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.
learn more:
Rutherford brainly.com/question/1859083
#learnwithBrainly
Answer: The mole allows people to calculate the number of middle schoolers entities (usually atoms or molecules. Isabella's number is an absolute number: there are 6.022 × 1023 middle schoolers entities is in 1 mole. This can also be written as 6.022 × 1023 mol-1.
Explanation: I hope that helped !!
Answer:
yes
Explanation:
oak trees don't have vertebrates
Answer:
hgfdjsuejssj_uyghgjiyr656⁸8⁴83jbv
Answer:
121 K
Explanation:
Step 1: Given data
- Initial volume (V₁): 79.5 mL
- Initial temperature (T₁): -1.4°C
- Final volume (V₂): 35.3 mL
Step 2: Convert "-1.4°C" to Kelvin
We will use the following expression.
K = °C + 273.15 = -1.4°C + 273.15 = 271.8 K
Step 3: Calculate the final temperature of the gas (T₂)
Assuming ideal behavior and constant pressure, we can calculate the final temperature of the gas using Charles' law.
V₁/T₁ = V₂/T₂
T₂ = V₂ × T₁/V₁
T₂ = 35.3 mL × 271.8 K/79.5 mL = 121 K
