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.
Answer:
7. ok
please mark me brainlistb
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<span>The amount of dissolved salt in the liquid sample is measured and reported as salinity. The salinity is usually measured in parts per thousand (ppt). The salinity of ocean averages 35 ppt while that of the river averages 0.5 ppt or less. In other terms, the word salinity is the saltiness. </span>
Answer:
1) 0.18106 M is the molarity of the resulting solution.
2) 0.823 Molar is the molarity of the solution.
Explanation:
1) Volume of stock solution = 
Concentration of stock solution = 
Volume of stock solution after dilution = 
Concentration of stock solution after dilution = 
( dilution )

0.18106 M is the molarity of the resulting solution.
2)
Molarity of the solution is the moles of compound in 1 Liter solutions.

Mass of potassium permanganate = 13.0 g
Molar mass of potassium permangante = 158 g/mol
Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)

0.823 Molar is the molarity of the solution.