The NMR is attached that is required to answer this question. We are told that we have a carboxylic acid and that there is a nitro group directly attached to an aromatic ring. We can begin by determining the substitution on the aromatic ring.
Looking at the NMR spectrum, we a peak that integrates to 1 H at 12 ppm which is characteristic of a carboxylic acid, which we already know is present. Next we have two equivalent doublets that both integrate to 2 H, giving us 4 hydrogens in total. These doublets are in the aromatic region and this type of coupling pattern is characteristic of a 1,4-substituted aromatic ring, so we know there is only one other group substituted on the ring. However, the molecular formula is C₉H₉NO₄, so there are still 2 carbons not accounted for, if we include our carboxlic acid. Therefore, the carboxylic acid must be attached to some alkyl group which is substituted onto the aromatic ring.
We have a doublet at 1.6 ppm that integrates to 3, which suggests this is a methyl group adjacent to a CH. We also have a quartet at 4.0 ppm with an integration of 1. This suggests it is a CH that is adjacent to 3 hydrogen, most likely the methyl group we just described.
Therefore, we have a CH attached to a CH3, so that CH requires two more bonds. The only pieces left to attach to it are the aromatic ring and the carboxylic acid functional group. This gives us the structure shown in the image provided.
5 plates is the highest amount that can be served
There’s only 5 sandwiches so 7 is automatically ruled out, there’s 14 corn cobs and 5 sandwiches only need 10 so it works out
To be able to compare the result with other experiments it has to be reported in moles.
number of moles = mass / molecular weight
number of moles of Mg(H₂PO₄)₂ = 600 / 218 = 2.75 moles
Answer:
Reactions 1, 3 and 5
Explanation:
First thing's first, let's ensure that all the reactions given are balanced. This is given as;
CO(g) + 1/2 O2(g )→ CO2(g)
Li(s) + 1/2 F2(l) → LiF(s)
C(s) + O2(g) → CO2(g)
CaCO3(g) → CaO + CO2(g)
2Li(s) + F2(g) → 2LiF(s)
For the condition to be valid;
- There is by convention 1 mol of product made. This means we eliminate reactions with more than one mole of compound formed. This eliminates reaction 5.
- The lements haveto be in their state at room temperature. Fluorine is a gas, not a liquid, at room temperature ans pressure, so 2 is not a correct answer.
This leaves us with reactions 1, 3 and 5 as the correct reactions that satisify the condition.