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Crazy boy [7]
3 years ago
6

6.An ice cube that has a mass of 20 g is in a sealed container. As the container is heated, the ice first melts, but eventually

it changes to water vapor.
A)The temperature at which the ice melts is ____________________.

B) Compare the motion of the particles in each state
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
8 0

Answer:

See explanation

Explanation:

a) Every substance melts at a particular temperature known as the melting point of the substance. The melting point of ice is 0°C. Hence ice melts at  0°C.

b) There are three states of matter; solid liquid and gas. The particles of matter in each stage of matter possess varying degrees of freedom.

In the solid state, the intermolecular forces between the particles of water that compose ice are very strong. Hence the particles do not translate, they only vibrate or rotate about their mean positions.

In the liquid state, the magnitude of intermolecular interaction is less than that of the solids hence the molecules can translate but not with a high kinetic energy.

In the gaseous state, there is a minimum intermolecular interaction between particles theoretically hence the particles are really free to move about with a very high kinetic energy.

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To name the compound written as CuCl2, you would write:
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Answer: option C. Copper (II) chloride

Explanation:

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Cu + 2(—1) = 0

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Collect like terms

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Consider the following reaction:
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Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

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4 years ago
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