Answer:
Part a: The IR spectrum is not sufficient due to the similar radical group presence in both product and reactants.
Part b: Mass Spectrum data is required in addition to the IR data for confirming the presence of recrystallized product.
Part c: The additional data is consistent with the expectations as the data is in support of that.
Explanation:
Part a
Acetanilide has a characteristic IR peak at around 3100-3200 cm^-1 (N-H Stretching) and 1500-1680 cm^-1 (C=O Stretching). Since a similar radical group (
) is present in the 4-bromoacetanilide, thus it is difficult to affirm the presence of the product solely on the basis of IR spectrum.
Part b
Mass Spectral data will affirm the presence of recrystallized product with certainty.
Part c
4-bromoacetanilide will have 
peak at two points (213 & 215) corresponding to the two Bromine isotopes (79 & 81).
Also parent peak at 171 (=43) will be in 49:51 due to presence of Bromine atoms.
<span>1. Translate, predict the products, and balance the equation above.
Li + Cu(NO3)2 = Li(NO3)2 + Cu
2. How many particles of lithium are needed to produce 125 g of copper?
125 g Cu ( 1 mol / 63.55 g ) (1 mol Li / 1 mol Cu ) ( 6.022 x 10^23 particles / 1 mol ) = 1.18x10^24 Li particles
3. How many grams of lithium nitrate are produced from 4.83E24 particles of copper (II) nitrate?
</span>4.83E24 particles of copper (II) nitrate ( 1 mol / 6.022x10^23 particles ) (1 mol Li(NO3)2 / 1 mol Cu(NO3)2 ) ( 130.95 g / 1 mol ) = 1043.77 grams Li(NO3)2
Hey there!:
Volume :
1.00 cm * 4.00 cm * 2.50 cm => 10 cm³
Density = 22.57 g/cm³
Therefore:
Mass = density * Volume
Mass = 22.57 * 10
Mass = 225.7 g
Answer: +35
explanation: 30+35£~£
