Answer:
V₂ =279.9 cm³
Explanation:
Given data:
Initial volume = 360 cm³
Initial temperature = 50°C
Initial pressure = 700 mmHg
Final volume = ?
Final temperature = 273 k
Final pressure = 1 atm
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Solution:
<em>We will convert the mmHg to atm.</em>
700/760 = 0.92 atm
<em>and °C to kelvin.</em>
50+273 = 323 K
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 0.92 atm × 360 cm³ × 273 K / 323 K ×1 atm
V₂ = 290417.6 atm .cm³. K / 323 k. atm
V₂ =279.9 cm³
<span>The vaporization of br2 from liquid to gas state requires 7.4 k/cal /mol.</span>
The cell notation is:
here in cell notation the left side represent the anodic half cell where right side represents the cathodic half cell
in anodic half cell : oxidation takes place [loss of electrons]
in cathodic half cell: reduction takes place [gain of electrons]
1) this is a galvanic cell
2) the standard potential of cell will be obtained by subtracting the standard reduction potential of anode from cathode
Therefore
3) as the value of emf is positive the reaction will be spontaneous as the free energy change of reaction will be negative
Δ
As reaction is spontaneous and there will be conversion of chemical energy to electrical energy it is a galvanic cell.
Answer:
The correct answer is "The coffee in the jug has more thermal energy than the coffee in the cup".
Explanation:
First I had to look for the problem to know the possible answers.
In this case, the coffee jug has a large amount of coffee at the same temperature. If we analyze that the decanter and the coffee are at the same temperature, we have a homogeneous thermal system. The cup is at room temperature, so by pouring coffee into it, the temperature of the coffee decreases to balance with the temperature of the cup. At this moment, the temperature of the cup-cafe system is lower than the jug-cafe system.
Thermal energy is the part of the internal energy of an equilibrated thermodynamic system that is proportional to its absolute temperature and increases or decreases by energy transfer.
In this way, we can ensure that the thermal energy of the cup-cafe system is lower than that of the jug-cafe system.
Have a nice day!
Answer:
Q = 3440Kj
Explanation:
Given data:
Mass of gold = 2kg
Latent heat of vaporization = 1720 Kj/Kg
Energy required to vaporize 2kg gold = ?
Solution:
Equation
Q= mLvap
It is given that heat required to vaporize the one kilogram gold is 1720 Kj thus, for 2 kg
by putting values,
Q= 2kg × 1720 Kj/Kg
Q = 3440Kj
