Why must indirect evidence be used to study the structure of atoms?

1 answer:

8 0

Indirect evidence must be used to study the structure of atoms, because the direct evidence affects the position and energy of the particles of the atomic structure.

For example, imagine you use a light ray to find the location of an electron in an atomic structure, the current location of the electron you are now seeing is as a result of the light ray, and the electron has changed it's initial position to a new position as a result of the effect of the light ray.

So indirect means would probably give more accurate answers about the atomic structure than that of the direct means.

An example of the indirect means would be for example setting up a mathematical equation or model of the atomic structure and trying to provide solution for the model or equation.

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A type of Lead (Pb) ion has a + 4 oxidation number. Sulfur (S) has a — 2 oxidation number. What would be the chemical formula fo

serious [3.7K]

Answer: The chemical formula for the compound of these two elements is $PbS_2$

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here metal lead is having an oxidation state of +4 called as $Pb^{4+}$ cation and sulphur non metal has oxidation state of -2 called as $S^{2-}$ . Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral $PbS_2$

The chemical formula for the compound of these two elements is $PbS_2$

4 0

3 years ago

1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$