Answer:
F G there you go the guy on top is right
Explanation:
F G
The chemical reaction is expressed as:
3Ba(NO3)2 + 2Na3PO4 = Ba3(PO4)2 + 6NaNO3
To determine the percent yield, we need to determine the theoretical yield of the reaction from the given amounts of the reactants. We do as follows:
0.3 mol 3Ba(NO3)2 ( 2 mol Na3PO4 / 3 mol Ba(NO3)2) = 0.2 mol Na3PO4
Therefore, the limiting reactant would be Ba(NO3)2 since it is consumed completely in the reaction.
Theoretical yield = 0.3 mol 3Ba(NO3)2 ( 1 mol Ba3(PO4)2 / 3 mol Ba(NO3)2) = 0.1 mol Ba3(PO4)2
Percent yield = actual yield / theoretical yield = 0.095 mol Ba3(PO4)2 / 0.1 mol Ba3(PO4)2 x 100 = 95%
Answer:
The SAE curriculum includes practical farming tasks conducted outside the scheduled classroom and laboratory period by students. SAEs offer a method for students in agricultural education to gain real-world work opportunities that they are most interested in in the field of agriculture. Supervised agricultural experience is an essential component of agricultural education, and all Agriculture, Food and Natural Resources (AFNR) courses are a necessary component.
Explanation: Hope it helps
<span>Cu + 2 AgNO</span>₃<span> = 2 Ag + Cu(NO</span>₃<span>)</span>₂
mole ratio:
1 mole Cu --------------> 2 moles Ag
? mole Cu --------------> 3.50 moles Ag
moles Cu = 3.50 x 1 / 2
moles Cu = 3.50 / 2
= 1.75 moles
Answer B
hope this helps!
