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2 years ago

why the total mass of the products would be less than the total weight of the reactant after a chemical reaction?

Chemistry

1 answer:

Lisa [10]2 years ago

4 0

The best explanation would be that Gases were released during the Chemical reaction, causing a loss of Mass.

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In the reaction below green chlorine gas and brown iodine chloride forms yellow iodine trichloride Cl2 (g) + ICl (l) ⇌ ICl3 (s)

Keith_Richards [23]

Answer:

B. It will become more green and brown

Explanation:

Le Chateliers principle states that when a constraint such as change in concentration, volume, pressure or temperature is imposed on a reaction system at the equilibrium, the equilibrium position will shift in such a way as to annul the constraint. Let us see how this applies to the equation under consideration.

If the system is at equilibrium and more ICl3 is added to the system, the equilibrium will shift towards the left hand side (more reactants are formed). This implies that the colour of the system will become more green and brown since there are more reactants now in the system.

6 0

3 years ago

PLEASE HELP ME ASAP! CHEMISTRY TUTOR SEE ATTACHED

Masteriza [31]

Answer:

$\large \boxed{\text{-827.4 kJ}}$

Explanation:

We have three equations:

1. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic) + O₂(g) ⟶ SO₂(g); ∆H = -296.8 kJ

3. PbO(s) + H₂S(g) ⟶ PbS(s) + SO₂(g); ∆H = -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s) + 3O₂(g) ⟶2PbO(s) + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.

7. 2S(s, rhombic) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

6. 2H₂S(g) + O₂(g) ⟶ 2S(s) + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s) + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g) ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

$\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}$

8 0

3 years ago

There are two common oxides of copper; one is 80% copper and the other is 89% copper by weight. Calculate the formula and name e

irakobra [83]

80% copper (Cu)

Cu: 80 : 63.546 = 1.259

O: 20 : 16 = 1.25

Cu:O = 1 : 1

the formula: CuO

89% copper (Cu)

Cu: 89 : 63.546 = 1.4

O: 11 : 16 = 0.6875

Cu:O = 2:1

the formula: Cu₂O

8 0

2 years ago

What terms refers to the process by which astronomers study distant objects by dissecting examining the spectrum of light an obj

mart [117]

Spectroscopy because it talks about the study of spectrum of light.

3 0

2 years ago

For the following exothermic reaction, A+B⇌C+D + Heat, if the temperature is increased, the reaction will experience a shift to

Elodia [21]

(Left) I’m pretty sure

6 0

3 years ago