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klio [65]
3 years ago
11

Completion

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
7 0

Answer:

  1. Compress
  2. Fixed
  3. Melts
  4. Melting Point
  5. Freezing Point
  6. High
  7. Crystalline
  8. Lattice
  9. Unit cell
  10. Amorphous solids

Explanation:

Solids tend to be dense and difficult to <u>compress.</u>

They do not  flow or take the shape of their containers, like liquids do, because  the particles in solids vibrate around <u>fixed</u> points.

When a solid  is heated until its particles vibrate so rapidly that they are no longer  held in fixed positions, the solid <u>melts</u>.  

<u>Melting point</u> is the  temperature at which a solid changes to a liquid. The melting and <u>freezing point</u>  of a substance are at the same temperature.

In general,  ionic solids tend to have relatively <u>high</u>  melting points, while  molecular solids tend to have relatively low melting points.

Most  solids are <u>crystalline</u>

The particles are arranged in a pattern known  as a crystal <u>lattice</u>

The smallest subunit of a crystal lattice is the <u>unit cell</u>  

Some solids lack an ordered internal structure   and are called <u>amorphous solids.</u>

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First let us generate a balanced equation for the reaction. This is illustrated below:

2C3H8O + 9O2 —> 6CO2 + 8H2O

Next we will calculate the molar mass and masses of C3H8O and H20. This is illustrated below:

Molar Mass of C3H8O = (3x12.011) + (8x1.00794) + 15.9994 = 36.033 + 8.06352 + 15.9994 = 60.09592g/mol.

Mass of C3H8O from the balanced equation = 2 x 60.09592 = 120.19184g

Molar Mass of H2O = (2x1.00794) + 15.9994 = 2.01588 + 15.9994 = 18.01528g/mol

Mass of H2O from the balanced equation = 8 x 18.01528 = 144.12224g

From the equation,

120.19184g of C3H8O produced 144.12224g of H20.

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Rate=4.77\times 10^{-19}M/s

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aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

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In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

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Rate constant = k=1.69\times 10^{-4}M^{-1}s^{-1}

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\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}

As, \text{Rate of reaction}=4.77\times 10^{-19}M/s

So, \text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s

Thus, the appearance of NO_3 is, 4.77\times 10^{-19}M/s

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