Answer: The 234.74 grams of sample should be ordered.
Explanation:
Let the gram of 114 Ag to ordered be 
The amount required for the beginning of experiment = 0.0575 g
Time requires to ship the sample = 4.2hour = 252 min(1 hr = 60 min)
Half life of the sample =
= 21 min
![\log[N]=\log[N_o]-\frac{\lambda t}{2.303}](https://tex.z-dn.net/?f=%5Clog%5BN%5D%3D%5Clog%5BN_o%5D-%5Cfrac%7B%5Clambda%20t%7D%7B2.303%7D)
![\log[0.0575 g]=\log[N_o]-\frac{0.033 min^{-1}\times 252 min}{2.303}](https://tex.z-dn.net/?f=%5Clog%5B0.0575%20g%5D%3D%5Clog%5BN_o%5D-%5Cfrac%7B0.033%20min%5E%7B-1%7D%5Ctimes%20252%20min%7D%7B2.303%7D)
The 234.74 grams of sample should be ordered.
4NH3+5O2 <=>4NO + 6H2O
Using the definition of Kp, we have
Kp=(Pno^4*Ph2o^6)/(Pnh3^4*Po2^5)
where Pno=partial pressure of NO, etc.
The numerical value for a given temperature can be evaluated when the actual partial pressures are known.
Maybe this example could help you to understand this problem.
https://image.slidesharecdn.com/121howmanyatoms-091201144624-phpapp02/95/12-1-how-many-atoms-17-728....
Answer:
ΔS° = - 47.2 J/mol.K
Explanation:
ΔS°= 4(S°mH3PO4) - 6(S°mH2O) - S°mP4O10
∴ S°mH2O(l) = 69.9 J/mol.K
∴ S°mP4O10 = 231 J/mol.K
∴ S°mH3PO4 = 150.8 J/mol.K
⇒ ΔS° = 4*(150.8) - 6*(69.9) - 231
⇒ ΔS° = - 47.2 J/mol.K
