1 moles -------- 6.02x10²³ atoms
?? moles ----- 7.4x10²¹ atoms
moles = 7.4x10²¹ / 6.02x10²³
= 0.01229 moles
hope this helps!
Volume = 22.4 dm3
n = 2 mol of H2
n = 1 mol of N2
Temperature = 273.15
All H2 reacts
reaction
N2 + 3H2 = 2NH3
1:3 ratio
Calculation:
N2 initial - N2 reacted = Final N2
1 - 2*(1/3) = 0.3333 mol of N2 left
H2 = 0 left
NH3 formed = 2/3*1 = 2/3 = 0.666
Total mol:
0.3333 + 0.666 = 1 mol
Apply the equation :
PV = nRT
P = nRT/V = 1*0.0082*(273.15)/(22.4) = 0.0999924 atm
PH2 = 0
PN2 = 1/3*0.0999924 = 0.0333308 atm
PNH3 = 2/3*0.0999924 = 0.0666616 atm
Answer is 0.0666616 atm
False .........................................
Answer:
37 mmol of acetate need to add to this solution.
Explanation:
Acetic acid is an weak acid. According to Henderson-Hasselbalch equation for a buffer consist of weak acid (acetic acid) and its conjugate base (acetate)-
![pH=pK_{a}(acetic acid)+log[\frac{mmol of CH_{3}COO^{-}}{mmol of CH_{3}COOH }]](https://tex.z-dn.net/?f=pH%3DpK_%7Ba%7D%28acetic%20acid%29%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7Bmmol%20of%20CH_%7B3%7DCOOH%20%7D%5D)
Here pH is 5.31, 
(acetic acid) is 4.74 and number of mmol of acetic acid is 10 mmol.
Plug in all the values in the above equation:
![5.31=4.74+log[\frac{mmol of CH_{3}COO^{-}}{10}]](https://tex.z-dn.net/?f=5.31%3D4.74%2Blog%5B%5Cfrac%7Bmmol%20of%20CH_%7B3%7DCOO%5E%7B-%7D%7D%7B10%7D%5D)
or, mmol of 
= 37
So 37 mmol of acetate need to add to this solution.
The part of the atom that is involved in chemical changes is A. electron. The electrons that are in the most outer shells are called valence electrons which are easily removed or shared to form bonds. Valence electrons are related to the number of valence electrons
