What does silver 505.

Consider the formation of ammonia in two experiments. (a) to a 1.00-l container at 727°c, 1.30 mol of n2 and 1.65 mol of h2 are

emmasim [6.3K]

When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant.

Let's do the ICE analysis

2 NH₃ ⇄ N₂ + 3 H₂
I 0 1.3 1.65
C +2x -x -3x
-------------------------------------
E 0.1 ? ?

The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 mol
Equilibrium N₂ = 1..3 - 0.05 = 1.25 mol

For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is

$K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}$

Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.

8 0

3 years ago

A 0.229-g sample of an unknown monoprotic acid is titrated with 0.112 m naoh. the resulting titration curve is shown here. part

Kazeer [188]

It would be 0.341 because if you add 0.229 and 0.112 it will be 0.341

4 0

4 years ago

Why is cl2 a subcript

vazorg [7]

Answer:

In Cl $_2$ , the 2 is a subscript because it indicates there are 2 of the same elements. The Lewis structure would display it as Cl-Cl.

On the other hand, a superscript would indicate a specific charge.

All subscripts show the amount of the specific element there is.

An example would be O $_2\\$ or N $_2$ , they both show that there are 2 of the same elements.

If the subscript is outside a parenthesis such as $Ca_3(PO_4)_2$ it indicates there are 2 $PO_4$ molecules.

5 0

3 years ago

I need HELP!!! Do the benefits of nuclear power outweigh the risks?

Georgia [21]

While there is no such thing as 100 percent safe, having nuclear energy is much safer than you think. It's thousands of times safer than conventional coal and other fossil-fuel-derived energy, not to mention the specter of environmental disaster from continued use of carbon-based energy sources.

3 0

3 years ago

g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron

Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O Oxidation

BrO₃⁻ → Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ Reduction

In order to balance the main reaction and balance the electrons we multiply (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0

4 years ago