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3 years ago

Why do scientist use models to study atoms

Chemistry

2 answers:

maxonik [38]3 years ago

6 0

Scientist use models to study atoms because you cannot either see or perceive directly what do atoms look like.

Models permit to explain in a simplified way how the atoms are constituted, and, so, figure out how they behave and how they are the bulding blocks of the matter (elements and compounds).

A model can be a drawing or structure with balls and sticks, which show in an idealized way what an atom is.

Since the scientists have discovered different facts about the atoms along the last 200 years the models of the atoms have also evolved.

The plum pudding model of the atom proposed by JJ Thompson was different of the nuclear model proposed by Rutherford, and Rutherford's was different from the planetary model proposed by Bohr, and this is different from the quantum model.

lesya [120]3 years ago

4 0

Models are used to study atoms because we can never really see an atom up close, and a model is like a super magnified version of the atom. Plus, its easier to study from them because we can pin point the different characteristics of the atom (Protons, Neutrons, etc.)

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Why don't elements in a group have consicutive atomic numbers?

slamgirl [31]

Answer:

Periods have consecutive atomic number as they are in a row and groups are vertical meaning they are in a column and groups have similar properties

7 0

3 years ago

Read 2 more answers

Consider the following intermediate chemical equations.

QveST [7]

Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$P_4(s)+6Cl_2\rightarrow 4PCl_3$ $\Delta H_1=-2439kJ$ (1)

$4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)$ $\Delta H_2=3438kJ$ (2)

Net chemical equation:

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H=?$ (3)

Adding 1 and 2 we get,

$4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2$ $\Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ$ (4)

Now dividing equation (4) by 4, we get

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2$

$\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ$ (4)

8 0

3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$