Answer:
The concentration is ![[Cu^{2+}]_a = 10^{-10.269}](https://tex.z-dn.net/?f=%20%20%20%5BCu%5E%7B2%2B%7D%5D_a%20%20%3D%2010%5E%7B-10.269%7D%20)
Explanation:
From the question we are told that
The voltage of the cell is 
Generally the reaction at the cathode is
the half cell voltage is V_c = 0.337 V
Generally the reaction at the anode is
the half cell voltage is V_a = -0.337 V
Gnerally the reaction of the cell is
At initial the voltage is V = 0 V
Generally the voltage of the cell at 25°C is
![E = V - \frac{0.0591}{n} log \frac{[Cu^{2+}] _a}{[Cu^{2+}]_c}](https://tex.z-dn.net/?f=E%20%3D%20%20V%20%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5BCu%5E%7B2%2B%7D%5D%20_a%7D%7B%5BCu%5E%7B2%2B%7D%5D_c%7D)
Here n is number of of electron and it is 2
So from the question we are told that one cell has a concentration 1.5 x 10-3 M
Let assume it is ![[Cu^{2+}]_c](https://tex.z-dn.net/?f=%5BCu%5E%7B2%2B%7D%5D_c)
So
![0.22= 0 - \frac{0.0591}{2} log \frac{[Cu^{2+}] _a}{ 1.5 * 10^{-3} }](https://tex.z-dn.net/?f=0.22%3D%20%200%20%20-%20%5Cfrac%7B0.0591%7D%7B2%7D%20log%20%5Cfrac%7B%5BCu%5E%7B2%2B%7D%5D%20_a%7D%7B%20%201.5%20%2A%2010%5E%7B-3%7D%20%7D)
=> ![-7.445 = log \frac{[Cu^{2+}] _a}{ 1.5 * 10^{-3} }](https://tex.z-dn.net/?f=-7.445%20%3D%20%20%20%20%20log%20%5Cfrac%7B%5BCu%5E%7B2%2B%7D%5D%20_a%7D%7B%20%201.5%20%2A%2010%5E%7B-3%7D%20%7D)
=> ![-7.445 = log [Cu^{2+}_a] - log [1.5*10^{-3}]](https://tex.z-dn.net/?f=-7.445%20%3D%20%20%20%20%20log%20%5BCu%5E%7B2%2B%7D_a%5D%20-%20log%20%5B1.5%2A10%5E%7B-3%7D%5D)
=> ![-7.445 + log [1.5*10^{-3} = log [Cu^{2+}_a]](https://tex.z-dn.net/?f=-7.445%20%20%2B%20log%20%5B1.5%2A10%5E%7B-3%7D%20%3D%20%20%20%20%20log%20%5BCu%5E%7B2%2B%7D_a%5D%20)
=> ![-7.445 - 2.824 = log [Cu^{2+}_a]](https://tex.z-dn.net/?f=-7.445%20%20-%202.824%20%3D%20%20%20%20%20log%20%5BCu%5E%7B2%2B%7D_a%5D%20)
Taking the antilog
=>
=> ![[Cu^{2+}]_a = 5.38 *10^{-11} \ M](https://tex.z-dn.net/?f=%20%20%20%5BCu%5E%7B2%2B%7D%5D_a%20%20%3D%205.38%20%2A10%5E%7B-11%7D%20%5C%20%20M%20%20)
Here we have to identify the sample which need more energy to heat the sample 1 degree Celsius.
Among the given elements magnesium will require more energy than the others to heat.
As per the definition of specific heat of a compound, the amount of heat required to increase the temperature of the material 1 degree Celsius is the specific heat of the material.
The given data are-
substance specific heat
Lead 0.129
Tin 0.21
Silver 0.235
Iron 0.449
Calcium 0.647
Granite 0.803
Aluminium 0.897
Magnesium 1.023
From the given data lead, magnesium, iron and aluminium have the specific heat 0.129, 1.023, 0.449 and 0.897 respectively. Thus magnesium will require more energy than the others to heat.
<span>The element that is used in light bulbs as a filament is tungsten - this is almost always the case in halogen and incandescent bulbs. Tungsten is chosen for this purpose because of the fact it can withstand temperatures of up to 4,500 degrees, as well as being incredibly flexible.</span>
Answer:
84g Thus, 84g of KNO3 must be dissolve in 100 grams of water to form a saturated solution at 50 oC.
and 136 of KNO3 must be dissolve in 100 grams of water to form a saturated solution at 70 oC.
Explanation:
