The temperature of a substance is related to the speed of the substance's particles. So, when the temperature of a substance changes,the speed of the particles also changes. So B.
<u>Answer:</u> The time taken by the reaction is 84.5 seconds
<u>Explanation:</u>
The equation used to calculate half life for first order kinetics:
where,
= half-life of the reaction = 9.0 s
k = rate constant = ?
Putting values in above equation, we get:
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
......(1)
where,
k = rate constant = 
t = time taken for decay process = 50.7 sec
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount of the reactant = ?
[A] = amount left after decay process = 0.0741 M
Putting values in equation 1, we get:
![0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}](https://tex.z-dn.net/?f=0.077%3D%5Cfrac%7B2.303%7D%7B50.7%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B0.0741%7D)
![[A_o]=3.67M](https://tex.z-dn.net/?f=%5BA_o%5D%3D3.67M)
Now, calculating the time taken by using equation 1:
![[A]=0.0055M](https://tex.z-dn.net/?f=%5BA%5D%3D0.0055M)
Putting values in equation 1, we get:
Hence, the time taken by the reaction is 84.5 seconds
Answer:
a) pH = 4.213
b) % dis = 2 %
Explanation:
Ch3COONa → CH3COO- + Na+
CH3COOH ↔ CH3COO- + H3O+
∴ Ka = 1.8 E-5 = ([ CH3COO- ] * [ H3O+ ]) / [ CH3COOH ]
mass balance:
⇒ <em>C</em> CH3COOH + <em>C</em> CH3COONa = [ CH3COOH ] + [ CH3COO- ]
<em>∴ C </em>CH3COOH = 3.40 mM = 3.4 mmol/mL * ( mol/1000mmol)*(1000mL/L)
∴ <em>C</em> CH3COONa = 1.00 M = 1.00 mol/L = 1.00 mmol/mL
⇒ [ CH3COOH ] = 4.4 - [ CH3COO- ]
charge balance:
⇒ [ H3O+ ] + [ Na+ ] = [ CH3COO- ] + [ OH- ]....is negligible [ OH-], comes from water
⇒ [ CH3COO- ] = [ H3O+ ] + 1.00
⇒ Ka = (( [ H3O+ ] + 1 )* [ H3O+ ]) / ( 3.4 - [ H3O+])) = 1.8 E-5
⇒ [ H3O+ ]² + [ H3O+ ] = 6.12 E-5 - 1.8 E-5 [ H3O+ ]
⇒ [ H3O+ ]² + [ H3O+ ] - 6.12 E-5 = 0
⇒ [ H3O+ ] = 6.12 E-5 M
⇒ pH = - Log [ H3O+ ] = 4.213
b) (% dis)* mol acid = <em>C</em> CH3COOH = 3.4
∴ mol CH3COOH = 500*3.4 = 1700 mmol = 1.7 mol
⇒ % dis = 3.4 / 1.7 = 2 %
1,38×10²² = 0,138×10²³
0,138×10²³ ----- 1,5g
6,02×10²³ ------ X
X = (1.5×6,02×10²³)/0,138×10²³
X = 65,435 g/mol
It's ZINC (Zn)
:•)
