Answer: The diagrams show gases that are stored in two separate but similar containers. If both gases are at the same temperature, which one has the greater pressure? Gas 2 because it has more particles that are colliding.
Explanation:
The number of electrons in the outermost shell of an atom determines its physical and chemical properties. All elements are most stable with 8 electrons (an octet) in their outer shell.
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Answer : The pH of the solution is, 3.41
Explanation :
First we have to calculate the moles of 
.
Now we have to calculate the value of 
.
The expression used for the calculation of is,
Now put the value of 
in this expression, we get:
The reaction will be:
Initial moles 0.375 0.100 0.375
At eqm. (0.375-0.100) 0 (0.375+0.100)
= 0.275 = 0.475
Now we have to calculate the pH of solution.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[F^-]}{[HF]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
Now put all the given values in this expression, we get:
![pH=3.17+\log [\frac{(\frac{0.475}{1.50})}{(\frac{0.275}{1.50})}]](https://tex.z-dn.net/?f=pH%3D3.17%2B%5Clog%20%5B%5Cfrac%7B%28%5Cfrac%7B0.475%7D%7B1.50%7D%29%7D%7B%28%5Cfrac%7B0.275%7D%7B1.50%7D%29%7D%5D)
Thus, the pH of the solution is, 3.41
Answer:
A₅B₄
Explanation:
Since we have one atom of element A at the center of each face of the unit cell, since the unit cell is a cubic cell, we have 6 faces. Since the atom on the face of the unit cell is shared with another cell, we have half of it in the unit cell is shared So, the number of atoms per face is 1/2 atom/face × 6 faces = 4 atoms on the faces of the unit cell.
Also, we have 1 atom at each corner of the cubic unit cell. Since there are 8 corner in the cubic unit cell. Also, each atom at the corner is shared with 8 unit cells, so we have 1/8 atom per corner. So, the number of atoms per unit cell is 1/8 atom/corner × 8 corners = 1 atoms at the corners of the unit cell.
So, in total we have 4 + 1 = 5 atoms of element A in the unit cell.
Also, there are 4 atoms of element B in the unit cell.
So, the ratio of atoms of element A to element B is 5 : 4.
A:B = 5:4
So, the empirical formula of the compound containing elements A and B is A₅B₄
The average atomic mass of N2 is 14.0067. The atomic masses of the 2 stable isotopes of nitrogen, N-14 and N-15, are14.003074002 and 15.00010897 amu, respectively. Calculate the percent abundance of each nitrogen isotope. Ex.: The average atomic weight of copper is 63.5amu .
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