Answer: 
Explanation:
Volume of metal = volume of water displaced = (30.0 - 25.0) ml = 5.0 ml
Density of metal = 5.50 g/ml
Mass of metal =
Volume of water = 25.0 ml
Density of metal = 1.0 g/ml
Mass of metal =

As we know that,

.................(1)
where,
q = heat absorbed or released
= mass of metal = 27.5 g
= mass of water = 25.0 g
= final temperature = ?
= temperature of metal = 
= temperature of water = 
= specific heat of lead = ?
= specific heat of water= 
Now put all the given values in equation (1), we get
![27.5g\times c_1\times (41.0-153)^0C=[25.0g\times 4.814J/g^0C\times (41.0-25.0)^0C]](https://tex.z-dn.net/?f=27.5g%5Ctimes%20c_1%5Ctimes%20%2841.0-153%29%5E0C%3D%5B25.0g%5Ctimes%204.814J%2Fg%5E0C%5Ctimes%20%2841.0-25.0%29%5E0C%5D)

Thus the specific heat of the unknown metal sample is 
Answer: Option (A) is the correct answer.
Explanation:
The name of
ion is hydroxide.
Name of
ion is phosphite.
Name of
ion is phosphate.
Name of
ion is ammonia.
Name of
ion is acetate.
Name of
ion is oxalate.
Thus, we can conclude that the symbols and names for the ions given correctly is OH-:hydroxide