Answer:
True statment
2) Styrofoam would make a good calorimeter
3) Insulating material would make a good calorimeter
Explanation:
The calorimeter is one which is insulated that is which will not absorb or let the heat to escape from it. the calorimeter is used to measure the heat change during a process so if it will allow to exchange heat with surrounding it will deviate the readings or observence.
Copper is a good conductor of heat so we cannot use it make a calorimeter.
Hence
1) Copper would make a good calorimeter : False
2) Styrofoam would make a good calorimeter: True
Styrofoam is a bad conductor or insulator so it can be and it is used for calorimeter.
3) Insulating material would make a good calorimeter
: True
4) A good calorimeter should easily absorb heat : false
Hope this might help ask me if u have any doubts
Answer:
C
Explanation:
Large chlorine atoms can not fit within the atoms of boron
Answer: Option (d) is the correct answer.
Explanation:
An equation in which electrolytes are represented in the form of ions is known as an ionic equation.
Strong electrolytes easily dissociate into their corresponding ions. Hence, they form ionic equation.
is a strong acid and 
is a strong bases, therefore, both of them will dissociate into ions.
Thus, total ionic equation will be as follows.
Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
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.
