Answer:
CO
Explanation:
Henry's law constant reflects the solubility of a gas in water. The larger the kH, the more soluble is the gas. There is a rule that states that "like dissolves like", meaning polar is soluble in polar and nonpolar is soluble in nonpolar. Since water is polar, we have to consider the nature of these gases.
<em>Xe</em> nonpolar
<em>Cl₂</em> nonpolar
<em>CO</em> polar
<em>CO₂</em> nonpolar
<em>CH₃CH₃</em> nonpolar
CO is the only polar gas, so it has the largest kH.
Independent: Gasoline with different amounts of lead
Dependent: Amount of pollution produced
Answer;
C. unchanged rock and mineral fragments
Explanation;
A large number of landforms and features found in desert environments are formed as the result of weathering. Weathering is defined as the breakdown and deposition of rocks by weather acting in situ
The two main types of weathering which occur in deserts are Mechanical weathering, which is the disintegration of a rock by mechanical forces that do not change the rock's chemical composition and Chemical weathering, which is the decomposition of a rock by the alteration of its chemical composition.
By contrast much of the weathered debris in deserts has resulted from mechanical weathering. Chemical weathering, however, is not completely absent in deserts. Over long time spans,clays and thin soils do form.
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
It actually depends on the percentage of the concentration give. Percentages can be expressed as %mass/mass, %volume/volume or %mass/volume. To keep things simple, let's just assume that it is in %volume/volume. Thus, 13% of 520 mL is pure acid.
Volume of pure acid = 520*0.13 = 67.6 mL
