Let's think, if you have a candle ( that is not blown out ) the physical properties are the candles mass and hence ( hence of the candle is the stiffness of the candle), weight, length, density, surface friction ( force resisting the relative motion of solid surface), and the energy content. You then, need to go to bed, so, therefore, you want to blow the candle out. Once you blow the candle out, the candle is evidently going to have at least a couple of different physical properties, than before it was blown out. The physical properties are a different color, the length of the candle, the texture, you could also apply the mass of the candleholder, and then, the mass of the candleholder and the candle, last but not least, the mass of just the candle. Once you observe the candle, you should be able to plug in those observations into the physical properties. As to, because you asked' what are the physical properties of a candle that has been blown out... We are going to assume that we did observe the candle, and the length of the candle in cm, after being blown out is 30cm. (12 inches; customary). Next, that the color of the candle is the same (let us say the original color is taffy pink). We can then say that the texture of the candle is waxy and the top and smooth as you get to the bottom ( the texture depends on how long the candle was burning, but we are saying that we lit the candle, and then immediately blew the flame out ) . We now have the mass of the candleholder, which will scientificity stay the same. Now, for the mass of the candleholder and the candle, that all depends of how long you let it burn ( remember, we are saying we lit the wick and then immediately blew the fame out ). So, the candle really didn't change is mass, so, therefore, wouldn't affect the mass of the candleholder including the candle. That also goes to the mass of the candle.
Answer: The pentose phosphate pathway (PPP) is localized to the cytosol because fatty acid synthesis uses the NADPH generated by the PPP.
Explanation:
The pentose phosphate pathway is mainly catabolic and provides an alternative glucose oxidizing pathway for the generation of NADPH that is required for reductive biosynthetic reactions such as those of cholesterol biosynthesis, bile acid synthesis, steroid hormone biosynthesis, and fatty acid synthesis.
Fatty acid biosynthesis occurs in the cytosol and requires the reducing equivalent NADPH in large amounts. <em>The main source of generating NADPH in animal cells, the pentose phosphate pathway is therefore, localized in the cytosol in order to furnish a strongly reducing environment for fatty acid biosynthesis to proceed.</em>
Answer:
acetic acid and phosphoric acid
Explanation:
After refluxing the reaction mixture ( synthesis of isoamyl acetate ) what is likely present in the solution is acetic acid and phosphoric acid, this due to the fact that if the reaction time between the reactants was less than the refluxing time which is 25 minutes,
there will be no reactant ( 3-methylbutanol )left in the reaction mixture
Answer:
(a) a = 5.08x10⁻⁸ cm
(b) r = 179.6 pm
Explanation:
(a) The lattice parameter "a" can be calculated using the following equation:
<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>
<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>
Hence the lattice parameter is:

![a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm](https://tex.z-dn.net/?f=%20a%20%3D%20%5Csqrt%5B3%5D%7B1.32%20%5Ccdot%2010%5E%7B-22%7D%20cm%5E%7B3%7D%7D%20%3D%205.08%20%5Ccdot%2010%5E%7B-8%7D%20cm%20)
(b) We know that the lattice parameter of a FCC structure is:

<em>where r: is the atomic radius of Th</em>
Hence, the atomic radius of Th is:
I hope it helps you!