The pool being cooler at the deep end. Also the air closer to the roof of a house being cooler.
First, we determine the mass of each element from the data collected. We can get the mass of molybdenum Mo from the difference between the mass of crucible and molybdenum and the mass of crucible:
Mass of molybdenum = 39.52 – 38.26 = 1.26 g Mo
We can calculate for the mass of molybdenum oxide from the difference between the mass of crucible and molybdenum oxide and the mass of crucible:
Mass of molybdenum oxide = 39.84 – 38.26 = 1.58g
We can now compute for the mass of oxygen O by subtracting the mass of molybdenum from the mass of molybdenum oxide:
Mass of oxygen in molybdenum oxide = 1.58 – 1.26 = 0.32g O
To convert mass to moles, we use the molar mass of each element.
1.26 g Mo * 1 mol Mo / 95.94 g Mo = 0.0131 mol Mo
0.32 g O * 1 mol O / 15.999 g O = 0.0200 mol O
0.0131 mol is the smallest number of moles. We divide each mole value by this number:
0.0131 mol Mo / 0.0131 = 1
0.0200 mol O / 0.0131 = 1.53
Multiplying these results by 2 to get the lowest whole number ratio,
0.0131 mol Mo / 0.0131 = 1 * 2 = 2
0.0200 mol O / 0.0131 = 1.5 * 2 = 3
Thus, we can write the empirical formula as Mo2O3.
In a closed system, heat should be conserved which means that the heat produced in the calorimeter is equal to the heat released by the combustion reaction. We calculate as follows:
Heat of the combustion reaction = mC(T2-T1)
= 1 (1.50) (41-21)
= 30 kJ
Answer is 0.289nm.
Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.
wt % of Fe in Fe-V alloy = 85%
wt % of V in Fe-V alloy = 15%
We need to calculate edge length of the unit cell having bcc structure.
Using density formula,
For calculating edge length,
For calculating 
, we use the formula
Similarly for calculating 
, we use the formula
From the periodic table, masses of the two elements can be written
Specific density of both the elements are
Putting and 
formula's in edge length formula, we get
![a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}](https://tex.z-dn.net/?f=a%3D%5Cleft%20%5B%5Cfrac%7BZ%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B%28wt%5C%25%29_%7BFe%7D%7D%7BM_%7BFe%7D%7D%2B%5Cfrac%7B%28wt%5C%25%29_%7BFe%7D%7D%7BM_%7BFe%7D%7D%7D%20%20%5Cright%20%29%7D%7BN_A%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B%28wt%5C%25%29_V%7D%7B%5Crho_V%7D%2B%5Cfrac%7B%28wt%5C%25%29_V%7D%7B%5Crho_V%7D%7D%20%20%5Cright%20%29%7D%20%20%5Cright%20%5D%5E%7B1%2F3%7D)
![a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}](https://tex.z-dn.net/?f=a%3D%5Cleft%20%5B%5Cfrac%7B2atoms%2F%5Ctext%7Bunit%20cell%7D%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B85%5C%25%7D%7B55.85g%2Fmol%7D%2B%5Cfrac%7B15%5C%25%7D%7B50.941g%2Fmol%7D%7D%20%20%5Cright%20%29%7D%7B%286.023%5Ctimes10%5E%7B23%7Datoms%2Fmol%29%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B85%5C%25%7D%7B7.874g%2Fcm%5E3%7D%2B%5Cfrac%7B15%5C%25%7D%7B6.10g%2Fcm%5E3%7D%7D%20%20%5Cright%20%29%7D%20%20%5Cright%20%5D%5E%7B1%2F3%7D)
By calculating, we get
