1) Conversion of an isotope one chemical element or an isotope into another chemical element is called as nuclear transmutation.
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2) In a nuclear transmutation reactions</span> can be achieved either due to radioactive decay or due to nuclear reactions.
3) In this technique, it is possible to convert a stable element into radioactive atom by bombarding in with high speed particles. The initial stable nuclei is referred as parent nuclei, the fast moving particle is referred as projectile while new element which is formed is called as daughter element.
4) In the present reaction:
<span>1 1 H+ 1 0 n -> 2 1 H
1 1H is a parent nuclei which is bombarded with the fast moving projectile
(1 0 n) to generate a new daughter nuclei (2 1H). </span>
Answer:
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Explanation:
ppm (parts per million) concentration is defined as the mass (in milligrams) of a substance dissolved in one liter of solution.
In our case we have:
mass of MgBr₂ = 12.41 g
volume of water (which is equal to the final solution volume) = 2.55 L
Now we devise the following reasoning:
if 12.41 g of MgBr₂ are dissolved in 2.55 L of water
then X g of MgBr₂ are dissolved in 1 L of water
X = (1 × 12.41) / 2.55 = 4.867 g of MgBr₂
if in 184 g (1 mole) of MgBr₂ we have 160 g of Br⁻
then in 4.867 g of MgBr₂ we have Y g of Br⁻
Y = (4.867 × 160) / 184 = 4.232 g of bromide (Br⁻)
4.232 g of bromide (Br⁻) = 4234 mg of bromide (Br⁻)
concentration of bromide (Br⁻) = 4234 mg/L = 4234 ppm
Answer:
-0.1767°C (Option A)
Explanation:
Let's apply the colligative property of freezing point depression.
ΔT = Kf . m. i
i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1
m = molality (mol of solute / 1kg of solvent)
We have this data → 0.095 m
Kf is the freezing-point-depression constantm 1.86 °C/m, for water
ΔT = T° frezzing pure solvent - T° freezing solution
(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1
T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C
Answer:
q1..no.2 and 4 are aromatic
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.
