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.
<u>Answer:</u> The empirical formula for the given compound is 
<u>Explanation:</u>
We are given:
Percentage of H = 5.80 %
Percentage of O = 23.02 %
Percentage of N = 20.16 %
Percentage of Cl = 51.02 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of H = 5.80 g
Mass of O = 23.02 g
Mass of N = 20.16 g
Mass of Cl = 51.02 g
To formulate the empirical formula, we need to follow some steps:
- <u>Step 1:</u> Converting the given masses into moles.
Moles of Hydrogen = 
Moles of Oxygen = 
Moles of Nitrogen = 
Moles of Chlorine = 
- <u>Step 2:</u> Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.44 moles.
For Hydrogen = 
For Oxygen = 
For Nitrogen = 
For Chlorine = 
- <u>Step 3:</u> Taking the mole ratio as their subscripts.
The ratio of H : O : N : Cl = 4 : 1 : 1 : 1
Hence, the empirical formula for the given compound is 
1)
HI(aq) → H⁺(aq) + I⁻(aq)
So this is an Arrhenius acid because it releases H⁺.
2)
LiOH(s) → Li⁺ + OH⁻
So this is an Arrhenius base because it releases OH⁻