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ch4aika [34]
3 years ago
8

If an ice cube melts at 0oC but water also freezes at 0oC, what is the difference between melting and freezing in terms of (a) t

he energy involved in the phase changes and (b) particles of the substances during the phase changes?
Chemistry
1 answer:
zalisa [80]3 years ago
6 0
The energy involved in melting in freezing is heat, At 0 degrees Celsius ice is at an equilibrium where the heat energy absorbed during melting (endothermic) and heat energy dispersed during freezing (exothermic). When ice melts its absorbs heat energy to change states, when water freezes to ice it dispersed heat energy so that the particles can move slow enough to be almost frozen, but still very small vibrations throughout the ice. Particles in the frozen state are packed tightly and help together by a crystal lattice and IMF's, liquid is free flowing with free floating particles that can move freely from one another particle. 
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Calculate the mass ratio of S to O in SO, and then calculate the mass ratio of S to O in SO2, given that the mass of S is approx
ziro4ka [17]

Explanation:

From the knowledge of law of multiple proportions,

mass ratio of S to O in SO:

mass of S : mass of O

= 32 : 16

= 32/16

= 2/1

mass ratio of S to O in SO2:

= mass of S : 2 × mass of O

= 32 : 2 × 16

= 32/32

= 1/1

ratio of mass ratio of S to O in SO to mass ratio of S to O in SO2:

= 2/1 ÷ 1/1

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Thus, the S to O mass ratio in SO is twice the S to O mass ratio in SO2.

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3 years ago
N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?
just olya [345]

Answer:

\large \boxed{\text{197.4 g}}

Explanation:

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

Mᵣ:     28.01               17.03

            N₂ + 3H₂ ⟶ 2NH₃

m/g:                          240.0

(a) Moles of NH₃

\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}

(b) Moles of N₂

\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}

(c) Mass of N₂

\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}

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