Answer:
5. the scattering of α particles by a metal foil
Explanation:
This is the classical Rutherford's experiment in which he bombarded a thin foil of gold with alpha particles which are positively charged helium nucleus.
He did observed that most of the particles passed through the foil relatively undeflected or if they were deflected it was by a very small angle.
Once in a while the alpha particle rebounded completely. An analogy is the one typically mentioned that it was as if we throw a ball at a piece of paper and it rebounds toward us.
This observations led Rutherford to conclude that the nucleus of the atom is very small positely charged and that the atom is relatively empty with electrons of very small masses. His model is referred as the Plum Pudding model and later Bohr modified it to the planetary model.
B moisture evaporating from an ocean
The required mole ratio of NH₃ to N₂ in the given chemical reaction is 2:4.
<h3>What is the stoichiometry?</h3>
Stoichiometry of the reaction gives idea about the number of entities present on the reaction before and after the reaction.
Given chemical reaction is:
4NH₃ + 3O₂ → 2N₂ + 6H₂O
From the stoichiometry of the reaction it is clear that:
4 moles of NH₃ = produces 2 moles of N₂
Mole ratio NH₃ to N₂ is 2:4.
Hence required mole ratio is 2:4.
To know more about mole ratio, visit the below link:
brainly.com/question/504601
Answer: The E for Silver-silver Chloride electrode = 0.287 V
Explanation:
Silver/Silver Chloride (Ag/AgCl) with a value for E° that is actually +0.222 V or approximately 0.23 V has the actual potential of the half-cell prepared in this way as +0.197 V vs SHE, (Standard Hydrogen Electrode) which arises because in addition to KCl, there is the contribuion of AgCl to the chloride activity, which isn't exactly unity.
Therefore, the E for the Ag/AgCl electrode would approximately equal 0.287 V
Answer:
Mass of barium sulfate = 8.17 g
Explanation:
Given data:
Mass of sodium sulfate = 4.98 g
Mass of barium sulfate produced = ?
Solution:
Na₂SO₄ + Ba(NO₃)₂ → BaSO₄ + 2NaNO₃
Moles of sodium sulfate:
Number of moles = mass/molar mass
Number of moles =4.98 g / 142.04 g/mol
Number of moles = 0.035 mol
Now we will compare the moles pf sodium sulfate and with barium sulfate.
Na₂SO₄ : BaSO₄
1 : 1
0.035 : 0.035
Mass of barium sulfate:
Mass = number of moles × molar mass
Mass = 0.035 mol ×233.4 g/mol
Mass = 8.17 g
