<span>12810 atomic mass units
Since a monomer is the basic building block of a polymer, we just need to multiply the mass of the monomers by the number of monomers used. So
105 atomic mass units * 122 = 12810 atomic mass units</span>
Answer:
We'll have 1 mol Al2O3 and 3 moles H2
Explanation:
Step 1: data given
Numer of moles of aluminium = 2 moles
Number of moles of H2O = 6 moles
Step 2: The balanced equation
2Al + 3H2O → Al2O3 + 3H2
Step 3: Calculate the limiting reactant
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
Aluminium is the limiting reactant. It will completely be consumed (2 moles).
H2O is in excess. There will react 3/2 * 2 = 3 moles
There will remain 6 - 3 = 3 moles
Step 4: Calculate moles products
For 2 moles Al we need 3 moles H2O to produce 1 mol Al2O3 and 3 moles H2
For 2 moles Al we'll have 2/1 = 1 mol Al2O3
For 2 moles Al We'll have 3/2 * 2 = 3 moles H2
We'll have 1 mol Al2O3 and 3 moles H2
45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.
a) Find the <em>length in millimetres</em>
Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing
b) Convert <em>millimetres to metres
</em>
Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
a) K = 5.3175
b) ΔG = 3.2694
Explanation:
a) ΔG° = - RT Ln K
∴ T = 25°C ≅ 298 K
∴ R = 8.314 E-3 KJ/K.mol
∴ ΔG° = - 4.140 KJ/mol
⇒ Ln K = - ( ΔG° ) / RT
⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))
⇒ Ln K = 1.671
⇒ K = 5.3175
b) A → B
∴ T = 37°C = 310 K
∴ [A] = 1.6 M
∴ [B] = 0.45 M
∴ K = [B] / [A]
⇒ K = (0.45 M)/(1.6 M)
⇒ K = 0.28125
⇒ Ln K = - 1.2685
∴ ΔG = - RT Ln K
⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )
⇒ ΔG = 3.2694
