There are 3 ions in one formula unit of (NH4)2CO3.
Ions present in Ammonium carbonate.(NH4)2CO3
(NH4)2CO3 is called Ammonium carbonate. The formula indicates that in one mole of ammonium carbonate, There are
- Two moles of ammonium ions,NH₄⁺ with valency of +1.
- 1 mole of Carbonate ions CO₃²⁻ with valency of -2.
When 1 mole of ammonium carbonate is dissolved in water it gives
(NH₄)₂CO₃ ₊ H₂O---->2NH₄⁺ ₊ CO₃²⁻
Ions present are 2NH₄⁺ and CO₃²⁻
Learn more on (NH4)2CO3 here:brainly.com/question/2272720
Answer : The energy required is, 574.2055 KJ
Solution :
The conversions involved in this process are :
Now we have to calculate the enthalpy change or energy.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= energy required = ?
m = mass of ice = 1 kg = 1000 g
= specific heat of solid water = 
= specific heat of liquid water = 
n = number of moles of ice = 
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[1000g\times 4.18J/gK\times (0-(-10))^oC]+55.55mole\times 6010J/mole+[1000g\times 2.09J/gK\times (95-0)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B1000g%5Ctimes%204.18J%2FgK%5Ctimes%20%280-%28-10%29%29%5EoC%5D%2B55.55mole%5Ctimes%206010J%2Fmole%2B%5B1000g%5Ctimes%202.09J%2FgK%5Ctimes%20%2895-0%29%5EoC%5D)
(1 KJ = 1000 J)
Therefore, the energy required is, 574.2055 KJ
How did what affect it? What’s the subject
The force acting between two charged particles A and B is 5.2 x 105 newtons. Charges A and B are 2.4 x 102 meters apart. If the charge on particle A is 7.2 x 108 coulombs, what is the charge of particle B? (k 7 9.0 x 109 newton meters?/coulomb)
