Answer:
In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams
Explanation:
This compound is the lead (IV) phosphate.
Grams that occupy 1 mole, means the molar mass of the compound
Pb = 207.2 .3 = 621.6 g/m
P = 30.97 .4 = 123.88 g/m
O = (16 . 4) . 4 = 256 g/m
621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m
Because they believed that heavenly bodies’ circular motions where caused by their being attached to unseen resolving solid spheres
Answer:
a. 1.23 V
b. No maximum
Explanation:
Required:
a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?
b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?
The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E°cell = E°red, cat - E°red, an
If E°cell must be at least 1.10 V (E°cell > 1.10 V),
E°red, cat - E°red, an > 1.10 V
E°red, cat - 0.13V > 1.10 V
E°red, cat > 1.23 V
The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.
Answer:
[Cl⁻] = 0.016M
Explanation:
First of all, we determine the reaction:
Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓ + Mg(NO₃)₂(aq)
This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:
PbCl₂(s) ⇄ Pb²⁺ (aq) + 2Cl⁻ (aq) Kps
Initial x
React s
Eq x - s s 2s
As this is an equilibrium, the Kps works as the constant (Solubility product):
Kps = s . (2s)²
Kps = 4s³ = 1.7ₓ10⁻⁵
4s³ = 1.7ₓ10⁻⁵
s = ∛(1.7ₓ10⁻⁵ . 1/4)
s = 0.016 M
Answer:
amino acids and urea
Explanation:
Amino acids are the building blocks of all proteins. Proteins comprise not only structural components such as muscle, tissue and organs, but also enzymes and hormones essential for the functioning of all living things. Urea is a byproduct of protein digestion.
