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):
![\\ \frac{diameter\;biscuit}{diameter\;atom(Bi)}](https://tex.z-dn.net/?f=%20%5C%5C%20%5Cfrac%7Bdiameter%5C%3Bbiscuit%7D%7Bdiameter%5C%3Batom%28Bi%29%7D)
For this, it is necessary to know the values in meters for any of these diameters:
![\\ 1m = 10^{3}mm = 1e+3mm](https://tex.z-dn.net/?f=%20%5C%5C%201m%20%3D%2010%5E%7B3%7Dmm%20%3D%201e%2B3mm)
![\\ 1m = 10^{12}pm = 1e+12pm](https://tex.z-dn.net/?f=%20%5C%5C%201m%20%3D%2010%5E%7B12%7Dpm%20%3D%201e%2B12pm)
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.
![\\ 320pm*\frac{1m}{10^{12}pm} = 3.20*10^{-10}m](https://tex.z-dn.net/?f=%20%5C%5C%20320pm%2A%5Cfrac%7B1m%7D%7B10%5E%7B12%7Dpm%7D%20%3D%203.20%2A10%5E%7B-10%7Dm)
<h3>Diameter of a biscuit in meters</h3>
![\\ 51mm*\frac{1}{10^{3}mm} = 51*10^{-3}m = 5.1*10^{-2}m](https://tex.z-dn.net/?f=%20%5C%5C%2051mm%2A%5Cfrac%7B1%7D%7B10%5E%7B3%7Dmm%7D%20%3D%2051%2A10%5E%7B-3%7Dm%20%3D%205.1%2A10%5E%7B-2%7Dm%20)
<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:
![\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1*10^{-2}m}{3.20*10^{-10}m}](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%20%5Cfrac%7B5.1%2A10%5E%7B-2%7Dm%7D%7B3.20%2A10%5E%7B-10%7Dm%7D)
![\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%20%5Cfrac%7B5.1%7D%7B3.20%7D%5Cfrac%7B10%5E%7B-2%7D%7D%7B10%5E%7B-10%7D%7D%5Cfrac%7Bm%7D%7Bm%7D)
![\\ Ratio_{\frac{biscuit}{atom}}= \frac{5.1}{3.20}\frac{10^{-2}}{10^{-10}}\frac{m}{m}](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%20%5Cfrac%7B5.1%7D%7B3.20%7D%5Cfrac%7B10%5E%7B-2%7D%7D%7B10%5E%7B-10%7D%7D%5Cfrac%7Bm%7D%7Bm%7D)
![\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{-2+10}](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%201.5%2A10%5E%7B-2%2B10%7D)
![\\ Ratio_{\frac{biscuit}{atom}}= 1.5*10^{8}=1.5e+8](https://tex.z-dn.net/?f=%20%5C%5C%20Ratio_%7B%5Cfrac%7Bbiscuit%7D%7Batom%7D%7D%3D%201.5%2A10%5E%7B8%7D%3D1.5e%2B8)
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.
The others are wrong it's B. Condensation
4.The other light bulb will stay on and glow brightly.
1 has a higher ionization dismal aoa
Answer:
E) 2.38
Explanation:
The pH of any solution , helps to determine the acidic strength of the solution ,
i.e. ,
- Lower the value of pH , higher is its acidic strength
and ,
- Higher the value of pH , lower is its acidic strength .
pH is given as the negative log of the concentration of H⁺ ions ,
hence ,
pH = - log H⁺
From the question ,
the concentration of the solution is 0.0042 M , and being it a strong acid , dissociates completely to its respective ions ,
Therefore , the concentration of H⁺ = 0.0042 M .
Hence , using the above equation , the value of pH can be calculated as follows -
pH = - log H⁺
pH = - log ( 0.0042 M )
pH = 2.38 .