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NARA [144]
3 years ago
15

Dispersion definition in chemistry and an example

Chemistry
2 answers:
Neko [114]3 years ago
4 0

Answer:

dispersion, in chemistry, mixture in which fine particles of one substance are scattered throughout another substance. A dispersion is classed as a suspension , colloid , or solution . Generally, the particles in a solution are of molecular or ionic size; those in a colloid are larger but too small to be observed with an ordinary microscope; those in a suspension can be observed under a microscope or with the naked eye. A coarse mixture (e.g., sand mixed with sugar) is usually not thought of as a dispersion.

Aliun [14]3 years ago
3 0

Answer:

Mixture in which fine particles of one substance are scattered throughout another substance.

Example:

The fat molecules suspended in milk provide a mode of delivery of important fat-soluble vitamins and nutrients from the mother to newborn.

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We can also perform a similar calculation for the mass defect and binding energy for nuclear reactions using the masses of the a
sukhopar [10]

Answer:

See Explanation

Explanation:

\frac{235}{92} U + \frac{1}{0} n ---->\frac{137}{52} Te + \frac{97}{40} Zr +2\frac{1}{0} n

Hence the mass defect is;

[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]

=  236.0526 - 235.85361

= 0.19899 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg

Binding energy = Δmc^2

Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J

ii) \frac{10}{5}B + \frac{1}{0}n-----> \frac{7}{3} Li + \frac{4}{2} He + Energy

Hence the mass defect is;

[10.01294 + 1.00867] - [7.01600 + 4.00260]

= 11.02161 - 11.0186

= 0.00301 amu

Since 1 amu = 1.66 * 10^-27 Kg

0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg

Binding energy = Δmc^2

Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J

7 0
2 years ago
From what philosophical tradition is chemistry said to have emerged?
EastWind [94]
Do you have the options?
7 0
3 years ago
you see a structural formula and which the symbols for elements are connected by a long Dash you can assume that the chemical bo
sergejj [24]

When you see a long dash joining two elements in the structural formula of a chemical molecule you can assume that the chemical bonds in the compound are simple bonds.

In the representation of chemical molecules long dashes are used to represent simple bonds since these bonds are longer than double and triple bonds, which are represented by = and ≡ respectively.

8 0
3 years ago
Read 2 more answers
Considere un elemento "X" que posee como valencias 2, 4 y 6. En su nomenclatura tradicional, considerando la valencia cuatro, es
Ierofanga [76]

Answer:

"ite"

Explanation:

Se sabe que los elementos del grupo 16 de la tabla prioritaria muestran estados de oxidación o valencias de 2,4 y 6 respectivamente, dependiendo del compuesto formado.

Por ejemplo, el azufre forma los siguientes compuestos;

sulfato de sodio (el azufre tiene una valencia de 6)

Sulfito de sodio (el azufre tiene una valencia de 4)

Sulfuro de sodio (el azufre tiene una valencia de 2)

Por lo tanto, en compuestos en los que exhiben una valencia de 4, la terminación común es "ite"

8 0
3 years ago
Read 2 more answers
At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:
Butoxors [25]

<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

Initial concentration of iodine gas = \frac{0.45}{2}=0.225M

Initial concentration of chlorine gas = \frac{0.45}{2}=0.225M

For the given chemical equation:

2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

                      I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091

<u>Initial:</u>         0.225      0.225

<u>At eqllm:</u>   0.225-x    0.225-x     2x

The expression of K_c for above equation follows:

K_c=\frac{[ICl]^2}{[Cl_2][I_2]}

Putting values in above equation, we get:

9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

6 0
3 years ago
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