In an electrolytic cell, the electrode that acts as a source of electrons to the solution is called the <u>cathode</u>; the chemical change that occurs at this electrode is called <u>reduction</u>.
<h3>Define Electrolyte:-</h3>
An electrolyte is a material that separates into charged ions when it is in contact with water. Cations are positively charged ions. Anions are ions that are negatively charged. A substance that may conduct an electric current when melted or dissolved in water is known as an electrolyte.
<h3>Electrochemical cell </h3>
There are three main categories of electrochemical cells. the galvanic cell, the concentration cell, and the electrolytic cell. These cells all share the same four fundamental components. These are the elements
- The electrolyte serves as the conduit for current flow between the anode and the cathode. In an aqueous solution, it normally is homogeneous, but in moist soil, the concentration or kind of dissolved compounds may vary locally.
- The anode, which can conduct electricity and is in contact with the electrolyte, corrodes when it combines with the chemicals in the electrolyte.
- A metal also contacts the electrolyte at the cathode. It is protected from corrosion rather than corroded.
- Anode and cathode are connected by the conductor, which also completes the circuit.
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Answer: hundred cubic meters of carbon dioxide initially at 150◦C and 50 bar is to beisothermally compressed in a frictionless piston-and-cylinder device to a final pressure of 300 bar.Calculatei The volume of the compressed gas.ii The work done to compress the gas.iii The heat flow on compression.assuming carbon dioxide(a) Is an ideal gas.(b) Obeys the principle of corresponding states of Sec. 6.6(c) Obeys the Peng-Robinson equation of state.SolutionWe haveT1= 150◦C,P1= 50 bar,T2= 150◦C,P2= 300 bar. (1 and 2 denote the initialand final conditions in this ’snapshot’ problem, respectively - we have sometimes called themt1andt2)(a) If CO2is an idea gas, we havePV=NRT.The number of moles can be calculated from theinitial conditions:N1=P 1 V 1 RT 1 = (6 × 10 6 Pa)(100 m 3 ) (8 . 314 J/(mol K))(150 + 273 K) = 142127 mol = 142 . 1 kmol i. Since we know N 1 = N 2 , T 2 , P 2 V 2 = N 2 RT 2 P 2 = (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) 30 × 10 6 Pa = 16 . 66 m 3 ii. Since there is no shaft work, and since the gas is isothermally compressed , we only have pressure-volume work: W = - Z V 2 V 1 PdV = - Z V 2 V 1 NRT V dV = - NRT ln V 2 V 1 W = - (142127 mol)(8 . 314 J/(mol K))(150 + 273 K ) ln 16 . 66 100 = 8 . 958 × 10 8 J iii. Energy balance (integral form) for the closed system is: U 2 - U 1 = Q + W Back from homework 2, for an ideal gas, stating from equation 6.2-21, dU = C V dT + " T ∂P ∂T V - P # dV reduces to: dU = C V dT However, the process is isothermal, so dT = 0 Which gives: dU = Δ U = 1 N Δ U Therefore 0 = Q + W → Q = - W Q = - 8 . 958 × 10 8 J 3
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Explanation:As revealed above, the stimuli connections are clearly stated
Answer:
Explanation:
To convert from grams to moles, we must use the molar mass. This can be found on the Periodic Table. First, find the molar mass of iron and chlorine.
- Fe: 55.84 g/mol
- Cl: 35.45 g/mol
Check the formula. There is a subscript of 3 after Cl, so there are 3 atoms of chlorine in 1 molecule. Multiply iron's molar mass by 3, then add iron's molar mass.
- FeCl₃: 55.84 + 3(35.45) = 55.84+106.35=162.19 g/mol
Use this number as a ratio.
Multiply by the given number of grams.
Flip the ratio so the grams of iron (III) chloride cancel.
The original measurement of grams has 5 significant figures, so our answer must have the same. For the number we calculated, that is the ten thousandth place.
258.45 grams is approximately 1.5935 moles of iron (III) chloride.
Answer:
1 that is 100%
Explanation:
here it is given that Fermi energy of the of the metal = 5 e.V
work function= 4 e.V
we have to find the probability that an incident electron will tunnel out if E=109 V/m
WORK FUNCTION : work function is defined as the energy required to withdraw an electron completely from the metal surface.
now the energy per unit volume in electric field is given by
E= 
E=
E=
in electron volt E=4.427\times 10^{6}joule × 6.24\times 10^{18}
E=
hence the applied energy per unit volume is greater than the work work function of electron so there is probability of 100% that electron will tunnel out
