All categories

3 years ago

Calculate the weight of Fe3O4 in 100.0g Fe2O3.

Chemistry

1 answer:

avanturin [10]3 years ago

4 0

Answer:

96.66 g of Fe₃O₄

Explanation:

In order to calculate the weight of Fe₃O₄ in 100.0 g of Fe₂O₃ we need elements in common between both substances, for instance, the mass of Fe in each one. We will use the molar mass of Fe = 55.84g/mol The conversion factors we need are:

159.69 g of Fe₂O₃ contain 2 × 55.84 = 111.68 g of Fe

231.54 g of Fe₃O₄ contain 3 × 55.84 = 167.52 g of Fe

Then, we can use proportions:

$100.0gFe_{2}O_{3}.\frac{111.68gFe}{159.69gFe_{2}O_{3}} .\frac{231.54gFe_{3}O_{4}}{167.52gFe} =96.66gFe_{3}O_{4}$

You might be interested in

Which of these limiting factors is MOST likely to affect low-growing plants in a tropical forest? Group of answer choices air wa

Georgia [21]

Answer:

air, water, sunlight

Explanation:

A limiting factor may be defined as a environmental condition or a resource that can limit the plant growth as well as distribution or the abundance of an organism or its population within the ecosystem. The ability of any plant species to grow and spread throughout any geographic area is the direct result of the adaption to its biotic and abiotic components of that region.

Some of the factor that affect the plant growth are : sunlight, air, proper temperature, moisture, nutrients, wind,etc.

The lack of the any one of the above essential component will determine the health of the plant.

8 0

4 years ago

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

6 0

3 years ago

A radioactive substance of mass 768g has a half life of 3 years. After how many years does it leave only 6g undecayed?

vagabundo [1.1K]

Answer:

The answer is 21 years.

Explanation:

7×3

8 0

3 years ago

Read 2 more answers

What would happen to the rate of a reaction with rate law rate = k [NO]^2[H2] if

Ede4ka [16]

The rate of a reaction would be one-fourth.

<h3>Further explanation</h3>

Given

Rate law-r₁ = k [NO]²[H2]

Required

The rate of a reaction

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

Can be formulated:

Reaction: aA ---> bB

$\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}$