<span>If a neutral atom becomes negatively charged, it has undergone reduction.
Reduction is the process through which a neutral atom gain an electron (thus reducing its oxidation number) and turns into a negative ion (also known as : anion)</span>
Answer:
71.92 kPa
Explanation:
Using the combined gas law equation;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (kPa)
P2 = final pressure (kPa)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
T1 = 50°C = 50 + 273 = 323K
V1 = 105L
T2 = -25°C = -25 + 273 = 248K
P2 = 105.4 kPa
P1 = ?
V2 = 55.0 L
Using P1V1/T1 = P2V2/T2
P1 × 105/323 = 105.4 × 55/248
105P1/323 = 5797/248
0.325P1 = 23.375
P1 = 23.375 ÷ 0.325
P1 = 71.92 kPa
When ethanoic acid react with methanol in the presence of H⁺ or heat it gives methyl ethanoate and water.
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now write the chemical equation
CH₃COOH + CH₃OH ⇄ CH₃COOCH₃ + H₂O
Here
CH₃COOH is Ethanoic acid (Acetic acid)
CH₃OH is Methanol (Methyl Alcohol)
CH₃COOCH₃ is Methyl ethanoate (Methyl Acetate)
H₂O is water
Thus from the above conclusion we can say that When ethanoic acid react with methanol in the presence of H⁺ or heat it gives methyl ethanoate and water.
Learn more about the Balanced Chemical Equation here: brainly.com/question/26694427
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Answer:
k = -0.0525 s⁻¹
Explanation:
The equaiton for a first order reaction is stated below:
ln[A]=−kt+ln[A]₀.
[A] = 5.50 x 10⁻³ M
[A]₀ = 7.60 x 10⁻² M
t = 85.0 - 35.0 = 50.0 s
The rate constant is represented by k and can be calculated substituting the values given above:
k = (ln[A]₀ - ln[A])/t
k = (ln5.50 x 10⁻³ M - ln7.60 x 10⁻² M)/50.0s
k = -0.0525 s⁻¹
Answer:
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2
Explanation:
Hello,
In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[Base]}{[Acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%20%29)
We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:
a. ![\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.497M%7D%7B0.365M%7D%3D1.36)
b. ![\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.217M%7D%7B0.521M%7D%3D0.417)
c. ![\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.713M%7D%7B0.821M%7D%3D0.868)
d. ![\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.116M%7D%7B0.121M%7D%3D0.959)
Therefore, the d. solution has the best buffering capacity.
Regards.