Answer:
k is 3,18*10⁻² s⁻¹ at 75°C
Explanation:
following Arrhenius equation:
k= k₀*e^(-Ea/RT)
where k= rate constant , k₀= frequency factor , Ea= activation energy , R= universal gas constant T=absolute temperature
then for T₁=25°C =298 K
k₁= k₀*e^(-Ea/RT₁)
and for T₁=75°C = 348 K
k₂= k₀*e^(-Ea/RT₂)
dividing both equations
k₂/k₁= e^(-Ea/RT₂+Ea/RT₁ )
k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )]
replacing values
k₂= k₁*e^[-Ea/R*(1/T₂-1/T₁ )] = 4,7*10⁻³ s⁻¹ *e^[-33.6*1000 J/mol /8.314 J/molK*(1/ 348 K -1/298 K )] = 3,18*10⁻² s⁻¹
thus k is 3,18*10⁻² s⁻¹ at 75°C
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So, if 1 mole occupies 22.4 L, the imediate conclusion is that a bigger number of moles will occupy more than 22.4 L, and a smaller number of moles will occupy less than 22.4 L. In your case, 3 moles of gas will occupy 3 times more volume than 1 mole of gas
no need to show work i am a teacher at a college
Answer:
It's Al 3+ + 3e - → Al
Explanation:
Because three negatively charged electrons are needed to balance the three positive charges on the aluminum ion
Answer:
Explanation:
<u>The alkaline solution: </u>
40 ml of 0.1 M KOH
<u>The acid solution: </u>
30 ml of 0.1 M HCl
The neutralization reaction:
For 1 mol of HCl, 1 mol of KOH is consumed. If 0.003 mol of HCl are added, 0.003 mol of KOH reacted.
After titration:
Concentration (don't forget to add the volumes)
Calcualtion of the pOH:
![pOH=-log([OH^-])=-log(0.0143)=1.84](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.0143%29%3D1.84)
For the pH:
