>
<em></em>
I hope it helps!
