The given statement is true .
<h3>What is Rutherford’s gold foil experiment?</h3>
- A piece of gold foil was hit with alpha particles, which have a favorable charge. Most alpha particles went right around. This showed that the gold particles were mostly space.
- The Rutherford gold leaf investigation supposed that most (99%) of all the mass of an atom is in the middle of the atom, that the nucleus is very small (105 times small than the length of the atom) and that is positively captured.
- For the distribution experiment, Rutherford enjoyed a metal sheet that could be as thin as practicable. Gold is the most malleable of all known metals. It can easily be converted into very thin sheets. Hence, Rutherford established a gold foil for his alpha-ray scattering experimentation.
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If the concentration of acetyl chloride is increased ten times the rate of reaction is increased ten times.
The conversion of acetyl chloride to methyl acetate is a substitution reaction. Recall that a substitution reaction is one in which a moiety in a molecule is replaced by another.
In this reaction, the CH3O- ion replaces the chloride ion. In the first step, the CH3O- ion attacks the substrate in a slow step. This creates a tetrahedral intermediate. Loss of the chloride ion yields the methyl acetate product.
The rate determining step is the formation of the tetrahedral intermediate. Since the reaction is first order in the acetyl chloride, if its concentration is increased ten times the rate of reaction is increased ten times.
Learn more: brainly.com/question/5624100
Answer:
Ba(OH)2 + H2SO3 = BaSO3 + H2O -
Explanation:
Answer: The enthalpy change for formation of butane is -125 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B4%5Ctimes%20H_f_%7BCO_2%7D%2B5%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%20H_f_%7BO_2%7D%5D)
![-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]](https://tex.z-dn.net/?f=-2877%3D%5B%284%5Ctimes%20-393%29%2B%285%5Ctimes%20-286%29%5D-%5B1%5Ctimes%20H_f_%7BC_4H_%7B10%7D%7D%2B%5Cfrac%7B13%7D%7B2%7D%5Ctimes%200%5D)
Thus enthalpy change for formation of butane is -125 kJ/mol
Explanation:
Since, it is given that critical temperature of Argon is 150.9 K and critical pressure of Argon is 48.0 atm.
It is known that gas phase of neon occurs at 50 K. As the boiling point of Ar is more than the boiling point of neon which means that there is strong intermolecular force of attraction between argon molecules as compared to neon molecules.
This is also because argon is larger in size. As a result, induced dipole-induced dipole forces leads to more strength in Ar as compared to Ne.
