Answer:
P O 2 = 5.21 atm P C O 2 = 4.79 atm
Explanation:
Hope it helps!
Potential because it hasnt released yet
The reaction mixture of problem 1 includes <span>10mL of 4.0 M acetone = 10 mL 1.0 M HCl = 10 mL 0.0050 M I2 = 20 mL H2O. if this is true then the procedure is the following:
In order to find the total volume of reaction then you need to do like this:
</span><span>V = 10 mL + 10 mL + 10 mL + 20 mL = 50 mL </span>
<span>[I2]o = (0.005 mol I2 / 1 L I2 solution) (10 mL I2 solution) / 50 mL = 0.001 M </span>
<span>To first order, the reaction rate is 0.001 M / 230 s = 4,3 e-6 M/s
Then if you want to find the rate yoe need to use the following formula:
</span><span> k [CO(CH3)2]^a [I2]^b [HCl]^c
</span>So: <span>4,3 e-6 = k (4 M * 10 mL / 50 mL)^a (1e-3 M)^b (1 M * 10 mL / 50 mL)^c
</span>
Answer:
3) ester
Explanation:
Esterification is the process in which alkanol and alkanoic acids reacts in the presence of a catalyst and heat. The product is usually an ester(alkanoate) and water.
For the reaction between ethanoic acid and 1-butanol, the product is butylethanoate and water as shown below:
CH₃COOH + C₄H₈OH → CH₃COOC₄H₈ + H₂O
<span>Answer
is: Ka for propinoic acid is 6,57·10</span>⁻⁵.<span>
Chemical reaction: C</span>₂H₅COOH(aq)
+ H₂O(l) ⇄ C₂H₅COO⁻(aq)
+ H₃O⁺(aq).<span>
n(C</span>₂H₅COOH) = 0,04 mol.<span>
V(C</span>₂H₅COOH) = 750 mL = 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,04 mol ÷ 0,75 L.<span>
c(C</span>₂H₅COOH) = 0,053 mol/L = 0,053 M.<span>
[C</span>₂H₅COO⁻]
= [H₃O⁺] = 1,84·10⁻³ M = 0,00184 M.<span>
[HCN] = 0,053 M - 0,00184 M = 0,0515 M.
Ka = [C</span>₂H₅COO⁻] · [H₃O⁺] /
[C₂H₅COOH].<span>
Ka = (0,00184 M)² / 0,0515 M.
Ka = 6,57·10</span>⁻⁵.
