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Free_Kalibri [48]
3 years ago
6

The diagrams below represent the four states of matter.

Chemistry
2 answers:
UkoKoshka [18]3 years ago
6 0
Diagram 1 exhibits the nature of the particles within the state that forms after a solid melts. Solids melt into liquids and the particles within a liquid have a greater spacing than the particles in a solid. Moreover, these particles are free to slip over one another, which means that liquids do not have a definite shape; however, the particles are still confined in by intermolecular forces, which means that the volume of a liquid is definite.
anastassius [24]3 years ago
3 0

Answer: Diagram 1 (with the blue balls).

Explanation:

1) After the solids melt they are liquids.

Liquid state is characterized by the particles being close but yet free to move, they can slip over each other, which is what permit the liquids flow and take the shape of the containter.

2) Solid is the state of the matter in which the particles are stuck together, they cannot move with respect to the others, but they are in fixed positions. In this state the particles vibrate and the matter has a definite shape and volume.

This state is represented by the diagram 3 (the red balls).

3) In gas state the particles are quite separated, free to move in and occupy all the space. In this state the particles are moving at great speed. The volume of the gas is the volumen of the container.

This state is represented by the diagram 2(the yellow balls).

4) Plasma is the state of the matter similar to gas state but the particles are ions. Those are particles that have charge, either positive or negative.

This is represented by the diagram 4 (the balls with positive and negative symbols).

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\large \boxed{\text{2.20 g Pb}}

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They gave us the masses of two reactants and asked us to determine the mass of the product.

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1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

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\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

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