First we will calculate free energy change:
ΔG₀ = ΔH₀ - (T * ΔS₀)
= - 793 kJ - (298 * - 0.319 kJ/K) = - 698 kJ
We know the relation between free energy change and cell potential is:
ΔG₀ = - n F E⁰ where
F = Faraday's constant = 96485 C/mol
n = 2 (given by equation that the electrons involved is 2)
ΔG₀ = - 2 x 96485 x E⁰
- 698 kJ = - 2 x 96485 x E⁰
E⁰ = (698 x 1000) / (2 x 96485) = 3.62 volts
Answer:
MnO4 + 4 H2C2O4 = Mn + 8 CO2 + 4 H2O
Answer:
- <em>Hydration number:</em> 4
Explanation:
<u>1) Mass of water in the hydrated compound</u>
Mass of water = Mass of the hydrated sample - mass of the dehydrated compound
Mass of water = 30.7 g - 22.9 g = 7.8 g
<u>2) Number of moles of water</u>
- Number of moles = mass in grams / molar mass
- molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol
- Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol
<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>
- The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g
- Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).
- Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol
<u>4) Ratio</u>
- 0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂
Which means that the hydration number is 4.
- formula for density is mass divided by volume
therefore density of butter = 10.0g divided by 11.6ml = 0.8620689 g/cm³ ≈ 0.862 g/cm³ (3sf)
Answer:
for the reaction is -186.75 J/K
Explanation:
Change in entropy (
) for the given reaction under standard condition is given by-
= ![[3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]](https://tex.z-dn.net/?f=%5B3%5Ctimes%20S_%7Brhombic%7D%5E%7B0%7D_%7B%28s%29%7D%5D%2B%5B2%5Ctimes%20S_%7BH_%7B2%7DO%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B2%5Ctimes%20S_%7BH_%7B2%7DS%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B1%5Ctimes%20S_%7BSO_%7B2%7D%7D%5E%7B0%7D_%7B%28g%29%7D%5D)
So
=
= -186.75 J/K