Answer:
Fe(NO3)3 + 3 NaOH ===》Fe(OH)3 + 3 NaNO3
9 g of hydrogen - 42 g of nitrogen
5 g of hydrogen - x g of nitrogen

The mass of nitrogen in the second sample is 23.33 g.
Answer:
if it gains or loses neutrons
Answer:
The enthalpy change during the reaction is -199. kJ/mol.
Explanation:

Mass of solution = m
Volume of solution = 100.0 mL
Density of solution = d = 1.00 g/mL

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

where,
m = mass of solution = 100 g
q = heat gained = ?
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:


Now we have to calculate the enthalpy change during the reaction.

where,
= enthalpy change = ?
q = heat gained = 2.242 kJ
n = number of moles fructose = 

Therefore, the enthalpy change during the reaction is -199. kJ/mol.
Answer:
5 g / ml
Explanation:
Convert the values given to g and ml
12.5 kg = 12500 g
2.5 L = 2500 ml
12500 g / 2500 ml = 5 g/ml