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wariber [46]
2 years ago
10

Calculate the equilibrium constant for the reaction using the balanced chemical equation and the concentrations of the substance

s at equilibrium.
Use the

appropriate significant figures in reporting the answers.

CO(g) + H2O(g) ⇌ CO2(g) + H2(g) [CO] = 0.0590 M; [H2O] = 0.00600 M;

[CO2] = 0.0410 M; [H2] = 0.0410 M

K =______________-
Chemistry
1 answer:
Oliga [24]2 years ago
3 0

Answer:

4.75 is the equilibrium constant for the reaction.

Explanation:

CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Equilibrium concentration of reactants :

[CO]=0.0590 M,[H_2O]=0.00600 M

Equilibrium concentration of products:

[CO_2]=0.0410 M,[H_2]=0.0410 M

The expression of an equilibrium constant is given by :

K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}

K_c=\frac{0.0410 M\times 0.0410 M}{0.0590M\times 0.00600 M}

K_c=4.75

4.75 is the equilibrium constant for the reaction.

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Which statement best describes the formula equation cl1(g) + 2kbr(aq) —> 2kcl(aq)+br2(i)
Zigmanuir [339]

Answer:

The chlorine gas and potassium bromide solution react to form liquid  bromine and potassium chloride solution.

Explanation:

Chemical equation:

Cl₂(g) + KBr (aq) → KCl (aq) + Br₂(l)

Balanced chemical equation:

Cl₂(g) + 2KBr (aq) → 2KCl (aq) + Br₂(l)

This equation showed that the chlorine gas and potassium bromide solution react to form liquid  bromine and potassium chloride solution.

Chlorine is more reactive than bromine it displace the bromine from potassium and form potassium chloride solution.

The given equation is balanced and completely hold the law of conservation of mass.

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Explanation:

This law was given by french chemist  Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

8 0
3 years ago
Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H
lianna [129]

Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

<u>Step 1:</u> Data given

Mass of 1.83 grams of Al

Mass of NaOH = 4.30 grams

Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

<u>Step 2:</u> The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

<u>Step 3:</u> Calculate moles of Al

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

<u>Step 4:</u> Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

<u>Step 6</u>: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

<u>Step 7</u>: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams

6 0
2 years ago
Quick electron emissions are called
Montano1993 [528]
Defined as a phenomenon of liberation of electron from the surface that is stimulated by temperature elevation, radiation, or by strong electric field.
3 0
2 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li
meriva

Answer:

The standard enthalpy change for the reaction at 25^{0}\textrm{C} is -2043.999kJ

Explanation:

Standard enthalpy change (\Delta H_{rxn}^{0}) for the given reaction is expressed as:

\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]

Where \Delta H_{f}^{0} refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

\Delta H_{rxn}^{0}=[3mol\times (-393.509kJ/mol)]+[4mol\times (-241.818kJ/mol)]-[1mol\times (-103.8kJ/mol)]-[5mol\times (0kJ/mol)]=-2043.999kJ

4 0
2 years ago
Changing _________________ can significantly alter the area that could be affected by a chemical release.
Harman [31]
The answer is D, location, because you'd have moved spots that could be altered.
6 0
3 years ago
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