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mr Goodwill [35]
2 years ago
13

A sample of 79.3 g of hydrogen bromide is reacted with 80.0 g of barium hydroxide to produce barium bromide and water. Using the

balanced equation below, predict which is the limiting reactant and the maximum amount in moles of barium bromide that can be produced
2HBr + Ba(OH)2 → BaBr2 + 2H2O

A. Hydrogen bromide, 0.490 Moles
B. Hydrogen bromide, 0.565 moles
C. Barium Hydroxide, 0.467 Moles
D. Barium Hydroxide, 0.624 moles
Chemistry
2 answers:
Ad libitum [116K]2 years ago
8 0
A. Hydrogen bromide 0.490
icang [17]2 years ago
6 0
A Hydrogen bromide, 0.490 moles ???
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<u>Answer:</u> For the given equation, only iron has the value of \Delta H_f equal to 0 kJ.

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Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

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\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

For the given chemical reaction:

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The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(Fe(s))})+(3\times \Delta H^o_f_{(CO_2(g))})]-[(3\times \Delta H^o_f_{(CO(g))})+(2\times \Delta H^o_f_{(Fe_2O_3(s))})]

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Catalase is an enzyme that speeds up the breakdown of hydrogen peroxide in cells. During this reaction oxygen is given off. A sc
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Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by estima
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Answer:

See explanation below

Explanation:

To do this, we will use the following expression to calculate the [H⁺]:

[H⁺] = Kw / [OH⁻]

[H⁺] is the same as [H₃O⁺]. So we have the [OH⁻] so, let's replace every value into the above expression to calculate the hydronium concentration and say if it's acidic, basic or neutral. This can be known because if the [H⁺] > 1x10⁻⁷ M the solution is acidic. If it's [H⁺] < 1x10⁻⁷ M the solution is basic, and if it's [H⁺] = 1x10⁻⁷ M the solution is neutral.

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b) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁹ = 1.11x10⁻⁶ M. Acidic.

c) [H⁺] = 1x10⁻¹⁴ / 8x10⁻¹⁰ = 1.25x10⁻⁵ M. Acidic.

d) [H⁺] = 1x10⁻¹⁴ / 7x10⁻¹³ = 0.0143 M. Acidic.

e) [H⁺] = 1x10⁻¹⁴ / 2x10⁻² = 5x10⁻¹³ M. Basic.

f) [H⁺] = 1x10⁻¹⁴ / 9x10⁻⁴ = 1.11x10⁻¹¹ M. Basic.

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Part B.

In this part, we'll use the following expression and replace the given values:

[OH⁻] = Kw / [H⁺]

Replacing the values:

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