The percent by mass of copper in CuBr2 is Use mc004-1.jpg. 28.45%. 44.30%. 63.55%. 71.55%.
2 answers:
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This is an incomplete question, here is a complete question.
A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?
Answer : The molecular of the compound is,
Explanation :
First we have to calculate the mass of benzene.
Now we have to calculate the molar mass of unknown compound.
Given:
Mass of unknown compound (solute) = 6.45 g
Mass of benzene (solvent) = 43.95 g = 0.04395 kg
Formula used :
where,
= change in freezing point =
= freezing point of solution
= freezing point of benzene
Molal-freezing-point-depression constant for benzene =
m = molality
Now put all the given values in this formula, we get
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 42.9 g
Mass of H = 2.4 g
Mass of N = 16.7 g
Mass of O = 38.1 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of N = 14 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of C =
Moles of H =
Moles of N =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
For N =
For O =
The ratio of C : H : N : O = 3 : 2 : 1 : 2
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula =
The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
Molecular formula =
Therefore, the molecular of the compound is,
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
