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3 years ago

16. When a reaction is at equilibrium

Chemistry

1 answer:

Ronch [10]3 years ago

8 0

Answer:

The forward reaction rate is equal to the reverse reaction rate

Explanation:

In many cases, a lot of people might think that equilibrium is a state when a reaction stops. This is not the case, however. Notice that equilibrium reactions are represented by double arrows indicating that two reactions take place simultaneously: both a forward and a reverse reaction.

When equilibrium is reached, the rate of a forward reaction becomes equal to the rate of a reverse reaction. The two reactions don't stop and still take place, just at equal rates. This is the reason why the concentrations of reactants and products are kept constant at equilibrium: the rates of the two reactions are balanced.

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

5 0

2 years ago

Examine the equation.

Alisiya [41]

Explanation:

The reaction expression is given as;

2H₂ $_{g}$ + O₂ $_{g}$ → 2H₂O $_{l}$

From the balance reaction expression:

2 mole of hydrogen gas combines with 1 mole of oxygen gas on the reactant side;

This produces 2 mole of water on the product side of the expression.

The product is in liquid form.

This reaction is a synthesis reaction because a single product is formed from two reactants.

5 0

2 years ago

Complete the charge balance equation for an aqueous solution of h2co3 that ionizes to hco−3 and co2−3.

Zielflug [23.3K]

The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]

<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.

The equation for aqueous solution of H₂CO₃ is

H₂CO₃ → H₂O + CO₂

The charge balance equation is

[HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]

Thus from the above conclusion we can say that The charge balance equation for an aqueous solution of H₂CO₃ that ionizes to HCO₃⁻ and CO₃⁻² is [HCO₃⁻] = 2[CO₃⁻²] + [H⁺] + [OH⁻]

Learn more about the Balanced Chemical equation here: brainly.com/question/26694427
#SPJ4

8 0

1 year ago

Consider the formation of hcn by the reaction of nacn (sodium cyanide) with an acid such as h2so4 (sulfuric acid): 2nacn(s)+h2so

Artemon [7]

2NaCN(s) + H₂SO₄(aq) --> Na₂SO₄(aq) + 2HCN(g)

The molar ratio between NaCN : HCN is 2:2 or 1:1

Mass of HCN = 16.7 g

Molar mass of HCN = 1 + 12 + 14 = 27 g/mol

Molar mass of NaCN = 49 g/mol

Therefore, the mass of NaCN is

16.7 g of HCN x 49 g/mol of NaCN / 27 g/mol of HCN = 30.3 grams of NaCN