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3 months ago

What was John Dalton's contribution to the development of the atomic theory?

1 answer:

6 0

John Dalton made some hypothesis about the structure of atom. He proposed that matter is composed of great number of indivisible particles called atoms they can neither be destroyed nor be created.

<h3>What is atomic theory?</h3>

There are different theories regarding the structure and electronic properties of an atom. Many scientists contributed to the modern theory of atomic structure in which John Dalton was first to mention the word atom.

According to Dalton' theory, matter is composed of indivisible particles called atoms. Atoms can neither be created nor be destroyed. All the atoms of the same element are identical in all aspects.

Atoms of different elements are different and the compounds are formed by the combination of atoms. Dalton's theory provided a sound basis for the laws of chemical combination and also several properties of gases and liquids known at that time.

However, he could not explain the reason for chemical combination of atoms and did not give any idea about the existence of isotopes and isobars.

Hence, the main aspects of Dalton's theory was the indivisibility of atoms and the chances of chemical combination.

To learn more about Dalton's theory, find the link below:

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In going across a row of the periodic table, protons and electrons are added and ionization energy generally increases. In going

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Answer:

Explanation:

Ionization energy:

It is the minimum amount of energy required to remove the electron from isolated gaseous atom to make the ion.

As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell.

When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required. Where as,

When we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

As the size of atom increases the ionization energy from top to bottom also decreases because it becomes easier to remove the electron because of less nuclear attraction and as more electrons are added the outer electrons becomes more shielded and away from nucleus.

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3 years ago

The temperature is -14°C the air pressure in an automobile tire is 149K PA if the volume does not change what is the pressure af

Jet001 [13]

Answer: The pressure after the tire is heated to 17.3°C is 167 kPa

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=149kPa\\T_1=-14^0C=(273-14)=259K\\P_2=?=27.5psi\\T_2=17.3^0C=(273+17.3)=290.3K$

Putting values in above equation, we get:

$\frac{149}{259}=\frac{P_2}{290.3}\\\\P_2=167kPa$

Hence, the pressure after the tire is heated to 17.3°C is 167 kPa

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2 years ago

In which color coded area would you find an element that behaves both like a metal and a non-metal?

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2 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

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2 years ago

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2 years ago