Answer: 0.3794 moles of Iodine gas are produced.
Explanation:
Volume of iodine gas produced at STP =8.5 L
At STP, the 1 mol of gas occupies volume = 22.4 L
So, 8.5 L of volume will be occupied by:
0.3794 moles of Iodine gas are produced.
0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
*-*-*-*-*-*-*-*-*-*-*-*
C = n/V
n = 0,35×0,175
n = 0,06125 mol
mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol
1 mol --------- 164g
0,06125 ---- X
X = 10,045g
To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
Answer:
ΔG = 16.218 KJ/mol
Explanation:
- dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate
∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol
∴ R = 8.314 J/K.mol
∴ T = 298 K
∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]
⇒ Q = 0.00300 / 0.100 = 0.03
⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))
⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )
