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3 years ago

CH3COOH mc015-1.jpg CH3COO– (aq) + H+(aq) What will happen to the chemical equilibrium of the solution if CH3COONa is added?

1 answer:

andrey2020 [161]3 years ago

7 0

The equilibrium will shift to the left or the backward reaction since addition of CH3COONa will add more CH3COO- ions to the solution. The formation of reactants are promoted.

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A sample of a compound that contains only the elements C, H, and N is completely burned in O₂ to produce 44.0 g of CO₂, 45.0 g o

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Explanation:

In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:

(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C

Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:

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(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H

The ratio of H to C in the compound is:

(4.994...mol H)/(0.99977... mol C) = 5 H:C

Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.

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2 years ago

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2 years ago

If 125g of KClO3 is heated, what is the total mass of the products?

Andrej [43]

Given parameters:

Mass of KClO₃ = 125g

Unknown:

Total mass of the products = ?

When KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

2KClO₃ → 2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

Find the number of moles of the given reactant:

Number of moles = $\frac{mass}{molar mass}$

molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5g/mol

So number of moles of KClO₃ = $\frac{125}{122.5}$ = 1.02moles

2. Now, using this number of moles, find the number of moles of the products using this value;

2 moles of KClO₃ produced 2 moles of KCl

1.02 moles of KClO₃ will also produce 1.02moles of KCl

2 moles of KClO₃ produced 3 moles of O₂

1.02 moles of KClO₃ will produce $\frac{1.02 x 3} {2}$ mole = 1.53 moles of O₂

3. Now find the masses of each product;

Mass = number of moles x molar mass

molar mass of KCl = 39 + 35.5 = 74.5g/mol

molar mass of O₂ = 16 x 2 = 32g/mol

Mass of KCl = 74.5 x 1.02 = 75.99g

Mass of O₂ = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

= 124.95g

This value is approximately the same as that of mass of KClO₃

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3 years ago