Aluminum is 26.982 grams per mole, so 26.982/19.9 will give you the moles 1.3558794
Answer:
Passive transport
Explanation:
In Passive transport the energy is not essential in other for transportation to take place.it is in active transport where you see the need of energy for the process to take place.Passive transport involves the action of facilitated diffusion and osmosis.
Answer:
1528.3L
Explanation:
To solve this problem we should know this formula:
V₁ / T₁ = V₂ / T₂
We must convert the values of T° to Absolute T° (T° in K)
21°C + 273 = 294K
70°C + 273 = 343K
Now we can replace the data
1310L / 294K = V₂ / 343K
V₂ = (1310L / 294K) . 343K → 1528.3L
If the pressure keeps on constant, volume is modified directly proportional to absolute temperature. As T° has increased, the volume increased too
The right answer is noble gases.
The noble gases, or rare gases, are the chemical elements of group 18 (formerly "group VIIIA" or even "group 0") of the periodic table. These are helium He, neon 10Ne, argon 18Ar, krypton 36Kr, xenon 54Xe and radon 86Rn, the latter being radioactive.
This property means that they can not bind with other atoms to form molecules or lose electrons to transform into ions, hence their name noble or inert gases (they are not very active and do not do not mix). In nature, all elements "want" to become stable.
Answer:
We identify nucleic acid strand orientation on the basis of important chemical functional groups. These are the <u>phosphate</u> group attached to the 5' carbon atom of the sugar portion of a nucleotide and the <u>hydroxyl</u> group attached to the <u>3'</u> carbon atom
Explanation:
Nucleic acids are polymers formed by a phosphate group, a sugar (ribose in RNA and deoxyribose in DNA) and a nitrogenous base. In the chain, the phosphate groups are linked to the 5'-carbon and 3'-carbon of the ribose (or deoxyribose) and the nitrogenous base is linked to the 2-carbon. Based on this structure, the nucleic acid chain orientation is identified as the 5'-end (the free phosphate group linked to 5'-carbon of the sugar) and the 3'-end (the free hydroxyl group in the sugar in 3' position).
