Which describes Ernest Rutherford’s experiment?
2 answers:
Answer: The correct option is C.
Explanation: Ernest Rutherford proposed a gold-foil experiment.
In his experiment, he took a Gold foil & fired - particles (carrying +4 charge). He, then put a photogenic film around the gold foil to have a look where the particles are going. He thought that the alpha-particles will pass through the gold-foil because it was very thin.
But to his surprise he observed that some particles bounced back. This observation will either mean that the alpha-particles are hitting something hard or hitting something that carries positive charge because positive charges repel each other.
Therefore, he made proposed that the nucleus is carrying some positive charge that consists of protons.
Hence, the correct option is C.
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Answer: The rate of the loss of is 0.52M/s
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate of disappearance of =
Rate of formation of =
Rate of formation of =
Thus Rate of disappearance of =
Answer:
Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1
Explanation:
The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:
Vrms =
where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol
For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R
Vrms = √(3 * R *4T)/0.04 = √300RT
For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R
Vrms = √(3 * R *T)/0.001 = √3000RT
Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316
Ratio of Vrms of argon to that of hydrogen = 0.316 : 1