Sodium chloride reacts with copper sulfate to produce sodium sulfate and copper chloride. mc011-1.jpg This equation represents a
2 answers:
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Answer : The correct option is, (D) 89.39 KJ/mole
Explanation :
First we have to calculate the moles of HCl and NaOH.
The balanced chemical reaction will be,
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH
Thus, the number of neutralized moles = 0.3562 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water =
Now we have to calculate the heat absorbed during the reaction.
where,
q = heat absorbed = ?
= specific heat of water =
m = mass of water = 274 g
= final temperature of water = 325.8 K
= initial temperature of metal = 298 K
Now put all the given values in the above formula, we get:
Thus, the heat released during the neutralization = -31.84 KJ
Now we have to calculate the enthalpy of neutralization.
where,
= enthalpy of neutralization = ?
q = heat released = -31.84 KJ
n = number of moles used in neutralization = 0.3562 mole
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 89.39 KJ/mole
Answer:
The unit cell edge length for the alloy is 0.405 nm
Explanation:
Given;
concentration of Ag, = 78 wt%
concentration of Pd, = 22 wt%
density of Ag = 10.49 g/cm³
density of Pd = 12.02 g/cm³
atomic weight of Ag, = 107.87 g/mol
atomic weight of iron, = 106.4 g/mol
Step 1: determine the average density of the alloy
Step 2: determine the average atomic weight of the alloy
Step 3: determine unit cell volume
for a FCC crystal structure, there are 4 atoms per unit cell; n = 4
Step 4: determine the unit cell edge length
Vc = a³
= 0.405 nm
Therefore, the unit cell edge length for the alloy is 0.405 nm