Answer:
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
Explanation:
2 NO (g) + O₂ (g) ⇄ 2NO₂ (g)
Let's apply the thermodynamic formula to calculate the ΔG
ΔG = ΔG° + R .T . lnQ
We don't know if the gases are at equilibrium, that's why we apply Q (reaction quotient)
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln Q
How can we know Q? By the partial pressures (Qp)
P NO = 0.450atm
PO₂ = 0.1 atm
PNO₂ = 0.650 atm
Qp = [NO₂]² / [NO]² . [O₂]
Qp = 0.650² / 0.450² . 0.1 = 20.86
ΔG = - 69 kJ/mol + 8.31x10⁻³ kJ/K.mol . 298K . ln 20.86
ΔG = -61.5 kJ/mol (<u>Spontaneous process</u>)
Explanation:
The physical properties of organic compounds typically of interest include both quantitative and qualitative features. Quantitative information includes a melting point, boiling point, and index of refraction. Qualitative properties include odor, consistency, solubility, and color.
Answer:
92gm
Explanation:
Atomic mass of Mg=24g=1 mole of Mg
∴ 24g =1 mole of Mg contain 6.022×10^23 atom
∴ 6gm contains 246.022×1023×6
=4×6.022×10^23 atoms
Now according to question, there are 6.022×1023 atoms of Na
23gm of Na contains 6.022×10^23 atoms
∴6.022×4×10^23 atoms of Na weighs 23×6.022×10^23×4/6.022×10^23⇒92gm
Answer:
449730.879 cal/g
Explanation:
Given data:
Mass of sample = 4.9 g
Change in temperature = 2.08 °C (275.23 k)
Heat capacity of calorimeter = 33.50 KJ . K⁻¹
Solution:
C(candy) = Q/m
Q = C (calorimeter) × ΔT
C(candy) = C (calorimeter) × ΔT / m
C(candy) = 33.50 KJ . K⁻¹ × 275.23 K / 4.90 g
C(candy) = 9220.205 KJ / 4.90 g
C(candy) = 1881.674 KJ / g
It is known that,
1 KJ /g = 239.006 cal/g
1881.674 × 239.006 = 449730.879 cal/g