<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
Answer:
Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.
Explanation:
Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.
Answer : The value of reaction quotient, Q is 0.0625.
Solution : Given,
Concentration of
= 2.00 M
Concentration of
= 2.00 M
Concentration of
= 1.00 M
Reaction quotient : It is defined as a concentration of a chemical species involved in the chemical reaction.
The balanced equilibrium reaction is,

The expression of reaction quotient for this reaction is,
![Q=\frac{[Product]^p}{[Reactant]^r}\\Q=\frac{[NH_3]^2}{[N_2]^1[H_2]^3}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BProduct%5D%5Ep%7D%7B%5BReactant%5D%5Er%7D%5C%5CQ%3D%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5E1%5BH_2%5D%5E3%7D)
Now put all the given values in this expression, we get

Therefore, the value of reaction quotient, Q is 0.0625.
Possibly, Yes if it is proven
Answer:
83.24 mmHg.
Explanation:
- <em>The vapor pressure of the solution (Psolution) = (Xmethanol)(P°methanol).</em>
where, Psolution is the vapor pressure of the solution,
Xmethanol is the mole fraction of methanol,
P°methanol is the pure vapor pressure of methanol.
- We need to calculate the mole fraction of methanol (Xmethanol).
<em>Xmethanol = (n)methanol/(n) total.</em>
where, n methanol is the no. of moles of methanol.
n total is the total no. of moles of methanol and urea.
- We can calculate the no. of moles of both methanol and urea using the relation: n = mass/molar mass.
n of methanol = mass/molar mass = (56.9 g)/(32.04 g/mol) = 1.776 mol.
n of urea = mass/molar mass = (7.38 g )/(60.06 g/mol) = 0.123 mol.
∴ Xmethanol = (n)methanol/(n) total = (1.776 mol)/(1.776 mol + 0.123 mol) = 0.935.
<em>∴ Psolution = (Xmethanol)(P°methanol)</em> = (0.935)(89.0 mmHg) =<em> 83.24 mmHg.</em>