Number 1 plssssssssssssss

In a reaction A --> B, when C is added it reacts irreversibly with A. What can we say about C? A. C will make the reaction sp

pav-90 [236]

The answer is C. For C, it can react irreversibly with A. That means when adding C, the concentration of A will decrease. For reaction A-->B. if A decrease, then the reaction rate will decrease.

8 0

3 years ago

The structure of a solid is due to its _____.

iren2701 [21]

Compactness because the solids aka metals are bond together by metallic bond

3 0

4 years ago

Read 2 more answers

Iron(III) oxide and hydrogen react to form iron and water, like this: (s)(g)(s)(g) At a certain temperature, a chemist finds tha

stepan [7]

The question is incomplete, here is the complete question:

Iron(III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 5.4 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, iron, and water at equilibrium has the following composition:

Compound Amount

$Fe_2O_3$ 3.54 g

$H_2$ 3.63 g

Fe 2.37 g

$H_2O$ 2.13 g

Calculate the value of the equilibrium constant for this reaction. Round your answer to 2 significant digits

Answer: The value of equilibrium constant for the given reaction is $2.8\times 10^{-4}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

For hydrogen gas:

Given mass of hydrogen gas = 3.63 g

Molar mass of hydrogen gas = 2 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

$\text{Molarity of hydrogen gas}=\frac{3.63}{2\times 5.4}\\\\\text{Molarity of hydrogen gas}=0.336M$

For water:

Given mass of water = 2.13 g

Molar mass of water = 18 g/mol

Volume of solution = 5.4 L

Putting values in above equation, we get:

$\text{Molarity of water}=\frac{2.13}{18\times 5.4}\\\\\text{Molarity of water}=0.0219M$

The given chemical equation follows:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2O]^3}{[H_2]^3}$

The concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression. So, the concentration of iron and iron (III) oxide is not present in equilibrium constant expression.

Putting values in above equation, we get:

$K_c=\frac{(0.0219)^3}{(0.336)^3}\\\\K_c=2.77\times 10^{-4}$