Ernest Rutherford
J. J Thomson
Explanation:
<u>Ernest Rutherford</u>
In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.
Experiment
In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.
Discovery and reflection on the atomic theory
To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.
<u>J. J Thomson</u>
Experiment
In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.
Discovery and reflection on the atomic theory
From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:
- they move in a straight line
- they possess kinetic energy
- they attract positive charges and repels negative charges
Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.
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Answer:
Explanation:
As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.
Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>
So, you must just convert the given temperature, 32°F, to kelvin scale.
You can do that in two stages.
- First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80
- Second, convert 0°C to kelvin:
T (K) = T(°C) + 273.15 K= 273.15 K
Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K
Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.
Answer:
Obtención. El carbono se encuentra - frecuentemente muy puro - en la naturaleza, en estado elemental, en las formas alotrópicas diamante y grafito. El material natural más rico en carbono es el carbón (del cual existen algunas variedades). Grafito: Se encuentra en algunos yacimientos naturales muy puro.
Answer:
pH = 8.314
Explanation:
equil: S S 3S
∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24
⇒ 6.0 E-24 = ( S )*( 3S )³
⇒ 6.0 E-24 = 27S∧4
⇒ 2.22 E-25 = S∧4
⇒ ( 2.22 E-25 )∧(1/4) = S
⇒ S = 6.866 E-7 M
⇒ [ OH- ] = 3*S =2.06 E-6 M
⇒ pOH = - Log [ OH- ]
⇒ pOH = - Log ( 2.06 E-6 )
⇒ pOH = 5.686
∴ pH = 14 - pOH
⇒ pH = 8.314
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I think the reaction that represents a balanced, double replacement chemical reaction is B which is <span>Rb2O + Cu(C2H3O2)2 → 2RbC2H3O2 + CuO </span>
