Who is credited with developing a method that led to the determination of standard relative atomic masses

Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem

belka [17]

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 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound Amount

Fe₂O₃ 3.95 g

H₂ 4.77 g

Fe 4.38 g

H₂O 2.00 g

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

<u>Answer:</u> The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

<u>Explanation:</u>

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

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

<u>For hydrogen gas:</u>

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

<u>For water:</u>

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

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

The expression of equilibrium constant for above equation follows:

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

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

6 0

4 years ago

What is the equilibrium condition in which a gas holds all the water vapor molecules that it can?

faust18 [17]

The equilibrium state of a gas when it can hold all the molecules of water vapor is known as saturation.

<h3>Which three equilibrium conditions apply?</h3>

A solid substance subjected to three forces with divergent lines of action is in equilibrium if all three of the following three circumstances hold true:

Coplanar action lines are present (in the same plane)
The paths of action have come together (they cross at the same point)
These forces add up to a vector with a total of zero.

<h3>What are equilibrium's first and second conditions?</h3>

An object must satisfy two requirements of equilibrium in order to maintain static equilibrium.

First, there must be no net force acting on the item.
Second, there must be no net torque pulling on the object.

learn more aboute quilibrium condition here

<u>brainly.com/question/517289</u>

#SPJ4

7 0

2 years ago

Question 22

ruslelena [56]

Answer:

O2, oxygen.

Explanation:

Hello.

In this case, for the undergoing chemical reaction, we need to compute the moles of CO2 yielded by 85 g of CH4 (molar mass = 16 g/mol) and by 320 g of O2 (molar mass 32 g/mol) via the following mole-mass relationships:

$n_{CO_2}^{by\ CH_4}=85gCH_4*\frac{1molCH_4}{16gCH_4} *\frac{1molCO_2}{1molCH_4} =5.3molCO_2\\\\n_{CO_2}^{by\ O_2}=320gO_2*\frac{1molO_2}{32gO_2} *\frac{1molCO_2}{2molO_2} =5molCO_2$

Considering the 1:2:1 among CH4, O2 and CO2. Therefore, since 320 g of O2 yield the smallest amount of CO2 we infer that the limiting reactant is O2.

Best regards.

5 0

3 years ago

Part B

mafiozo [28]

Answer:

The Changes I see is the Hot glass have less air bubbles than the Cold Glass. The Cold Glass has more air bubbles than the Hot Glass.

I hope this helps. I tried my best!! Good Luck!!

3 0

3 years ago

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During the bromination of methane, the free radical CH3• is generated. A possible terminating step of this reaction is the forma

aalyn [17]

The answer is (CH₃)₂CH-CH(CH₃)₂.

That is the free radical generated in bromination of propane is (CH₃)₂CH° and CH₃CH₂CH₂° , but due to more stability of secondary free radical , (CH₃)₂CH° is more stable than that of CH₃CH₂CH₂°, so reaction proceeds via (CH₃)₂CH° radical. and than at termination the the product formed will be (CH₃)₂CH-CH(CH₃)₂.

3 0

3 years ago

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