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11111nata11111 [884]
3 years ago
12

if the initial concentration of acetic acid is .200 M and the equilibrium concentration of H3O+ is .0019 M, calculate Ka for ace

tic acid
Chemistry
1 answer:
Blizzard [7]3 years ago
4 0

Answer:

The first one

Explanation:

- K must be calculated with the concentrations AT equilibrium, not initial concentrations.

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Solid calcium oxide reacts with sulfur trioxide gas to produce solid calcium sulfate
antiseptic1488 [7]

<u>Answer:</u> The chemical equation for the given reaction is given below.

<u>Explanation:</u>

When solid calcium oxide reacts with sulfur trioxide gas to produce solid calcium sulfate, the chemical equation follows:

CaO(s)+SO_3(g)\rightarrow CaSO_4(s)

This is the reaction when a metal oxide reacts with non-metal oxide to produce a salt.

By Stoichiometry of the reaction:

1 mole of solid calcium oxide reacts with 1 mole of sulfur trioxide gas to produce 1 mole of solid calcium sulfate.

5 0
3 years ago
A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat
loris [4]

Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

Therefore we have:

571.3 g of H2SO4 in 1329 g of the solution

Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987

Mass percentage of H2SO4 (%w/w) is 42.99 %

3 0
3 years ago
Equation of ethane to ethanoic acid
VMariaS [17]
Feb 16, 2012 ... How do you convert ethane to ethanoic acid? (with equation please). First you chlorinate it in presence of light. C2H6 + Cl2 ---hv -> C2H5Cl + ...

Preparation of ethanoic acid from ethane This free online course covers the preparation of ethanoic acid from ethane. The Contents tab displays all the lessons ...


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8 0
2 years ago
A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0
2 years ago
A mixture of sodium bicarbonate and ammonium bicarbonate is 75.9% bicarbonate by mass. what is the mass percent of sodium bicarb
Lorico [155]
Answer: 24.1%,  under below assumptions.

Justification:

The question is quite ambiguous, because one of the data is not clearly stated. It says that the mixture consists on two compounds:

- sodium bicarbonate, and
- ammonium bicarbonate

.After, it says that it is 75.9 % bicarbonate, but it does not specify which bicarbonate, it might be the sodium bicarbonate or the ammonium bicarbonate. It is apparent that you omitted that information by error.

Given that later, the question is <span>what the mass percent of sodium bicarbonate is in the mixture, it is supposed that the 75.9% content is of ammonium bicarbonate.

With that said, you can calculate the mass percent of sodium bicarbonate, because there are only two compounds and so you know that both add up the 100% of the mixture.

In formulas:

100% = %m/m sodium bicarbonate + %m/m ammonium bicarbonate = 100%

=> % m/m sodium bicarbonate = 100% - % m/s ammonium bicarbonate

=> % sodium bicarbonate = 100% - 75.9% = 24.1%

Answer: 24.1%
</span>
8 0
3 years ago
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