Answer:
Three times with 5 ml will yield more
Explanation:
Let x represent the amount yield
Kd = (x/15) / ((50-x) / 15) where Kd = 1.5
1.5 = (x/15) / ((50-x) / 15)
x / (50 - x) = 1.5
x = 75 - 1.5x
x + 1.5x = 75
2.5x = 75
x = 75 / 2.5 = 30 mg
when extraction three times
1st extraction
(x1/5) / ((50 - x1) / 15) = 1.5
3x1 / 50 - x1 = 1.5
3x1 = 75 - 1.5x1
3x1 + 1.5x1 = 75
4.5x1 = 75
x1 = 75 / 4.5 = 16.67 gm
second extraction
(x2/ 5) / (33.33 - x2 ) / 15) = 1.5
3x2 / ( 33.33 - x2) = 1.5
3x2 = 1.5(33.33 - x2)
3x2 = 49.995 - 1.5x2
3x2 + 1.5x2 = 49.995
4.5x2 = 49.995
x2 = 49.995 / 4.5
x2 = 11.11 mg
Third extraction
(x3/5) / ((22.22 - x3) / 15) = 1.5
3x3 = 1.5 ( 22.22 - x3)
3x3 + 1.5x3 = 33.33
4.5x3 = 33.33
x3 = 33.33 / 4.5 = 7.41 mg
total extraction = x1 + x2 + x3 =16.67 + 11.11 + 7.41 = 35.19 mg
the three times extraction using 5ml yields 5.19 mg more
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
Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ.
Answer:
krypton is a cation
magnesium is also a cation
aluminium is a anion
and , silicon is cation
