Molar mass of c(no3)2

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

I need the answer for the first question 1.

sp2606 [1]

metal salt acid hydroyon and why cuz I guessed and I haven't t learned this yet and I need points

4 0

3 years ago

What is the pH when the hydrogen ion concentration is 1 x 10-3 M?

Fynjy0 [20]

Answer:

e. 3

Explanation:

In order to solve this problem we need to keep in mind the definition of pH:

pH = -log[H⁺]

As stated by the problem, the hydrogen ion concentration, [H⁺], is 1x10⁻³ M.

As all required information is available, we now can <u>calculate the pH</u>:

pH = -log(1x10⁻³ M)

pH = 3

The correct option is thus e.

6 0

2 years ago

Commercial concentrated aqueous ammonia is 28% nh3 by mass and has a density of 0.90 g/ml. what is the concentration of ammonia

patriot [66]

Answer:- 14.9 M

Solution:- Given commercial sample of ammonia is 28% by mass. Let's say we have 100 grams of the sample. Then mass of ammonia would be 28 grams.

Density of the solution is given as 0.90 grams per mL.

From the mass and density we could calculate the volume of the solution as:

$100g(\frac{1mL}{0.90g})$

= 111 mL

Let's convert the volume from mL to L as molarity is moles of solute per liter of solution.

$111mL(\frac{1L}{1000mL})$

= 0.111 L

Now, we convert grams of ammonia to moles on dividing the grams by molar mass. Molar mass of ammonia is 17 gram per mole.

$28g(\frac{1mole}{17g})$

= 1.65 mole

To calculate the molarity we divide the moles of ammonia by the liters of solution:

$molarity=\frac{1.65mole}{0.111L}$

= 14.9 M

So, the molarity of the given commercial sample of ammonia is 14.9 M.

5 0

3 years ago

Use the periodic table to determine the electron configuration for iodine (i). express your answer in condensed form.

Dovator [93]

Iodine electron configuration is:

1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 5S^2 4d^10 5P^5
when Krypton is the noble gas in the row above iodine in the periodic table,
we can change 1S^2 2S^2 2P^6 3S^2 3P^6 4S^2 3d^10 4P^6 by the symbol
[Kr] of Krypton.

So we can write the electron configuration of Iodine:
[Kr] 5S^2 4d^10 5P^5

8 0

2 years ago

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