The ratio of aluminum bronze components is:
92.0 Cu / 8.0 Al
The ratio of Cu and Al avilable is: 73.5 Cu / 42.2 Al
Then, it is evident that Al is in excess and Cu is the limitant material.
So, you need two work with the open proportion 92.0/ 8.0 = 73.5Cu / x
=> x = 73.5 * 8 / 92 = 6.39
Then, you can use 6.4 grams of Al and 73.5 grams of Cu to prepare 6.39g + 73.5g = 79.89.grams of Bronze.
I hope this helps.
Answer:
207.24.
Explanation:
The mass numbers of the 4 isotopes are 82+ 122 = 204, 82 + 125 = 207,
82 + 124 = 206 and 208.
The approximate atomic mass is worked out by adding the products of their relative abundances and mass numbers:
0.014*204 + 0.221*207 + 0.241*206 + 0.524*208
= 207.24.
Answer:
rate = k[A][B] where k = k₂K
Explanation:
Your mechanism is a slow step with a prior equilibrium:
(The arrow in Step 1 should be equilibrium arrows).
1. Write the rate equations:
2. Derive the rate law
Assume k₋₁ ≫ k₂.
Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.
In an equilibrium, the forward and reverse rates are equal:
k₁[A][B] = k₋₁[C]
[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)
rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]
The rate law is
rate = k[A][B] where k = k₂K
Answer:
1.54 atm
Explanation:
By Dalton's Law Of partial pressure,
Total Pressure = Sum of all partial pressures
So,P= P1 + P2 + P3
Therefore, P=0.23+0.42+0.89
=1.54 atm