Answer:
1.) AgNO₃
2.) 0.563 moles AgBr
Explanation:
The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).
AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)
Molarity (M) = moles / liters
100 mL = 1 L
AgNO₃
45.0 mL / 100 = 45.0 L
1.25 M = ? moles / 0.450 L
? moles = 0.563 moles
KBr
75.0 mL / 100 = 0.750 L
0.800 M = ? moles / 0.750 L
? moles = 0.600 moles
In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.
Technically speaking, yes you can. Using a microscope though.
Answer:
i think Aluminum (Al) oxidized, zinc(Zn) reduced
Answer:
B. Solvent
Explanation:
In osmosis, water always moves from low solute concentration to high solute concentration. SOLUTE NEVER MOVES AS IT CANNOT PASS THE SELECTIVELY PERMEABLE MEMBRANE. alot of caps but need to stress this concept cuz otherwise this concept gets very confusing
Answer:
2.067 L ≅ 2.07 L.
Explanation:
- The balanced equation for the mentioned reaction is:
<em>CS₂(g) + 3O₂(g) → CO₂(g) + 2SO₂(g),</em>
It is clear that 1.0 mole of CS₂ react with 3.0 mole of O₂ to produce 1.0 mole of CO₂ and 2.0 moles of SO₂.
- At STP, 3.6 L of H₂ reacts with (?? L) of oxygen gas:
It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.
<u><em>using cross multiplication:</em></u>
1.0 mol of O₂ represents → 22.4 L.
??? mol of O₂ represents → 3.1 L.
∴ 3.1 L of O₂ represents = (1.0 mol)(3.1 L)/(22.4 L) = 0.1384 mol.
- To find the no. of moles of SO₂ produced from 3.1 liters (0.1384 mol) of hydrogen:
<u><em>Using cross multiplication:</em></u>
3.0 mol of O₂ produce → 2.0 mol of SO₂, from stichiometry.
0.1384 mol of O₂ produce → ??? mol of SO₂.
∴ The no. of moles of SO₂ = (2.0 mol)(0.1384 mol)/(3.0 mol) = 0.09227 mol.
- Again, using cross multiplication:
1.0 mol of SO₂ represents → 22.4 L, at STP.
0.09227 mol of SO₂ represents → ??? L.
∴ The no. of liters of SO₂ will be produced = (0.09227 mol)(22.4 L)/(1.0 mol) = 2.067 L ≅ 2.07 L.
