PH= −log
10
[H
+
]
= −log
10
(0.001)
= −log
10
(10
−3
)
= −(−3)log
10
10
pH=3.
01
Explanation:
substance Q could be <em><u>oxygen (O2)</u></em>
substance R could be <em><u>carbon</u></em><em><u> </u></em><em><u>d</u></em><em><u>i</u></em><em><u>o</u></em><em><u>x</u></em><em><u>i</u></em><em><u>d</u></em><em><u>e</u></em><em><u> </u></em><em><u>(</u></em><em><u>C</u></em><em><u>O</u></em><em><u>2</u></em><em><u>)</u></em>
Answer:
Physical properties include: appearance, texture, color, odor, melting point, boiling point, density, solubility, polarity, and many others. That is your answer! Thanks! :)
Explanation:
Some
of the solutions exhibit
colligative properties. These properties depend on the amount of solute
dissolved in a solvent. These properties include freezing point depression, boiling
point elevation, osmotic pressure and vapor pressure lowering. Calculations
are as follows:
<span>
ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>
Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
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