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:
3.2 g O₂
Explanation:
To find the mass of O₂, you need to (1) convert grams H₂O to moles H₂O (via molar mass), then (2) convert moles H₂O to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given value (3.6 g).
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
2 H₂O -----> 2 H₂ + 1 O₂
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
3.6 g H₂O 1 mole 1 mole O₂ 31.996 g
---------------- x --------------- x --------------------- x --------------- = 3.2 g O₂
18.014 g 2 moles H₂O 1 mole
Since there is one carbon with 4 Fluorines attached to it, and both compounds are no metals, we use the covalent method for naming,
Here we ignore the prefix for the first element if it is 1. Mono. Then pay attention to the second one, it would be tetra, because tetra means 4. Here there are 4 fluorines.
Drop ine and place ide
CF4 = carbon tetrafluoride.
Answer:
the measured amount of product that is made from a given amount of reactant
Explanation:
google & quizlet
The answer is D. Okay l hope this helps
