A scientist is at the hospital visiting a friend's young daughter, who has been

Which will not appear in the equilibrium constant expression for the reaction below?

n200080 [17]

Answer:

[C] carbon solid

Explanation:

Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.

5 0

2 years ago

Marcie is hiking in the mountains and she discovers an interesting shiny rock.As she picks it up she wonders whether it could be

pogonyaev

Answer:

Marcie pick up a rock

Explanation:

A rock is a natural occuring no living thing that has a definite structure from the physical characteristics Marcie saw in the rock she picked it tends to rock because unlike minerals rocks has definite structure.

3 0

2 years ago

With the help of balanced chemical equation explain what happens when:(1)Zinc is put in dilute hydrochloric acid (2)Zinc is put

UkoKoshka [18]

Find the attached document.

6 0

2 years ago

Read 2 more answers

A) Write the word equation for the reaction of barium nitride (Ba3N2) with potassium.

artcher [175]

Answer:

Explanation:

In order to balance it, we need to have the same number of atoms of each element on both sides of the equation. There are two atoms of nitrogen on the left, so we need to put 2 in front of K₃N. Now, we have six atoms of potassium on the right, so we need to put 6 in front of K on the left. Finally, there are three atoms of barium on the left, so we put 3 before Ba on the tight. Which means:

Ba₃N₂ + 6K = 2K₃N + 3Ba

Now, we can do the work. First, we determine the molar mass of each reactant ( from the periodic table). Molar mass of the barium is 137, potassium 39 and nitrogen 14.

Ba₃N₂₂ has molar mass of 3Ba and 2N, which means 3 • 137 + 2 • 14 = 439. That means that one mole of Ba₃N₂ weights 439 grams.

We are given grams of reactants, but in order to find the limiting and the excess reactant, we need to transfer it into moles.

We are given 66.5 grams of Ba₃N₂ and we know that 439 grams equals 1 mole. We want to know how many moles there are in 66.5 grams, so the answer is 66.5 / 439 = 0.15 moles.

Let's do the same for potassium. We are given 29 grams of K and we know that 1 mole has 39 grams. We want to know how many moles of K are there in 29 grams, so the answer is 29 / 39 = 0.74 moles.

We now know that 0.15 moles of Ba₃N₂ reacted with 0.74 moles of K. From the balanced equation we see that 1 mole of Ba₃N₂ reacts with 6 moles of K, so the ratio has to be 1:6.

Now let's find limiting and excess reactant. That means that in our reaction, there are more (or less) of one reactant then needed.

We know that we had 0.15 moles of Ba₃N₂ reacting. Let's pretend we don't know the moles of K and let's see with which amount of K should 0.15 moles of barium nitride react, if the ratio is 1:6.

0.15 moles of Ba₃N₂ : x moles of K = 1:6

x = 0.9 moles of K

So, for the completed reaction we need to have 0.9 moles of K, but we previously calculated that we had 0.74. That means that there is less K then needed, so potassium is our limiting reactant, which obviously means that Ba₃N₂ is our excess reactant.

Now, we need to find how many moles of Ba₃N₂ there needed to be for a completed reaction

x moles of Ba₃N₂ : 0.74 moles of K = 1:6

x = 0.124 moles of Ba₃N₂

So we needed to have 0.124 moles, but we had 0.15 of Ba₃N₂, which is 0.15 - 0.124 = 0.026 moles in excess.

If we want to find how many grams that is, we only multiply it with molar mass of Ba₃N₂:

0.026 • 439 = 11.4 grams

That means that only 66.5 - 11.4 = 55.1 grams of Ba₃N₂ reacted.

3 0

2 years ago

In a study of the decomposition of the compound XX via the reaction

Leya [2.2K]

Answer:

( About ) 0.03232 M

Explanation:

Based on the units for this reaction it should be a second order reaction, and hence you would apply the integrated rate law equation "1 / [X] = kt + 1 / [ $X_o$ ]"

This formula would be true for the following information -

{ $X_o$ = the initial concentration of X, k = rate constant, [ X ] = the concentration after a certain time ( which is what you need to determine ), and t = time in minutes }

________

Therefore, all we have left to do is plug in the known values. The initial concentration of X is 0.467 at a time of 0 minutes, as you can tell from the given data. This is not relevant to the time needed in the formula, as we need to calculate the concentration of X after 18 minutes ( time = 18 minutes ). And of course k, the rate constant = 1.6

1 / [X] = ( 1.6 )( 18 minutes ) + 1 / ( 0.467 ) - Now let's solve for X

1 / [X] = 28.8 + 1 / ( 0.467 ),

1 / [X] = 28.8 + 2.1413...,

1 / [X] = 31,

[X] = 1 / 31 = ( About ) 0.03232 M

Now for this last bit here you probably are wondering why 1 / 31 is not 0.03232, rather 0.032258... Well, I did approximate one of the numbers along the way ( 2.1413... ) and took the precise value into account on my own and solved a bit more accurately. So that is your solution! The concentration of X after 18 minutes is about 0.03232 M

3 0

2 years ago