He would study it so he could see how it interacted with the living things around it.
Answer:
1.5e+8 atoms of Bismuth.
Explanation:
We need to calculate the <em>ratio</em> of the diameter of a biscuit respect to the diameter of the atom of bismuth (Bi):
For this, it is necessary to know the values in meters for any of these diameters:
Having all this information, we can proceed to calculate the diameters for the biscuit and the atom in meters.
<h3>Diameter of an atom of Bismuth(Bi) in meters</h3>
1 atom of Bismuth = 320pm in diameter.
<h3>Diameter of a biscuit in meters</h3>
<h3>Resulting Ratio</h3>
How many times is the diameter of an atom of Bismuth contained in the diameter of the biscuit? The answer is the ratio described above, that is, the ratio of the diameter of the biscuit respect to the diameter of the atom of Bismuth:
In other words, there are 1.5e+8 diameters of atoms of Bismuth in the diameter of the biscuit in question or simply, it is needed to put 1.5e+8 atoms of Bismuth to span the diameter of a biscuit in a line.
Answer:
A) CH3CH2SH
Explanation:
Dispersion forces are weak attractions found between non-polar and polar molecules. The attractions here can be attributed to the fact that a non-polar molecule sometimes become polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant. If this happens, the molecule has a temporary dipole. This dipole can induce the neighbouring molecules to be distorted and form dipoles as well. The attractions between these dipoles constitute the Dispersion Forces.
Therefore; the greater the molar mass of a compound or molecule, the higher the Dispersion Force. This implies that the compound or molecule with the highest molar mass have the largest dispersion forces.
Now; for option (A)
CH3CH2SH
The molar mass is :
= (12 + (1 × 3 ) +12 + (1 ×2) + 32+1)
= (12 + 3+ 12 + 2 + 32 + 1)
= 62 g/mol
For option (B)
CH3NH2
The molar mass is:
= (12 + (1 × 3 ) +14 + (1 × 2)
= (12 + 3 + 14 + 2)
= 31 g/mol
For option (C)
CH4
The molar mass is :
= 12 + (1 × 4)
= 12 + 4
= 16 g/mol
For option (D)
CH3CH3
The molar mass is :
= 12 + ( 1 × 3 ) + 12 + ( 1 × 3)
= 12 + 3 + 12 + 3
= 30 g/mol
Thus ; option (A) has the highest molar mass, as such the largest dispersion force is A) CH3CH2SH
Answer:
1000L
Explanation:
the 1 is a sig fig and since the 0 is between the 1 and 4 its also a significant number. to round them off you look at the 0,then look back at the 4 since its less than 5 u round down. then u replace the 43 with 0's
