Answer: a) Anode: 
Cathode: 
b) Anode : Cr
Cathode : Au
c) 
d) 
Explanation: -
a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
At cathode: 
At anode: 
b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.
At anode which is a negative terminal, oxidation occurs which is loss of electrons.
Gold acts as cathode ad Chromium acts as anode.
c) Overall balanced equation:
At cathode:
(1)
At anode:
(2)
Adding (1) and (2)

d)
= -0.74 V
= 1.40 V

Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAu%5E%7B3%2B%7D%5D%7D%7B%5BCr%5E%7B3%2B%7D%5D%5E%7D)
where,
n = number of electrons in oxidation-reduction reaction = 3
= standard electrode potential = 2.14 V
![E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.14-%5Cfrac%7B0.0592%7D%7B3%7D%5Clog%20%5Cfrac%7B%5B1.0%7D%7B%5B1.0%5D%7D)

Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.
Can you show the chart please
Atomic number
Atoms of different elements must have different atomic number. The atomic number of each element is different than other Elements.
Which element is a metalloid? <span>B. Silicon (Si)
</span>
Looking at the periodic table, which element is a metal? <span>D. Strontium (Sr)
</span>Argon (Ar) is a noble gas, silicon (Si) is a metalloid, hydrogen (H) is a non-metal, and strontium is a alkaline earth metal.
<span>
I have used the periodic table from </span><span>http://chemreference.com/elements/1/properties?display=periodic&trend=number to help me when I'm unsure.</span>
Answer:
The amount of heat required to raise the temperature of the sample from 298 to 385 Kelvin, is 16.6 kJ
Explanation:
<u>Step 1: </u>Given data
A 79.0 g sample of ethanol raises from 298 K to 385 K
The specific heat of ethanol is 2.42J/g°C
<u>Step 2:</u> Calculate the heat transfer
Q = m*Cp*ΔT
with m = the mass of the ethanol sample (in grams)
⇒ mass = 79 grams
with Cp = the specific heat capacity of ethanol (in J/g°C)
⇒ Cp = 2.42 J/g°C
with ΔT = the change of temperature (T2-T1)
⇒ ΔT = 385 K - 298K = 112 °C - 25 °C = 87
Q = 79 grams * 2.42 J/g°C * 87 = 16632.66 j = 16.6 kJ
The amount of heat required to raise the temperature of the sample from 298 to 385 Kelvin, is 16.6 kJ