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dedylja [7]
3 years ago
12

When elemental iron corrodes it combines with oxygen in the air to ultimately form red brown iron (III) oxide which we call rust

. (a) If a shiny ironnail with an initial mass of 23.2 g is weighed after being coated in a layer of rust, would you expect the mass to have increased, decreased, or remained the same? Explain. (b) If the mass of the iron nall increases to 24.1 g what mass of oxygen combined with the iron?
Chemistry
1 answer:
olga2289 [7]3 years ago
8 0

Answer:

a) Increase

b) Mass of oxygen = 0.9 g

Explanation:

The chemical reaction depicting the formation of rust is:

4Fe(s) + 3O_{2}(g)\rightarrow 2Fe_{2}O_{3}(s)

where rust is Fe2O3 i.e. iron(II) oxide or ferric oxide

a) When a piece of Fe is exposed to air or oxygen the surface gets coated with Fe2O3. This will increase the amount of matter and hence the mass.

b) Initial mass of iron , Fe= 23.2 g

Mass of iron coated with oxide, Fe2O3 = 24.1 g

Therefore mass of oxygen combined= 24.1 - 23.2 = 0.9 \ g

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What is percentage composition?
blondinia [14]

Answer:

The percent composition of a component in a compound is the percent of the total mass of the compound that is due to that component. To calculate the percent composition of a component in a compound: Find the molar mass of the compound by adding up the masses of each atom.

7 0
3 years ago
How is molarity measured?
Tanzania [10]
The answer to this question will be C
6 0
3 years ago
calculate the number of moles of sulfuric acid that is contained in 250 mL of 8.500 M sulfuric acid solution
san4es73 [151]

Answer : The moles of H_2SO_4 are, 2.125 mole.

Explanation : Given,

Molarity of H_2SO_4 = 8.500 M

Volume of solution = 250 mL  = 0.250 L    (1 L = 1000 mL)

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Moles of }H_2SO_4}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

8.500M=\frac{\text{Moles of }H_2SO_4}{0.250L}

\text{Moles of }H_2SO_4=2.125mol

Therefore, the moles of H_2SO_4 are, 2.125 mole.

5 0
3 years ago
What is the mole fraction of methanol in a solution that contains 6.0 mol of methanol and 3.0 mol of water? The formula for meth
Ivenika [448]
The mole fraction of a product is the number of moles of the product divided by the total number of moles of the solution.

Here moles of methanol = 6.0 moles

Moles of solution = 6.0 moles of methanol + 3.0 moles of water = 9.0 moles of solution

Mole fraction of methanol = 6.0 / 9.0 = 0.67

Answer: 0.67  
4 0
3 years ago
Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det
mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0
3 years ago
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