<u>Answer:</u>
<u>For a:</u> The edge length of the unit cell is 314 pm
<u>For b:</u> The radius of the molybdenum atom is 135.9 pm
<u>Explanation:</u>
To calculate the edge length for given density of metal, we use the equation:

where,
= density = 
Z = number of atom in unit cell = 2 (BCC)
M = atomic mass of metal (molybdenum) = 95.94 g/mol
= Avogadro's number = 
a = edge length of unit cell =?
Putting values in above equation, we get:
![10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm](https://tex.z-dn.net/?f=10.28%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%20%28a%29%5E3%7D%5C%5C%5C%5Ca%5E3%3D%5Cfrac%7B2%5Ctimes%2095.94%7D%7B6.022%5Ctimes%2010%5E%7B23%7D%5Ctimes%2010.28%7D%3D3.099%5Ctimes%2010%5E%7B-23%7D%5C%5C%5C%5Ca%3D%5Csqrt%5B3%5D%7B3.099%5Ctimes%2010%5E%7B-23%7D%7D%3D3.14%5Ctimes%2010%5E%7B-8%7Dcm%3D314pm)
Conversion factor used:
Hence, the edge length of the unit cell is 314 pm
To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

where,
R = radius of the lattice = ?
a = edge length = 314 pm
Putting values in above equation, we get:

Hence, the radius of the molybdenum atom is 135.9 pm
pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.
pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.
pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.
Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.
Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :
HCL(0.040M)----> H+(0.040M) +CL-(0.040M)
HBr(0.075M)----> H+(0.075M) +Br-(0.075M)
so 0.040M (H+ from HCL) + 0.075M (H+ from HBr) = 0.115M H+ in total.
pH is calculated as:
pH = -log[H+]
Substituting values in the equation:
log(0.115M)= 0.939 pH
pOH is calculated as:
14 - pH = pOH
Substituting values in the equation above:
14 - 0.939= 13.061 pOH
Therefore, pH is 0.939 and pOH is 13.061.
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Answer:
Highest pH(most basic)
Sr(OH)2(aq)
KOH (aq)
NH3(aq)
HF (aq)
HClO4(aq)
Lowest pH(most acidic)
Explanation:
The concentration of H+ ion will determine the pH of a solution. The pH actually reflects the ratio of H+ ion and OH- since both of them can combine into water. Solution with more H+ ion will have a lower pH and called acidic, while more OH- will have high pH and be called basic. Strong acid/base will be ionized more than weak acid/base.
Sr(OH)2(aq) = strong base, release 2 OH- ion per mole
KOH (aq) = Strong base, release 1 OH- per mole
NH3(aq) = weak base, release less than 1 OH- per mole
HF (aq) =strong acid, release 1 H+ per mole
HClO4(aq) = stronger acid, release 1 H+ per mole
Answer:
The enthalpy change in the the reaction is -47.014 kJ/mol.
Explanation:

Volume of water in calorimeter = 22.0 mL
Density of water = 1.00 g/mL
Mass of the water in calorimeter = m

Mass of substance X = 2.50 g
Mass of the solution = M = 2.50 g + 22 g = 24.50 g
Heat released during the reaction be Q
Change in temperature =ΔT = 28.0°C - 14.0°C = 14.0°C
Specific heat of the solution is equal to that of water :
c = 4.18J/(g°C)


Heat released during the reaction is equal to the heat absorbed by the water or solution.
Heat released during the reaction =-1.433 kJ
Moles of substance X= 
The enthalpy change, ΔH, for this reaction per mole of X:

The empirical formula of the given compound is
.
The correct option is B.
<h3>What is an empirical formula?</h3>
The empirical formula of a chemical compound is the simplest whole-number ratio of atoms contained in the substance.
Given,
1.0 g of S
1.5 g of O
To calculate the empirical formula, we will divide the masses of the elements by their atomic weight.
For sulfur

For oxygen

Now, divide the greater value of mole came by the smaller value

Thus, the empirical formula for the given compound is 1 for S and 3 for O

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