N = 4 to n = 3 is the right answer, so it' none of the above
Answer:
It is just sugar decomposing to a dark tar looking goo.
Explanation:
Cite: What happens after you boil kool-aid
For the answer to the questions above,
a) Ag2CO3(s) => Ag2O(s)+CO2(g)
<span>b) Cl2(g)+2(KI)(aq) => I2(s)+2(KCl)(aq) (coefficients are for balanced equation) </span>
<span>net ionic is Cl2(g)+2I- => I2(s)+2Cl-(aq) </span>
<span>c) I2(s)+3(Cl2)(g)=>2(ICl3)
</span>I hope I helped you with your problem
The molar mass of NH4NO3 in g/mol is 80g/mol.
HOW TO CALCULATE MOLAR MASS:
The molar mass of a compound can be calculated by summing the atomic masses of its constituent elements.
In ammonium nitrate (NH4NO3), there are nitrogen, hydrogen, and oxygen elements.
- Atomic mass of nitrogen = 14
- Atomic mass of oxygen = 16
- Atomic mass of hydrogen = 1
Molar mass of NH4NO3 = 14 + 1(4) + 14 + 16(3)
Molar mass of NH4NO3 = 80g/mol
- Therefore, the molar mass of NH4NO3 in g/mol is 80g/mol.
Learn more about molar mass at: brainly.com/question/8101390?referrer=searchResults
Answer: -227 kJ
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+ n_{H_2O}\times \Delta H_{H_2O})]-[(n_{C_2H_2}\times \Delta H_{C_2H_2})+(n_{O_2}\times \Delta H_{O_2})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%20n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BC_2H_2%7D%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-1255.8=[(2\times -393.5)+(1\times -241.8)]-[(1\times \Delta H_{C_2H_2})+(\frac{5}{2}\times 0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%282%5Ctimes%20-393.5%29%2B%281%5Ctimes%20-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%28%5Cfrac%7B5%7D%7B2%7D%5Ctimes%200%29%5D)
![-1255.8=[(-787)+(-241.8)]-[(1\times \Delta H_{C_2H_2})+(0)]](https://tex.z-dn.net/?f=-1255.8%3D%5B%28-787%29%2B%28-241.8%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_%7BC_2H_2%7D%29%2B%280%29%5D)

Therefore, the enthalpy change for
is -227 kJ.