Answer:
Various limitations of Mendeleev's periodic table are:-
Position of hydrogen - he couldn't assign a correct position to hydrogen as it showed properties of both alkali and halogens .
Position of isotopes - he considered that the properties of elements are a function of their atomic masses. Hence isotopes of a same element couldn't be placed.
In the d-block , elements with lower atomic number were placed before higher atomic number.
Explanation:
Answer:
Actinium is a chemical element with the symbol Ac and atomic number 89. A soft, silvery-white radioactive metal, actinium reacts rapidly with oxygen and moisture in air forming a white coating of actinium oxide that prevents further oxidation.
This is because it has a full outer valence shell! so there are 8 electrons and that means it doesn't have the urge the gain anymore
Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole
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 this law, the chemical equation can be treated as ordinary algebraic expression 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.
The formation reaction of 
will be,
The intermediate balanced chemical reaction will be,
(1) 
(2) 
(3) 
Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :
(1) 
(2) 
(3) 
The expression for enthalpy of formation of will be,
Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole
These are two questions and two answers
Question 1.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 9.11 × 10⁻³¹ kg
b) λ = 3.31 × 10⁻¹⁰ m
c) c = 3.00 10⁸ m/s
d) s = ?
<u>2) Formula:</u>
The wavelength (λ), the speed (s), and the mass (m) of the particles are reltated by the Einstein-Planck's equation:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Solve for s:
Substitute:
- s = 6.626×10⁻³⁴J.s / ( 9.11 × 10⁻³¹ kg × 3.31 × 10⁻¹⁰ m) = 2.20 × 10 ⁶ m/s
To express the speed relative to the speed of light, divide by c = 3.00 10⁸ m/s
- s = 2.20 × 10 ⁶ m/s / 3.00 10⁸ m/s = 7.33 × 10 ⁻³
Answer: s = 7.33 × 10 ⁻³ c
Question 2.
Answer:
Explanation:
<u>1) Data:</u>
a) m = 45.9 g (0.0459 kg)
b) s = 70.0 m/s
b) λ = ?
<u>2) Formula:</u>
Macroscopic matter follows the same Einstein-Planck's equation, but the wavelength is so small that cannot be detected:
- h is Planck's constant: h= 6.626×10⁻³⁴J.s
<u>3) Solution:</u>
Substitute:
- λ = 6.626×10⁻³⁴J.s / ( 0.0459 kg × 70.0 m/s) = 2.06 × 10 ⁻³⁴ m
As you see, that is tiny number and explains why the wave nature of the golf ball is undetectable.
Answer: 2.06 × 10 ⁻³⁴ m.
