The value of ΔG° at this temperature is -18034.18 J/mol
Calculation,
Given information
formation constant (Kf)= 1.7 × 
Universal gas constant (R) = 8.314 J/K• mol
Temperature = 25° C = 25 °C + 273 = 300 K
Formula used:
ΔG° = -RT㏑Kf
By putting the valur of R,T, Kf we get the value of ΔG°
ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 ×
ΔG° = -2494.2㏑ 1.7 × = -18034.18 J/mol
So, change in standard Gibbs's free energy is -18034.18 J/mol
Learn about formation constant
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C = 0.11 mol
V = 5.65 L
n = ???
n = C*V
n = 0.11 * 5.65
n = 0.622 mols
1 mol of CaCl2 = 40 + 2*35.5 = 111 grams
0.622 mol = x
x = 111 * 0.622
x = 69.0 grams CaCl2
Answer:
IUPAC Rules for Alkane Nomenclature
Find and name the longest continuous carbon chain.
Identify and name groups attached to this chain.
Number the chain consecutively, starting at the end nearest a substituent group.
Designate the location of each substituent group by an appropriate number and name.
Explanation:
<span>2Li⁺(aq) + Zn⁰(s) → 2Li⁰(s) + Zn²⁺(aq)
</span>2Li⁺(aq) + 2e⁻ → 2Li⁰(s)
Zn⁰(s) → Zn²⁺(aq) +2e⁻
2 electrons are transferred from atom of Zn⁰ to 2 ions of Li⁺.
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