Answer:
The solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Explanation:
Step 1: Given data
The volume of water sample = 46.0 mL.
The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.
Step 2: Calculate the solubility of X in water
46.00 mL of water sample contains 0.87 g of the mineral compound X.
To calulate how many grams of the mineral compound 1.0 mL of water sample contains:
0.87 g/46.0 mL = 0.0189 g.
This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Answ burh i dont even know
Explanation: i dont even know this
Answer:
20.5torr
Explanation:
Given parameters:
V₁ = 15L
P₁ = 8.2 x 10⁴torr
V₂ = 6 x 10⁴L
Unknown:
P₂ = ?
Solution:
To solve this problem we have to apply the claims of Boyle's law.
Boyle's law is given mathematically as;
P₁ V₁ = P₂V₂
where P₁ is the initial pressure
V₁ is the initial volume
P₂ is final pressure
V₂ is final volume
8.2 x 10⁴ x 15 = P₂ x 6 x 10⁴
P₂ = 20.5torr
Answer:
0.184 atm
Explanation:
The ideal gas equation is:
PV = nRT
Where<em> P</em> is the pressure, <em>V</em> is the volume, <em>n</em> is the number of moles, <em>R</em> the constant of the gases, and <em>T</em> the temperature.
So, the sample of N₂O₃ will only have its temperature doubled, with the same volume and the same number of moles. Temperature and pressure are directly related, so if one increases the other also increases, then the pressure must double to 0.092 atm.
The decomposition occurs:
N₂O₃(g) ⇄ NO₂(g) + NO(g)
So, 1 mol of N₂O₃ will produce 2 moles of the products (1 of each), the <em>n </em>will double. The volume and the temperature are now constants, and the pressure is directly proportional to the number of moles, so the pressure will double to 0.184 atm.
Answer:
See below
Step-by-step explanation:
You won't see much happening. The solution is saturated, so the salt will fall to the bottom of the container and sit there. It will not dissolve.
However, at the atomic level, Na⁺ and Cl⁻ ions are being pulled from the surface of the crystals and going into solution as hydrated ions. At other places, Na⁺ and Cl⁻ ions are returning to the surface of the crystals.
The process is
NaCl(s) ⇌ Na⁺(aq) + Cl⁻(aq)
The rates of the forward and reverse processes are equal, so you see no net change.
