<u>Answer:</u> Copper is getting oxidized and is a reducing agent. Silver is getting reduced and is oxidizing agent.
<u>Explanation:</u>
Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.
Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.
Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.
Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.
For the given chemical reaction:
The half reactions for the above reaction are:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u> 
From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.
Silver is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.
Answer:
Gas
Explanation:
Because Crude oil can usually be found in the ground as a liquid and in the air is gas can be kerosene.
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:
Molar mass of acetone = 58 g/mol
Number of moles = 
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have = 
To calculate the entropy change for the system, we use the formula:
Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
(Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,
Hello!
To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.
In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.
Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.
35.0° C - 25.0° C = 10.0° C
We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.
q = 125 g × 4.184 J/g °C × 10.0° C
q = 523 J/°C × 10.0° C
q = 5230 J
Therefore, it will take 5230 joules (J) to raise the temperature of the water.
