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:
c and o
Explanation:
Carbon and Oxygen cannot react to form an ionic compound because the two elements are non-metals.
- To form an ionic bond, a metal combines with a non-metal through electrostatic attraction of oppositely charged ions.
- This leads the bond formation.
- Carbon is a non-metal, with oxygen, they will prefer to form a covalent bond where they share their electrons
To solve this problem, we should recall that
the change in enthalpy is calculated by subtracting the total enthalpy of the reactants
from the total enthalpy of the products:
ΔH = Total H of products – Total H of reactants
You did not insert the table in this problem, therefore I
will find other sources to find for the enthalpies of each compound.
ΔHf CO2 (g) = -393.5 kJ/mol
ΔHf CO (g) = -110.5 kJ/mol
ΔHf Fe2O3 (s) = -822.1 kJ/mol
ΔHf Fe(s) = 0.0 kJ/mol
Since the given enthalpies are still in kJ/mol, we have to
multiply that with the number of moles in the formula. Therefore solving for ΔH:
ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0
kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>
ΔH = <span>795.2
kJ</span>
Answer:
92.344mL
Explanation:
acording to boyle's law that PV=constant then P1V1=P2V2
