Answer C
I could be wrong
Explanation
Answer:
102.26 moles of helium were required to Fill the Goodyear Blimp
Explanation:
To solve this question we need to use combined gas law:
PV = nRT
<em>Where P is pressure, V is volume of gas (2500L), n are moles of gas (Our incognite), R is gas constant (0.082atmL/molK) and T is absolute temperature</em>
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Assuming atmospheric condition we can write P = 1atm and T = 25°C = 298.15K
Replacing:
PV/RT = n
1atm*2500L / 0.082atmL/molK*298.15K = n
<h3>102.26 moles of helium were required to Fill the Goodyear Blimp</h3>
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There are TWO atoms in one molecule of hydrogen.
The question can be changed into a new form:
Which element has the most negative electron affinity, or attraction for electrons? halogens have the highest electron affinities, and thus are more attracted to the electrons in the Hydrogen atom than any element in their respective periods.
In this case all the following choices are in the same period, thus Cl or Chlorine is the answer as it is a halogen.
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.
