Answer:
the sun is the smallest object in our Solar System
Explanation:
Answer:
χH₂ = 0.4946
χN₂ = 0.4130
χAr = 0.0923
Explanation:
The total pressure of the mixture (P) is:
P = pH₂ + pN₂ + pAr
P = 443.0 Torr + 369.9 Torr + 82.7 Torr
P = 895.6 Torr
We can find the mole fraction of each gas (χ) using the following expression.
χi = pi / P
χH₂ = pH₂ / P = 443.0 Torr/895.6 Torr = 0.4946
χN₂ = pN₂ / P = 369.9 Torr/895.6 Torr = 0.4130
χAr = pAr / P = 82.7 Torr/895.6 Torr = 0.0923
Answer:
Following two compounds have Hydrogen Bond Interactions;
1) CH₃(CH₂)₂NH₂ (Propan-1-amine)<span>
2) </span>CH₃(CH₂)₂NH(CH₂)₄CH₃ (N-propylpentan-1-amine)
Explanation:
Hydrogen Bond Interactions are formed between those molecules which has hydrogen atoms covalently bonded to most electronegative atoms like Fluorine, Oxygen and Nitrogen. This direct attachment of Hydrogen to electronegative atom makes it partial positive resulting in hydrogen bonding with neighbor's partial negative most electronegative atom. So, in above selected compounds it can be seen that both compounds contain hydrogen atoms directly attached to Nitrogen atoms, Therefore, allowing them to form Hydrogen Bonding Interactions.
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
I'm not completely sure but if I did know I would definitely tell u
