Answer:
i dk how ur school works, but you'll most likely have to take that science again to get the credits u need to graduate
Explanation:
Answer : The activation energy of the reaction is, 
Solution :
The relation between the rate constant the activation energy is,
![\log \frac{K_2}{K_1}=\frac{Ea}{2.303\times R}\times [\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7BK_2%7D%7BK_1%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial rate constant = 
= final rate constant = 
= initial temperature = 
= final temperature = 
R = gas constant = 8.314 kJ/moleK
Ea = activation energy
Now put all the given values in the above formula, we get the activation energy.
![\log \frac{8.75\times 10^{-3}L/mole\text{ s}}{4.55\times 10^{-5}L/mole\text{ s}}=\frac{Ea}{2.303\times (8.314kJ/moleK)}\times [\frac{1}{468K}-\frac{1}{531K}]](https://tex.z-dn.net/?f=%5Clog%20%5Cfrac%7B8.75%5Ctimes%2010%5E%7B-3%7DL%2Fmole%5Ctext%7B%20s%7D%7D%7B4.55%5Ctimes%2010%5E%7B-5%7DL%2Fmole%5Ctext%7B%20s%7D%7D%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20%288.314kJ%2FmoleK%29%7D%5Ctimes%20%5B%5Cfrac%7B1%7D%7B468K%7D-%5Cfrac%7B1%7D%7B531K%7D%5D)
Therefore, the activation energy of the reaction is,
Answer:
79.8g/dm³
Explanation:
As you can see, the solution in the problem contains 0.5 moles of copper sulfate per dm³. To solve this question we must convert these moles to grams using its molar mass (Molar mass CuSO4 = 159.609g/mol) as follows:
0.5mol CuSO4/dm³ * (159.609g/mol) =
<h3>79.8g/dm³</h3>
Answer:
atoms are everything and they are use to create anything and everything
Answer:
Pp O2 = 82.944 KPa
Explanation:
heliox tank:
∴ %wt He = 32%
∴ %wt O2 = 68%
∴ Pt = 395 KPa
⇒ Pp O2 = ?
assuming a mix of ideal gases at the temperature and volumen of the mix:
∴ Pi = RTni/V
∴ Pt = RTnt/V
⇒ Pi/Pt = ni/nt = Xi
⇒ Pi = (Xi)*(Pt)
∴ Xi: molar fraction (ni/nt)
⇒ 0.68 = mass O2/mass mix
assuming mass mix = 100 g
⇒ mass O2 = 68 g
∴ molar mass O2 = 32 g/mol
⇒ moles O2 = (68 g)(mol/32 g) = 2.125 mol O2
⇒ mass He = 32 g
∴ molar mass He = 4.0026 g/mol
⇒ moles He = (32 g)(mol/4.0026 g) = 7.995 mol He
⇒ nt = nO2 + nHe = 2.125 mol + 7.995 mol = 10.12 moles
molar fraction O2:
⇒ X O2 = nO2/nt = (2.125 mol/10.12 mol) = 0.2099
⇒ Pp O2 = (X O2)(Pt)
⇒ Pp O2 = (0.2099)(395 KPa)
⇒ Pp O2 = 82.944 KPa
