The scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.
<h3>What is a cathode ray?</h3>
A cathode ray is a tube that contains negatively charged electrode( that is the cathode) which emits electrons when heated at a low pressure.
The cathode ray was used by the scientist, J.J. Thomson to find the ratio of charge to mass (e/m) of the electrons.
Therefore, the scientist that performed the cathode ray experiment leading to the discovery of electrons is J.J. Thomson.
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Charcoal with a carbon-14 activity of 0.60 compared to new wood has less than 5,730 years.
<h3>What is a radioactive isotope?</h3>
A radioactive isotope is an element in nature that emit radioactivity in a given period of time (e.g., the half-life for C14 is equal to 5,730 years).
Radioactive dating is a technique to measure the age of an element by measuring its radioactive activity.
In conclusion, charcoal with a carbon-14 activity of 0.60 compared to new wood has less than 5,730 yr.
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A. High energy radiation produced in the ozone layer. (:
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
