Answer:
The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. ... The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.
Explanation:
The limiting reagent is the reactant that is completely used up in a reaction, and thus determines when the reaction stops. ... The limiting reagent is the one that is totally consumed; it limits the reaction from continuing because there is none left to react with the in-excess reactant.
Answer:
D, Li2S
Explanation:
This is because Lithium, which is in group IA of the periodic table, has a charge of +1. Sulfur will have a charge of -2 because it is in group 6A in the periodic table, which means to balance these out, there needs to be 2 lithium ions which would result in a charge of +2. With Lithium now having a charge of +2 due to having two atoms in the compound, and sulfur already having a charge of -2 as one atom, these two cancel out meaning the compound is neutral.
Answer:
0.581 L or 581 mL
Explanation:
As stated in the question, the combined gas law is (P1*V1/T1) = (P2*V2/T2)
Write down the amounts you are given.
V1 = 0.152 L (I was taught to always convert milliliters to liters)
P1 = 717 mmHg
T1 = 315 K
V2 = ?
P2 = 463 mmHg
T2 = 777 K
The variable that is being solved for is final volume. Fill in the combined gas law equation with the corresponding amounts and solve for V2.
(717 mmHg*0.152 L) / (315 K) = (463 mmHg*V2) / (777 K)
0.346 = (463*V2) / (777)
0.346*777 = (463*V2) / (777)*777
268.842 = 463*V2
268.842/463 = (463*V2)/463
V2 = 0.581
Pressure and volume are indirectly proportional. This checks out because the volume increased while pressure decreased. Volume and temperature are directly proportional. This checks out because both volume and temperature increased. This is a good way to check your answers. You can also solve each side of the combined gas law equation to see if they are both the same.
