Answer:
The moles present in 60 g of calcium are 1.5 moles.
The molecular weight of unknown gas : 23.46 g/mol
<h3>Further explanation</h3>
Given
A vessel contains 10% of oxygen and 90% of an unknown gas.
diffuses rate of mixed gas = 86 s
diffuses rate of O₂ = 75 s
Required
the molecular weight of unknown gas (M)
Solution
The molecular weight of mixed gas :(M O₂=32 g/mol)
Graham's Law :
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
Answer: Rutherford.
Explanation:
It was the scientist Ernest Rutherford who, by 1911, performed the gold foil experiment in which α particles were shoot to a thin foild of gold.
That experiment showed that although most α particles passed through the thin gold foild, some of them were deviated in small angles and some other were bounced backward.
The conclusion of the experiment was that the atom contained a small dense positively charged nucleous and negative particles (electrons) surroundiing the nucleous. Being the space in between the nucleous and the electrons empty.
Before Rutherford's experiment the model of the atom was that of the plum pudding presented by J.J Thomson, in which the atom was a solid positively charged sphere with embeded negative charge uniformly distributed in it.
