The closer the atoms are the more the nitrogen will be come bigger or the oppisite
PH = −log [H+] = − log [5.4 × 10−3] ≈ 2.27 or 2.3.
or basically 2
A sorry if it’s wrong good luck
Answer:
Explanation:
We shall apply Arrhenius equation which is given below .
![ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ]](https://tex.z-dn.net/?f=ln%5Cfrac%7Bk_2%7D%7Bk_1%7D%20%3D%20%5Cfrac%7BE_a%7D%7BR%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D)
K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .
Putting the given values
![ln\frac{5.9}{1} = \frac{39500}{8.3} [\frac{1}{298} -\frac{1}{T_2} ]](https://tex.z-dn.net/?f=ln%5Cfrac%7B5.9%7D%7B1%7D%20%3D%20%5Cfrac%7B39500%7D%7B8.3%7D%20%5B%5Cfrac%7B1%7D%7B298%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D)
![.000373= [\frac{1}{298} -\frac{1}{T_2} ]](https://tex.z-dn.net/?f=.000373%3D%20%5B%5Cfrac%7B1%7D%7B298%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D)
T₂ = 335.27 K
= 62.27 °C
Answer:
V CH3OH = 262.5 mL
Explanation:
- %v/v = (v compound / v sln)×100
∴ %v/v = 21.0 %
∴ v sln = 1.25 L
⇒ v compound = ?
⇒ 21.0 % = ( v compound / 1.25)×100
⇒ 0.21 = v compound / 1.25 L
⇒ (0.21)*(1.25 L) = v compound
⇒ v compound = 0.2625 L = 262.5 mL