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lorasvet [3.4K]
3 years ago
5

What is a atom of chemistry

Chemistry
1 answer:
Yuliya22 [10]3 years ago
8 0

Answer:

Atom is the smallest particle of matter.

Elements are made up of same atoms.

Atom consist of electron, proton and neutron.

Explanation:

Atom was first discovered by John Dalton.

word "Atom" came from Greek word, that means something that could not split. he explained that atom is indivisible particle.

In the end of 18th century J.J. Thomson put forward a new concept of atom. he said that atom have negative charged particles called electrons but overall atom is neutral.

In 1909 Rutherford with his students discovered positive charged particles and nucleus and said that it is in the center of atom.

He explain a model of atom and said that electrons revolve around a hard core in the center called nucleus.

In 1913 Niels Bohr  explains the atomic spectra and put forward the concept of shells and sub-shells.

So overall Structure of an Atom is

  • consist of Neutrons in nucleus
  • +ve charged protons in nucleus
  • A -vely charged electrons revolving around the nucleus
  • the electron revolve in shells i.e. K, L, M, and N
  • Each shell divide in sub-Shell such as s, p,d and f.

Atom has a specific atomic mass and atomic number

Atomic number = number of protons or electron

Atomic mass = number of protons +  number of neutrons in nucleus.

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Molar mass  helium gas :

He = 4.0 g/mol

number of moles:  3.75 / 4.0 => 0.9375 moles

1 mole ------ 22.4 ( at STP)
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volume = 0.9375 x 22.4 / 1

Volume = 21 / 1

Volume = 21 L

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Answer:

B. their outer electron levels are filled

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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T
Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

  • To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 298 K

Putting values in above equation, we get:

12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

  • The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (glucose) = 89.18 g

M_{solute} = Molar mass of solute (glucose) = 180.16  g/mol

W_{solvent} = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC

Hence, the freezing point of solution is -0.974°C

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2 A certain gas of 25 g at 25°c and 0.65 atm occupies a volume of 23.52L Determine the molecule mass of the gas.​
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Answer:

{ \bf{PV= \frac{m}{M} RT}} \\  \\ { \tt{(0.65  \times 23.52)  =  \frac{25}{M}  \times 0.081 \times (25 + 273)}} \\  \\ M = 39.5 \: g

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