Identical objects are placed in Beaker A and Beaker B. The objects float as shown in the diagram. What can you conclude about th
e liquid in each of the beakers?
a
Question 6 options:
The liquid in Beaker B has a greater density than the liquid in Beaker A.
The liquid in Beaker A has a greater density than the liquid in Beaker B.
The liquid in Beaker A has a greater density than the objects. The liquid in Beaker B is less dense than the objects.
Both liquids are less dense than the objects.
I think the answers are B or D please help me!
a
Question 6 options:
The liquid in Beaker B has a greater density than the liquid in Beaker A.
The liquid in Beaker A has a greater density than the liquid in Beaker B.
The liquid in Beaker A has a greater density than the objects. The liquid in Beaker B is less dense than the objects.
Both liquids are less dense than the objects.
I think the answers are B or D please help me!
2 answers:
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Answer: The boiling point of water at 81 kPa will be 298.17 K.
Explanation:
The boiling point of water is 373 K at atmospheric pressure. So, to calculate the boiling point at 81 kPa, we use the law given by Gay-Lussac.
Gay-Lussac Law states that pressure is directly proportional to the temperature when volume and moles remain constant.
Mathematically,
where,
= Initial pressure and temperature
= Final pressure and temperature
We are given:
Putting values in above equation, we get:
Hence, the boiling point of water at 81 kPa will be 298.17 K.
Answer:
Iron is being oxidized at the anode and water is acting as the electrolyte.
Explanation:
When iron is exposed to oxygen and water , the rusting of iron takes place.
<u>The reaction taking place at anode : Oxidation of iron.</u>
The reaction taking place at cathode : Reduction of oxygen in the air.
The overall reaction:
The rust that is hydrated iron(III) oxide can form iron(II) ions which can react further with oxygen.
Thus, from the above reactions ,
Iron is being oxidized at the anode and water is acting as the electrolyte.