Answer:
The actual free energy = 78.54 X 10⁻² J
Explanation:
Given standard free energy = 23.8 kJ/mol
Free energy due to the presence of fructose 1,6-bisphosphate (1.4 X 10⁻⁵ M)
= 1.4 X 10⁻⁵ M * (23800 J/mol)
= 33.32 X 10⁻² J
Free energy due to the presence of glyceraldehyde 3-phosphate (3 X 10⁻⁶ M)
= 3 X 10⁻⁶ M * (23800 J/mol)
=7.14 X 10⁻² J
Free energy due to the presence of dihydroxyacetone phosphate (1.6 x 10⁻⁵M)
= 1.6 x 10⁻⁵M * (23800 J/mol)
= 38.08 X 10⁻² J
The actual free = 33.32 X 10⁻² J + 7.14 X 10⁻² J + 38.08 X 10⁻² J
The actual free energy = 78.54 X 10⁻² J
Answer:
1.51834×10^22
Explanation:
Number of atom=Mass/Molar Mass × avogadros constant
2.3g/91.224g/mol × 6.203×10^23
Answer:
7.92gml-1
Explanation:
water=25.20ml
water+iron=25.92ml
iron=5.7g
P=mass/volume (formula of density)
mass=5.7g
volume=25.92-25.20
=0.72ml
p=5.7/0.72
=7.92gml-1
Answer:
12.07 g.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>2NH₃(g) + CO₂(g) → H₂NCONH₂(g) + H₂O(g),</em>
It is clear that 2.0 moles of NH₃ react with 1.0 mole of CO₂ to produce 1.0 mole of H₂NCONH₂ and 1.0 moles of H₂O.
- Consider the reaction proceeds at STP conditions:
At STP, 9.0 L of NH₃ react with an excess of CO₂ gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of NH₃ represents → 22.4 L.
??? mol of NH₃ represents → 9.0 L.
∴ 9.0 L of NH₃ represents = (1.0 mol)(9.0 L)/(22.4 L) = 0.4018 mol.
- To find the no. of moles of urea (H₂NCONH₂) produced:
<u><em>Using cross multiplication:</em></u>
2.0 mol of NH₃ produce → 1.0 mol of H₂NCONH₂, from stichiometry.
0.4018 mol of NH₃ produce → ??? mol of H₂NCONH₂.
∴ The no. of moles of H₂NCONH₂ = (1.0 mol)(0.4018 mol)/(2.0 mol) = 0.201 mol.
- Now, we can find the mass of H₂NCONH₂ produced:
<em>mass = n * molar mass</em> = (0.201 mol) * (60.06 g/mol) = <em>12.07 g.</em>