If i had to answer i would choose a hope this helps
Answer:
Increasing the temperature of the copper made the final temperature increase and decreasing the temperature of the copper made the final temperature decrease. ... How does changing the initial mass of the copper affect how much heat energy it has? The more copper, the more heat energy.
Explanation:
The mass of Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g
<h3>What mass of Ba(IO3)2 can be dissolved in 500 ml of water at 25 degrees celcius?</h3>
The Ksp of Ba(IO3)2 = 1.57 × 10^-9
Molar mass of Ba(IO3)2 = 487 g/mol?
Dissociation of Ba(IO3)2 produces 3 moles of ions as follows:

![Ksp = [Ba^{2+}]*[IO_{3}^{-}]^{2}](https://tex.z-dn.net/?f=Ksp%20%3D%20%5BBa%5E%7B2%2B%7D%5D%2A%5BIO_%7B3%7D%5E%7B-%7D%5D%5E%7B2%7D)
![[Ba(IO_{3})_{2}] = \sqrt[3]{ksp} =\sqrt[3]{1.57 \times {10}^{ - 9} } \\ [Ba(IO_{3})_{2}] = 1.16 \times {10}^{-3} moldm^{-3}](https://tex.z-dn.net/?f=%5BBa%28IO_%7B3%7D%29_%7B2%7D%5D%20%3D%20%20%5Csqrt%5B3%5D%7Bksp%7D%20%3D%3C%2Fp%3E%3Cp%3E%5Csqrt%5B3%5D%7B1.57%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%209%7D%20%7D%20%5C%5C%20%20%5BBa%28IO_%7B3%7D%29_%7B2%7D%5D%20%3D%201.16%20%5Ctimes%20%20%7B10%7D%5E%7B-3%7D%20moldm%5E%7B-3%7D)
moles of Ba(IO3)2 = 1.16 × 10^-3 × 0.5 = 0.58 × 10^-3 moles
mass of Ba(IO3)2 = 0.58 × 10^-3 moles × 487 = 2.82 g
Therefore, mass Ba(IO3)2 that can be dissolved in 500 ml of water at 25 degrees celcius is 2.82 g.
Learn more about mass and moles at: brainly.com/question/15374113
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Answer: ΔH for the reaction is -277.4 kJ
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28products%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28reactant%29%5D)
![\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCCl_4%7D%5Ctimes%20%5CDelta%20H_%7BCCl_4%7D%29%2B%28n_%7BHCl%7D%5Ctimes%20B.E_%7BHCl%7D%29%20%5D-%5B%28n_%7BCH_4%7D%5Ctimes%20%5CDelta%20H_%7BCH_4%7D%29%2Bn_%7BCl_2%7D%5Ctimes%20%5CDelta%20H_%7BCl_2%7D%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-139%29%2B%281%5Ctimes%20-92.31%29%20%5D-%5B%281%5Ctimes%20-74.87%29%2B%281%5Ctimes%20121.0%5D)

Therefore, the enthalpy change for this reaction is, -277.4 kJ