Answer:
0.07172 L = 7.172 mL.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.
</em>
where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 273 K, Standard T).
<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>
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:
Anode half reaction;
Co(s) ----> Co^2+(aq) + 2e
Cathode half reaction;
2Ag^+(aq) + 2e-------> 2Ag(s)
Explanation:
A voltaic cell is an electrochemical cell that spontaneously produces electrical energy from chemical reactions. A voltaic cell comprises of an anode (where oxidation occurs) and a cathode (where reduction occurs). The both electrodes are connected with a wire . A salt bridge ensures charge neutrality in the anode and cathode compartments. Electrons flow from anode to cathode.
For the cell referred to in the question;
Anode half reaction;
Co(s) ----> Co^2+(aq) + 2e
Cathode half reaction;
2Ag^+(aq) + 2e-------> 2Ag(s)
Answer:
465mL
Explanation:
Volume of a solution, V =Mass of substance, m/(Molarity of the solution of the substance, M × molar mass of the substance, M.m)
Given in the question,
M=.132M
M.m=23+35.5 = 58.5g/mol
m=3.59g
V= 3.59/(.132×58.5)
V = 0.465L
Volume in mL = volume in L × 1000
= 0.465 × 1000 = 465mL
Therefore, 465mL of a .132M aqueous solution of sodium chloride, NaCl, must be taken to obtain 3.59 grams of the salt
