Answer:
Since Q > Ksp, a precipitate of AgCl will form.
Explanation:
Step 1: Data given
Molarity AgNO3 = 0.10 M
Molarity of NaCl = 0.075 M
Ksp AgCl = 1.77 * 10^-10
Step 2: The balanced equation
AgNO3 + NaCl → AgCl(s) + NaNO3(aq)
For 0.10 moles AgNO3 we have 0.10 moles Ag+ ( molarity = 0.10 M)
For 0.075 moles NaCl we have 0.075 moles Cl- (molarity = 0.075 M)
Step 3: Calculate Q
Q = [Ag+][Cl-]
Q = (0.10 M )(0.075 M ) = 0.0075
Ksp = 1.77 * 10^-10
Q >>> Ksp
Since Q > Ksp, a precipitate of AgCl will form.
I think the answer is B.
I know the answer cant be A because a lot of metal that is new can be shiny.
C is incorrect because malleable means that it can be permanently bent out of shape ( which metal can be )
D is incorrect because a lot of metal can conduct electricity for example copper and brass can both conduct electricity
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Answer: 25.89g of O2 (g)
Explanation:
We begin by finding the molar mass of compounds of interest in the given equation. 2 KClO3 ⟶ 2 KCl + 3 O 2
1 mole of KCLO3 = 39 .10 + 35.5 + (3x16) = 122.55g, 2Moles would be 2 x 122.5 = 245.1g
1 mole Oxygen gas = 32g, 3 moles would be, 3 x 32= 96g
According to the equation;
245.1g of KCLO3 produces 96g of O2 (g)
If 1g of KCLO3 produces 96÷245.1 of O2 (g) from heating.
66.1g of KCLO3 would produce; 96÷245.1 x 66.1 = 25.89g of O2 (g) from heating.
Reaction on positive electrode (cathode):
PbSO₄₍s₎ + 2H₂O₍l₎ → 2e⁻ + PbO₂₍s₎ + 4H⁺₍aq₎ + SO₄²⁻₍aq₎.
s - solid.
l - liquid.
aq - dissolve in water.
PbSO4 is converted to Pb at one electrode (anode) and to PbO₂ at the other (cathode). Lead battery can be recharged, during recharging, an external source of energy is used.