1. exercise because you are using exercise to affect the amount of concentration. 2. concentration because concentration is what is being measured
Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
m(CoCl₂) = 0,88 g.
n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
n(CoCl₂) = 0,88 g ÷ 130 g/mol
n(CoCl₂) = 0,0068 mol.
m(H₂O) = 1,62 g - 0,88 g.
m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O)
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
Answer:
Chemistry is the scientific study of the structure of substances, how they react when combined and how they behave under different situations.
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.