PV=nRT
Here
P1/T1= P2 / T2
1 torr=133 pascal
600 *133 /215 = 750 *133 / t2
T2= 268.75 K
Answer:
The correct answer is - palatine tonsils
.
Explanation:
Palatine tonsils or more normally tonsils are the pair of soft tissue masses of highlighted secondary lymphatic tissue present on the lateral wall of the oropharynx in the tonsilar fossa just below to the soft plate. It is the lymphatic tissue that is located to inferior to the soft plate.
Thus, the correct answer is - palatine tonsils.
The volume of the unit cell is 2.67 x 10⁻²⁸ m³.
<h3>What is the volume of a unit cell of a body-centered cubic crystal?</h3>
In a body-centered cubic unit cell, the volume occupied by the particles of the substance is about 68% of the total unit cell.
Assuming that a single atomic a sphere, the volume is:
Volume(atom) = 4/3 x π x r³
Volume(atom) = 4/3 x π x (169 x 10⁻¹²)³
Volume(atom) = 2.02 x 10⁻²⁹ m³
There are a total of 9 atoms in a body-centered unit cell, so the total volume occupied by atoms is:
2.02 x 10⁻²⁹ x 9
= 1.82 x 10⁻²⁸ m³
Volume of cell = (1.15 x 10⁻²⁸ ) / 0.68
Volume of cell = 2.67 x 10⁻²⁸ m³
Therefore, the volume of the unit cell is 2.67 x 10⁻²⁸ m³.
Learn more volume of unit cells at: brainly.com/question/1594030
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Mass percentage of sodium chloride(NaCl) in ocean waters = 3.5 %
That means 3.5 g sodium chloride(NaCl) is present for every 100 g of ocean water.
The given mass of sodium chloride(NaCl) is 45.8 g
Calculating the mass of ocean waters that would contain 45.8 g sodium chloride(NaCl):

= 1309 g ocean water
Therefore, 45.8 g sodium chloride is present in 1309 g ocean water.
Answer:
Compound A and compound B are constitutional isomers with molecular formula C3H7Cl.
When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination reaction predominates.
Explanation:
Constitutional isomers are the one which differs in the structural formula.
When compound A is treated with sodium methoxide, a substitution reaction predominates.
That means sodium methoxide is a strong base and a strong nucleophile.
But when it reacts with primary alkyl halides it forms a substitution product and when it reacts with secondary alkyl halide it forms mostly elimination product.
The reaction and the structures of A and B are shown below: