It would be 4.6cgL not sure tho because I didint do that good in this
Answer:
Second order
Explanation:
We could obtain the order of reaction by looking at the table very closely.
Now notice that in experiment 1 and 2, the concentration of [OH^-] was held constant while the concentration of [S8] was varied. So we have;
a situation in which the rate of reaction was tripled;
0.3/0.1 = 2.10/0.699
3^1 = 3^1
Therefore the order of reaction with respect to [S8] is 1.
For [OH^-], we have to look at experiment 2 and 3 where the concentration of [S8] was held constant;
x/0.01 = 4.19/2.10
x/0.01 = 2
x = 2 * 0.01
x = 0.02
So we have;
0.02/0.01 = 2^1
2^1 = 2^1
The order of reaction with respect to [OH^-] = 1
So we have the overall rate law as;
Rate = k[S8]^1 [OH^-] ^1
Overall order of reaction = 1 + 1 = 2
Therefore the reaction is second order.
Answer:
The reactants would appear at a higher energy state than the products.
Have a nice day!
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L
At higher temperature, and lower pressure.
