When nitric oxide reacts v hydrogen gas to kind ammonia gas and also water vapor, the equation because that the complying with reaction follows: Reactants the the reaction space nitric oxide and hydrogen gas and also all are current in gaseous phase.

You are watching: Nitric oxide gas reacts with hydrogen gas to form ammonia gas and water vapor

Products of the reaction are ammonia and also water vapor and all are present in gaseous phase.

Cyclohexane, C6H12, experience a molecular rearrangement in the visibility of AlCl3 to kind methylcyclopentane, CH3C5H9, follow

aramuseum.org: =0.03441\\ M\" alt=\"=0.03441\\ M\" align=\"absmiddle\" class=\"latex-formula\"> =0.24059\\ M\" alt=\"=0.24059\\ M\" align=\"absmiddle\" class=\"latex-formula\">

Explanation:

The provided equilibrium reaction and also the equilibrium concentrations are shown listed below as:- The Kc of one equilibrium reaction procedures relative amounts of the products and the reactants existing during the equilibrium.

It is the ratio of the concentration of the products and also the reactants each raised to your stoichiometric coefficients. The concentration the the liquid and also the gaseous species does not adjust and for this reason is not composed in the expression.

The expression for the Kc is:- }\" alt=\"K_c=\\frac\" align=\"absmiddle\" class=\"latex-formula\">

Given that:- Kc = 0.143

Thus, applying the values as:-    Thus, =0.075+x=0.075-0.04059=0.03441\\ M\" alt=\"=0.075+x=0.075-0.04059=0.03441\\ M\" align=\"absmiddle\" class=\"latex-formula\"> =0.200-x=0.200-(-0.04059)=0.24059\\ M\" alt=\"=0.200-x=0.200-(-0.04059)=0.24059\\ M\" align=\"absmiddle\" class=\"latex-formula\">

4 0
1 year ago

Which requires much more energy to dissolve, ionic or covalent compounds?covalentionic
gregori <183>
Ionic requires much more energy to dissolve
7 0
1 year ago

A main team metal to be studied and found to exhibition the complying with properties: the does not occur free in nature. It loses valence
ANEK <815>

aramuseum.org:

Barium (Ba)

Explanation:

Barium is a group 2 elements.

Valence shell digital configuration = 6s^2\" alt=\"6s^2\" align=\"absmiddle\" class=\"latex-formula\">

Its valence shell has only 2 electrons and by loosing this electrons, the attains inert gas configuration. Therefore, the looses electron easily.

It is extremely reactive and also because of its high reactive nature, never existing in nature in cost-free state.

In general, reactivity increases down the group in the periodic table.

This is due to the fact that down the group atomic size increases and also electrons becomes much from the nucleus, so outer electrons space not tightly hosted by the nucleus. Therefore, valence covering electrons loose easily which subsequently increases reactivity.

So, Ba is more reactive amongst all the given elements.

3 0
8 month ago

Balance the complying with skeleton reaction and identify the oxidizing and also reducing agents: incorporate the states of every reactants and also
rusak2 <61>

aramuseum.org: 4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)\" alt=\"4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)\" align=\"absmiddle\" class=\"latex-formula\">

-) Oxidizing agent: -) to reduce agent: Explanation:

The first step is separate the reaction into the semireactions:

A. Zn(OH)_4^-^2\" alt=\"Zn~->Zn(OH)_4^-^2\" align=\"absmiddle\" class=\"latex-formula\"> ~NH_3\" alt=\"NO_3^-~->~NH_3\" align=\"absmiddle\" class=\"latex-formula\">

If we want to balance in basic tool we have to follow the rules:

1. We adjust the oxygen through 2. We adjust the H through 3. We readjust the charge through Lets balance the very first semireaction A. : Zn(OH)_4^-^2~+~2e^-\" alt=\"Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-\" align=\"absmiddle\" class=\"latex-formula\">

Now, allows balance semireaction B: ~NH_3~+~9OH^-\" alt=\"NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-\" align=\"absmiddle\" class=\"latex-formula\">

Finally, we have actually to include the two semireactions:

_________________________________________ Zn(OH)_4^-^2~+~2e^-)\" alt=\"8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)\" align=\"absmiddle\" class=\"latex-formula\"> ~NH_3~+~9OH^-)\" alt=\"2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)\" align=\"absmiddle\" class=\"latex-formula\">

_________________________________________ 8Zn(OH)_4^-^2~+~16e^-)\" alt=\"(8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)\" align=\"absmiddle\" class=\"latex-formula\"> ~2NH_3~+~18OH^-)\" alt=\"(2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)\" align=\"absmiddle\" class=\"latex-formula\">

Cancel out the species on both sides: 8Zn(OH)_4^-^2~+~2NH_3\" alt=\"8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3\" align=\"absmiddle\" class=\"latex-formula\">

Simplifying the equation : 4Zn(OH)_4^-^2~+~NH_3\" alt=\"4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3\" align=\"absmiddle\" class=\"latex-formula\">

The is oxidized therefefore is the reducing agent. The is reduced because of this is the oxidizing agent.