<span>Answer:
Glass has a much higher specific heat than styrofoam so more of the energy produced will go into heating the glass. This quantity of heat will be lost (unaccounted for) hence the delta t in the calculation will be too small and the heat of neutralization will be lower.</span>
Answer:
nsnsjejd when ne nb w wnjwnw w wnjwnw wnjwnw wnjwnw wanna wanna e e wnjwnw e e she added ccccccccf
Answer:
a) The volume is 5.236x10⁻¹³L
b) The molarity of a single virus is 1.91x10¹² mol/L
c) The molarity for a 100 virus particles is 1.91x10¹⁴ mol/L
Explanation:
a) Given:
D = diameter of the cell = 10 μm
r = radius = 10/2 = 5 μm
The volume of the spherical cell is equal:
![V=\frac{4}{3} \pi r^{3} =\frac{4}{3} \pi *(5)^{3} =523.6\mu m^{3}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%20%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%2A%285%29%5E%7B3%7D%20%3D523.6%5Cmu%20m%5E%7B3%7D)
If 1 μm³ = 1x10⁻¹⁵L, then 523.6 μm³ = 5.236x10⁻¹³L
b) The molarity is:
![M=\frac{number-of-moles}{volume-of-solution}](https://tex.z-dn.net/?f=M%3D%5Cfrac%7Bnumber-of-moles%7D%7Bvolume-of-solution%7D)
For a single virus within the cell
![M=\frac{1}{5.236x10^{-13} } =1.91x10^{12} mol/L](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B1%7D%7B5.236x10%5E%7B-13%7D%20%7D%20%3D1.91x10%5E%7B12%7D%20mol%2FL)
c) For a 100 virus particles the molarity is:
![M=\frac{100}{5.236x10^{-13} } =1.91x10^{14} mol/L](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B100%7D%7B5.236x10%5E%7B-13%7D%20%7D%20%3D1.91x10%5E%7B14%7D%20mol%2FL)
Answer:
B. ethylamine.
Explanation:
- Since the compound contains NH₂, it is an amine compound.
- The alkyl group C₂H₅- contains 2 C atoms (ethyl group).
<em>So, the compound is ethylamine.</em>
<em />
Answer:
the oxygenmolecule(02or0-0) its self is free redicles or a-b redical whith two unpaired electrons that is electron configuration of the molecule has two unpaired electros occupying two degenerate molecular or bitals with the same spin orientation(up-upordown-dowan
Explanation:
in the lower atomsphere important redical are produced by the photodissociation if nitrogen dioxide to an oxygen atom and nitric oxide