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Stolb23 [73]
3 years ago
11

For the reaction between aqueous solutions of acetic acid (CH3COOH) and barium hydroxide, Ba(OH)2, 1. Write the balanced molecul

ar equation.
Chemistry
1 answer:
Crazy boy [7]3 years ago
8 0

Answer:

2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)

Explanation:

When acetic acid solution and barium hydroxide solution react together to give an aqueous solution of barium acetate and water

The balanced chemical reaction will be given by

2CH_3COOH(aq)+Ba(OH)_2(aq)\rightarrow Ba(CH_3COO)_2(aq)+2H_2O(l)

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Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC
Anastaziya [24]

Answer:

There are 1.287 grams of acetylene collected

Explanation:

Total gas pressure = 909 mmHg

Vapor pressure of water = 20.7 mmHg

Pressure of acetylene = 909 mmHg - 20.7 mmHg = 888.3 mmHg

1mmHg = 1 torr

22 ° C + 273.15 = 295.15 Kelvin

Ideal gas law ⇒ pV = nRT

⇒ with p = pressure of the gas in atm

⇒ with V = volume of the gas in L

⇒ with n = amount of substance of gas ( in moles)

⇒ with R = gas constant, equal to the product of the Boltzmann constant and the Avogadro constant (62.36 L * Torr *K^−1 *mol^−1)

⇒ with T = absolute temperature of the gas (in Kelvin)

888.3 torr * 1.024 L = n * 62.36 L * Torr *K^−1 *mol^−1 * 295.15 K

n = 0.04942 moles of C2H2

Mass of C2H2 = 0.04942 moles x 26.04 g/mole = 1.287 g

There are 1.287 grams of acetylene collected

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3 years ago
What is a Spontaneous charge
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In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water b
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Answer:In determining the energy of activation, why was it prudent to run the slowest trial done at room temperature in the hot water bath and the fastest trial done at room temperature in the cold water bath?

Explanation:

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2 years ago
Electron configuration for Lithium.
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What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances
belka [17]

The given question is incomplete. The complete question is:What is the relative atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances.

Isotope                    mass amu        Relative abundance

1                                77.9                     14.4

2                               81.9                     14.3

3                               85.9                      71.3

Express your answer to three significant figures and include the appropriate units.

Answer: 84.2 amu

Explanation:

Mass of isotope 1 = 77.9  

% abundance of isotope 1 = 14.4% = \frac{14.4}{100}=0.144

Mass of isotope 2 = 81.9

% abundance of isotope 2 = 14.3% = \frac{14.3}{100}=0.143

Mass of isotope 3 = 85.9

% abundance of isotope 2 = 71.3% = \frac{71.3}{100}=0.713

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(77.9\times 0.144)+(81.9\times 0.143)+(85.9\times 0.713)]

A=84.2amu

Therefore, the average atomic mass of a hypothetical element that consists isotopes in the indicated natural abundances is 84.2 amu

4 0
3 years ago
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