Biotic are living organisms and abiotic are non living things such as rocks water soil
Answer : The 
for this reaction is, -88780 J/mole.
Solution :
The balanced cell reaction will be,
Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half oxidation-reduction reaction will be :
Oxidation : 
Reduction : 
Now we have to calculate the Gibbs free energy.
Formula used :
where,
= Gibbs free energy = ?
n = number of electrons to balance the reaction = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = 0.46 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the for this reaction is, -88780 J/mole.
Based on the solubility observations, barium & aluminum could be distinguished by the addition of sodium chloride to the solutions.
<h3>What happens when NaCl is added to a solution?</h3>
- The ionic link that held sodium and chloride ions together is broken when water molecules force the ions apart.
- The sodium and chloride atoms are encircled by water molecules after the salt compounds are separated. After that, the salt dissolves and forms a homogenous solution.
- In order to keep patients from dehydrating, sodium chloride, an important nutrient, is employed in healthcare. It is employed as a spice to improve flavor and as a food preservative. Additionally, sodium chloride is employed in the production of polymers and other goods. Additionally, it is used to de-ice sidewalks and roadways.
- Adding water to sodium chloride results in a physical change because no new product is created.
Learn more about sodium chloride added to a solution refer to :
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Answer:
oxidation:Ti to Ti Reduction:O2 toO2
Explanation:
<em>oxidation loses electron while Reduction gains electron</em>
Answer:
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
Explanation:
<u>Step 1: </u>Data given
mass of water = 300 grams
initial temperature = 10°C
final temperature = 50°C
Temperature rise = 50 °C - 10 °C = 40 °C
Specific heat capacity of water = 4.184 J/g °C
<u>Step 2:</u> Calculate the heat
Q = m*c*ΔT
Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)
Q = 50208 Joule = 50.2 kJ
There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C
