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3 years ago

Which enthalpy change (î"h) is correct for a substance changing from the liquid phase to the solid phase?

1 answer:

7 0

I believe this should be the enthalpy of fusion. Enthalpy change is the amount of heat evolved or absorbed in a reaction. Enthalpy of fusion also called the heat of fusion resulting from providing heat energy or release of energy to a specific amount of a substance to change from solid state to liquid, without change in temperature.

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There are a lot of empty _____ between the particles

Furkat [3]

There are a lot of empty space between the particles

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2 years ago

The density of aluminium is 2.7 g/cm3. Find the mass in grams of a bar of aluminum measuring 2.7 cm by 1.2 cm by 14.3 cm.

Fofino [41]

Answer:125.0964g

Explanation:

Density = mass/volume

Density = 2.7 gcm-3

Volume=2.7*1.2*14.3=46.332

Mass=?

Mass= density* volume

= 2.7× 46.332=125.0964g

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3 years ago

you have accidentally broken a test tube and spilled a chemical on the bench. which of the following best describes what you sho

ANEK [815]

Tell the teacher, do NOT clean it up yourself.

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2 years ago

Which of the following is the best explanation for the fact that crushing increases the rate at which a solid dissolves in a liq

Oksana_A [137]

It’s D cause i know I know I know I know

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3 years ago

Given the following data: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39

nalin [4]

Answer:

ΔH° = -11 kj

Explanation:

Step 1: Data given

1) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ΔH = −23 kJ

2) 3 Fe2O3(s) + CO(g) → 2 Fe3O4(s) + CO2(g) ΔH = −39 kJ

3) Fe3O4(s) + CO(g) → 3 FeO(s) + CO2(g) ΔH = +18 kJ

Step 2: The balanced equation

FeO + CO → Fe + CO2

Step 3: Calculate ΔH for the reaction FeO(s) + CO(g) → Fe(s) + CO2(g).

To get this equation, we need to combine the 3 equations

We have to multiply the third equation by 2.

2Fe3O4(s) + 2CO(g) → 6 FeO(s) + 2CO2(g) ΔH = +54 kJ

This equation we add to the second equation

3Fe2O3(s) + CO(g) + 2Fe3O4(s) + 2CO(g) → 2Fe3O4(s) + CO2(g) + 6FeO(s) + 2CO2 (g)

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

ΔH° = 2*18 + (-39) = 36 - 39 = -3 kJ

This new equation we will divide by 3

3Fe2O3(s) + 3CO(g) → 3CO2(g) + 6FeO(s)

Fe2O3(s) + CO(g) → CO2(g) + 2FeO(s)

ΔH° =-3/3 = -1 kJ

Now we will substract this new equation from the first equation

Fe2O3(s) + 3CO(g) - Fe2O3(s) - CO(g) → 2Fe(s) + 3CO2(g) -2FeO(s) - CO2(g)

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

ΔH° = -23kJ +1kJ

ΔH° = -22 kj

The next equation we will divide by 2

2CO(g) + 2FeO(s) → 2Fe(s) + 2CO2(g)

CO(g) + FeO(s) → Fe(s) + CO2(g)

ΔH° = -22kJ /2

ΔH° = -11 kj

7 0

3 years ago