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:
True
Explanation:
Depends of the size and comparison
Answer:
<u>136.67 g of Na3PO4 i</u>s required to create 100 gram of NaOH.
Explanation:
The balanced equation:

1 mole Na3PO4 = 164 g/mole (Molar mass)
1 mole NaOH = 40 g/mole (Molar mass)
Now,
1 mole of Na3PO4 produce = 3 mole of NaOH
164 g/mol of Na3PO4 produce = 3(40) g/mol of NaOH
or
120 g/mol of NaOH is produced from = 164 g/mol of Na3PO4
1 g/mol of NaOH is produced from =

100 grams of NaOH is produced from =
gram of Na3PO4
calculate,
= 136.67 g
Answer: It's the circulatory system
Explanation: