Answer:
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Explanation:
Hello there!
In this case, since the integrated rate law for a second-order reaction is:
![[SO_3]=\frac{[SO_3]_0}{1+kt[SO_3]_0}](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B%5BSO_3%5D_0%7D%7B1%2Bkt%5BSO_3%5D_0%7D)
Thus, we plug in the initial concentration, rate constant and elapsed time to obtain:
![[SO_3]=\frac{1.44M}{1+14.1M^{-1}s^{-1}*0.240s*1.44M}\\\\](https://tex.z-dn.net/?f=%5BSO_3%5D%3D%5Cfrac%7B1.44M%7D%7B1%2B14.1M%5E%7B-1%7Ds%5E%7B-1%7D%2A0.240s%2A1.44M%7D%5C%5C%5C%5C)
![[SO_3]=0.25M](https://tex.z-dn.net/?f=%5BSO_3%5D%3D0.25M)
Best regards!
Answer:
The answer to your question is 27 g of Al
Explanation:
Data
mass of Al = ?
moles of Al₂O₃ = 0.5
The correct formula for the product is Al₂O₃
Balanced chemical reaction
4Al + 3O₂ ⇒ 2Al₂O₃
Process
1.- Calculate the molar mass of the product
Al₂O₃ = (27 x 2) + (16 x 3)
= 54 + 48
= 102 g
2.- Convert the moles of Al₂O₃ to grams
102 g ---------------- 1 mol
x ---------------- 0.5 moles
x = (0.5 x 102) / 1
x = 51 g of Al₂O₃
3.- Use proportions to calculate the mass of Al
4(27) g of Al --------------- 2(102) g of Al₂O₃
x --------------- 51 g
x = (51 x 4(27)) / 2(102)
x = 5508 / 204
x = 27 g of Al
both conserve mass.......
Answer:
Explanation:
1) The number of moles of jojoba oil wax ester = mass/molar mass = 5.0g/618.07g/mol = 8.09*10^-3 mol
Hence we need at least the number of moles of NaOH required = 2*8.09*10^-3 = 16.18 *10^-3 mol
We have 1.0mol/1000mL solution. Hence volume of 1.0M NaOH solution = (16.18*10^-3)/(1/1000) = 16.18mL
Thus the answer is 16mL.