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.
I think the correct answers from the choices listed above are the first, third and the last option. Ionic compounds are compounds that dissociates into ions when in aqueous solution. From the list, NH4Cl, KF and MgO are the ionic compounds. Hope this answers the question.
Answer:
Key Points. Ionic bonds are formed through the exchange of valence electrons between atoms, typically a metal and a nonmetal.
Explanation:
Since you have not included the chemical reaction I will explain you in detail.
1) To determine the limiting agent you need two things:
- the balanced chemical equation
- the amount of every reactant involved as per the chemical equation
2) The work is:
- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.
- determine the number of moles of each reactant with this formula:
number of moles = (mass in grams) / (molar mass)
- set the proportion with the two ratios (theoretical moles and actual moles)
- compare which reactant is below than the stated by the theoretical ratio.
3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:
i) Chemical equation: H₂ + O₂ → H₂O
ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O
iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂
iv) Covert 100 g of H₂ into number of moles
n = 100g / 2g/mol = 50 mol of H₂
v) Convert 100 g of O₂ to moles:
n = 100 g / 32 g/mol = 3.125 mol
vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂
vii) Compare the two ratios:
2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂
Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.
Therefore, the O₂ is the limiting reactant in this example.
Answer:
remaining still during the night.
Explanation:
