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.
Sp2 hybridization forms 1 sigma bond and 1 pi bond.
Answer:
I say the correct answers are primary and secondary and teriary.
Explanation:
I say you are right!!
Answer:
В.
The volume will decrease.
Explanation:
<span>4,800 pounds (2,200 kg) of ammonium nitrate fertilizer, nitromethane, and diesel fuel mixture.</span>