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Lynna [10]
3 years ago
11

Equation for the for formation of ethanoic acid​

Chemistry
1 answer:
vladimir2022 [97]3 years ago
5 0

Answer:

2 C4H10 + 5 O2 → 4 CH3CO2H + 2 H2O.

Explanation:

Light naphtha components are readily oxidized by oxygen or even air to give peroxides, which decompose to produce acetic acid according to the chemical equation, illustrated with butane .

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Please help me I will give a brainleist to anyone who answers​
tekilochka [14]

Answer:

4

Explanation:

4 0
3 years ago
Which of the following is used for chemical symbols today? a. drawings c. letters b. icons d. numbers
devlian [24]

Answer:

Letters

Explanation:

For example, today we use the periodic table which is full of elements named with 1 or 2 letters. Like how Helium is He and Sodium is Na. Hope this helps!!!

3 0
3 years ago
. How many ethyl groups are in the molecule 3, 3-diethylpentane?
Mamont248 [21]

Answer:

2 groups

Explanation:

The molecule 3, 3-diethylpentane has 2 ethyl groups because the prefix di- means that there are 2 homogeneous substituent groups present in the molecule.

6 0
2 years ago
How many grams of water can I convert from a liquid to a gas with 6768 joules?
vredina [299]

Answer:

The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g

Explanation:

The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid

The latent heat of evaporation of water, ΔH_v ≈ 2,230 J/g

The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows

Q = m × ΔH_v

Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;

6,768 J = m × 2,230 J/g

∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g

The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.

8 0
3 years ago
A mixture of XO2 (P = 3.00 atm) and O2 (P = 1.00 atm) is placed in a container. This elementary reaction takes place at 27 °C: 2
sukhopar [10]

Answer:

a) \triangle G^{0} = 7.31 kJ/mol

b) K_{-1} = 0.0594 m^{-1} s^{-1}

Explanation:

Equation of reaction:

                                     2 XO_{2} (g) + O_{2} (g) \rightleftharpoons 2XO_{3} (g)

Initial pressure                  3              1              0

Pressure change             2P           1P             2P

Total pressure = (3-2P) + (1-P) + (2P)

Total Pressure = 3.75 atm

(3-2P) + (1-P) + (2P) = 3.75

4 - P = 3.75

P = 4 - 3.75

P = 0.25 atm

Let us calculate the pressure of each of the components of the reaction:

Pressure of XO2 = 3 - 2P = 3 - 2(0.25)

Pressure of XO2 =2.5 atm

Pressure of O2 = 1 - P = 1 -0.25

Pressure of O2 = 0.75 atm

Pressure of XO3 = 2P = 2 * 0.25

Pressure of XO3 = 0.5 atm

From the reaction, equilibrium constant can be calculated using the formula:

K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }

K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}

Standard free energy:

\triangle G^{0} = - RT ln k_{eq} \\\triangle G^{0} = -(0.008314*300* ln0.0533)\\\triangle G^{0} = 7.31 kJ/mol

b) value of k−1 at 27 °C, i.e. 300K

K_{1} = 7.8 * 10^{-2} m^{-2} s^{-1}

K_{c} = K_{p}RT\\K_{c} = 0.0533* 0.0821 * 300\\K_{c} = 1.313 m^{-1}

K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2}  }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}

6 0
3 years ago
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