Answer:
The specific heat capacity of the metal is 0.843J/g°C
Explanation:
Hello,
To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.
Heat gained by the metal = heat loss by water + calorimeter
Data,
Mass of metal (M1) = 512g
Mass of water (M2) = 325g
Initial temperature of the metal (T1) = 15°C
Initial temperature of water (T2) = 98°C
Final temperature of the mixture (T3) = 78°C
Specific heat capacity of metal (C1) = ?
Specific heat capacity of water (C2) = 4.184J/g°C
Heat loss = heat gain
M2C2(T2 - T3) = M1C1(T3 - T1)
325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)
1359.8 × 20 = 512C1 × 63
27196 = 32256C1
C1 = 27196 / 32256
C1 = 0.843J/g°C
The specific heat capacity of the metal is 0.843J/g°C
Answer:
1.18 V
Explanation:
The given cell is:
Half reactions for the given cell follows:
Oxidation half reaction: 
Reduction half reaction: 
Multiply Oxidation half reaction by 2 and Reduction half reaction by 3
Net reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
To calculate the EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAl%5E%7B3%2B%7D%5D%5E2%7D%7B%5BFe%5E%7B2%2B%7D%5D%5E3%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +1.21 V
n = number of electrons exchanged = 6
Putting values in above equation, we get:
Answer:
that are formed from hybridized orbitals
Explanation:
The chemical concept of resonance is when a change in the position of the electrons occurs, without changing the position of the atoms.
The structure obtained in the resonance will not be any of the previous ones, but a hybrid of resonance between those structures.
Answer : The process is not spontaneous.
Explanation :
As, we know that:
Change in entropy = Change in entropy of system + Change in entropy of surrounding
As we are given in question, the entropy of surroundings decrease by the same amount as the entropy of the system increases.
For the given reaction to be spontaneous, the total change in entropy should be positive.
Given :
Entropy change of system = +125J/K
Entropy change of surroundings = -125J/K
Total change in entropy = Entropy change of system + Entropy change of surroundings
Total change in entropy = 125 J/K + (-125 J/K)
Total change in entropy = 0
The process is at equilibrium because the entropy change is equal to zero. So, the process is not spontaneous.
Cu = 63.546
N= 14.001 g/mol
O= 15.999 g/mol * 3 = 47.997
Copper (II) Nitrate has a MW of 125.544 g/mol
6.25 x 125.544
= 784.65 <--- is your answer, if there were was a multiple choice or not :)
