Answer:
M KIO3 = 1.254 mol/L
Explanation:
∴ w KIO3 = 553 g
∴ mm KIO3 = 214.001 g/mol
∴ volumen sln = 2.10 L
⇒ mol KIO3 = (553 g)×(mol/210.001 g) = 2.633 mol
⇒ M KIO3 = (2.633 mol KIO3 / (2.10 L sln)
⇒ M KIO3 = 1.254 mol/L
Answer: The correct answer is D. 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit.
Explanation:
Conversion of degree Celsius to Kelvin :
K=^oC+273
Conversion of degree Celsius to degrees Fahrenheit :
^oF=(\frac{9}{5}\times ^oC)+32
By using these two conversion factors, we get the three temperature readings all mean the same thing.
For option A :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option B :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option C :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
For option D :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
From the given options, only option (D) is correct.
Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
Hope this helps!
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
There’s lots of measurements. (m, kg, s, mol, cm, in, mm) etc
Answer:
CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4
Explanation:
