Make a ball of clay and embed small beads throughout it. The plum pudding model.
Answer:
V ∝ n
Step-by-step explanation:
Suppose that pressure and temperature are constant.
If you try to force more molecules of air into a balloon, the balloon will expand.
This is an example of <em>Avogadro's Law</em>: the volume of a gas is directly proportional to the number of moles (particles).
V ∝ n
Answer:
Yes
Explanation:
Denatured ethanol fuel is a polar solvent, which is soluble in water. A
Polar solvent is a compound with a charge separation in chemical bonds, such as alcohol, most acids, or ammonia. These have affinity with water and will dissolve easily. Denatured fuel ethanol has a flash point of -5 ° F and a vapor density of 1.5, indicating that it is heavier than air.
Consequently, ethanol vapors do not rise, similar to the gasoline vapors they are looking for lower altitudes. The specific gravity of denatured fuel ethanol is 0.79, which indicates that it is lighter than water and has a self-ignition temperature of 709 ° F and a boiling point of 165-175 ° F. Like gasoline, the most denatured fuel, the greatest danger of ethanol as an engine fuel component is its flammability.
It has a wider flammable range than gasoline (LEL is 3% and UEL is 19%).
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
B. Rotten orange is the correct answer. Hope this helps!
