Answer:
The expected ratio of half-lives for a reaction will be 5:1.
Explanation:
Integrated rate law for zero order kinetics is given as:
![k=\frac{1}{t}([A_o]-[A])](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%28%5BA_o%5D-%5BA%5D%29)
= initial concentration
[A]=concentration at time t
k = rate constant
if, ![[A]=\frac{1}{2}[A_o]](https://tex.z-dn.net/?f=%5BA%5D%3D%5Cfrac%7B1%7D%7B2%7D%5BA_o%5D)
, the equation (1) becomes:
![t_{\frac{1}{2}}=\frac{[A_o]}{2k}](https://tex.z-dn.net/?f=t_%7B%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cfrac%7B%5BA_o%5D%7D%7B2k%7D)
Half life when concentration was 0.05 M=
Half life when concentration was 0.01 M=
Ratio of half-lives will be:
![\frac{t_{\frac{1}{2}}}{t_{\frac{1}{2}}'}=\frac{\frac{[0.05 M]}{2k}}{\frac{[0.01 M]}{2k}}=\frac{5}{1}](https://tex.z-dn.net/?f=%5Cfrac%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7Bt_%7B%5Cfrac%7B1%7D%7B2%7D%7D%27%7D%3D%5Cfrac%7B%5Cfrac%7B%5B0.05%20M%5D%7D%7B2k%7D%7D%7B%5Cfrac%7B%5B0.01%20M%5D%7D%7B2k%7D%7D%3D%5Cfrac%7B5%7D%7B1%7D)
The expected ratio of half-lives for a reaction will be 5:1.
Answer:b
Explanation: it’s in the picture just go counterclockwise
Explanation:
<h3>H2SO4 + 2 NaOH = Na2SO4 + 2H20, </h3>
<h3>2 in sodium hydroxide </h3><h3>And 2 in water </h3>
Answer:
7.28 moles Ag°
Explanation:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Given 7.28 moles 7.28 moles
To determine limiting reactant, divide the mole values by the respective coefficient of balanced equation. The resulting smallest value is the limiting reactant. Note: this is a short cut method for determining limiting reactant only. Once the limiting reactant is determined one must use the given mole values of the limiting reactant to solve problem. That is ...
Limiting reactant determination:
Cu° + 2 AgNO₃ => Cu(NO₃)₂ + 2Ag°
Cu: 7.28 / 1 = 7.28
AgNO₃ : 7.28 / 2 = 3.64 => Limiting Reactant is AgNO₃
Solving Problem depends on AgNO₃; Cu will be in excess.
Since coefficients of AgNO₃ & Ag° are equal, then the moles AgNO₃ used equals moles Ag° produced and is therefore 7.28 moles Ag°.