Answer:
4.68x10²⁵ ions of Na⁺
Explanation:
First of all, we dissociate the salt:
NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
An aqueous solution of NaCl dissociates in chlorides anions and sodium cations. Ratio is 1:1, per 1 mol of NaCl, we have 1 mol of Na⁺
We determine the moles of salt: 4543.3 g . 1mol / 58.45 g = 77.7 moles
77.7 moles are the amount of NaCl, therefore we have 77.7 moles of Na⁺.
We count the ions:
1 mol fo Na⁺ has 6.02x10²³ ions
77.7 moles of Na⁺ must have (77.7 . 6.02x10²³) / 1 = 4.68x10²⁵ ions of Na⁺
Answer:
english pls so i can answer
Before proceeding, we should write the reaction equation to better understand what is happening:
2AgNO₃ + Na₂S → Ag₂S + 2NaNO₃
Now, we may apply the law of conservation of mass, due to which the total mass before a chemical reaction is equivalent to the total mass after a chemical reaction. Therefore:
Mass of silver nitrate + mass of sodium sulfide = mass of silver sulfide + mass of sodium nitrate
Mass of silver nitrate + 156.2 = 595.8 + 340
Mass of silver nitrate = 779.6 grams
40 drops of blood in a tube that holds 2 mL
Answer:
2.7 × 10⁻⁴ bar
Explanation:
Let's consider the following reaction at equilibrium.
SbCl₅(g) ⇄ SbCl₃(g) + Cl₂(g)
The pressure equilibrium constant (Kp) is 3.5 × 10⁻⁴. We can use these data and the partial pressures at equilibrium of SbCl₅ and SbCl₃, to find the partial pressure at equilibrium of Cl₂.
Kp = pSbCl₃ × pCl₂ / pSbCl₅
pCl₂ = Kp × pSbCl₅ / pSbCl₃
pCl₂ = 3.5 × 10⁻⁴ × 0.17 / 0.22
pCl₂ = 2.7 × 10⁻⁴ bar