<h2>Answer : Daniel was related to study of Earth and Space
</h2><h2>Whereas Sanya was closely related to chemistry</h2><h3>Explanation :</h3>
As in the given information, Daniel the scientist was studying the effect of tropical storms on coastal erosions. These studies comes under the field of Earth and Space. Earth because coastal erosions are involved in the study, and Space because storms are discussed which is something related to space.
In case of Sanya, it is clear that her study focuses on discharges from the industrial plant that has effect on freshwater swamps. Precisely, she is studying the pH concentration which comes in chemistry.
This prove that Daniel studies Earth and Space whereas Sanya studies Chemistry.
I dont see any options to choose from but im pretty sure its the electron count that the elements have
Answer:
Explanation:
The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized
Answer:
T final = 80°C
Explanation:
∴ Q = 18000 cal
∴ m H2O = 300 g
∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K
∴ T1 = 20°C = 293 K
∴ T2 = ?
⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)
⇒ (18000 cal)/(300 cal/K) = T2 - 293 K
⇒ T2 = 293 K + 60 K
⇒ T2 = 353 K (80°C)