Answer:
<h3>The answer is 11 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula

From the question
mass = 3025 g
volume = 275 mL
We have

We have the final answer as
<h3>11 g/mL</h3>
Hope this helps you
The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abbreviation (w/v) is also common. 262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution
<h3>Define Solute</h3>
A solute is a material that dissolves in a solution. The amount of solvent present in fluid solutions is greater than the amount of solute. The two most common examples of solutions in daily life are salt and water. Salt is the solute because it dissolves in water.
<h3>forms of ratios for product concentration or yield:-</h3>
- w/v:- Weight by volume or weight per volume are the terms used. Any solid compound's concentration in a liquid can be calculated using it. It is measurable in gm/ml.
- Weight by weight ratio is referred to as w/w.It is employed to determine the final yield of the compound obtained from the starting compound. as in —mg/—gm.
It provides the real yield of the substance or item.
- Volume/volume. It is used to specify a liquid's composition or percent in a liquid compound.
using w/v we can calculate the weight of sucrose:-
40.0% means 40 g sucrose/ 100 g solution
40.0g sucrose x (655/100)=grams of sucrose
262 grams of sucrose are needed to make 655 mL of a 40.0% (w/v) sucrose solution.
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Answer:
A . 2 O₃(g) + 2 NO ⇒ 2 O₂ (g) + 2 NO₂(g)
B . Yes
C. O and NO₃
Explanation:
A. The overall reaction is obtained by adding the individual steps in the reaction mechanism where we will get the reactants and product and the intermediates will cancel.
Thus, adding 1+ 2 +3 we get
2 O₃(g) + 2 NO ⇒ 2 O₂ (g) + 2 NO₂(g)
B. The reaction intermediates are those that are produced from the initial and/or subsequent steps and are consumed later on in the reaction mechanism, but are neither reactants nor products, they just participate.
From this definition it follows that O(g) and NO₃ are reaction intermediates.
C. O and NO₃
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.