Answer:
Q= 245 =2.5 * 10^2
Explanation:
ΔG = ΔGº + RTLnQ, so also ΔGº= - RTLnK
R= 8,314 J/molK, T=298K
ΔGº= - RTLnK = - 6659.3 J/mol = - 6.7 KJ/mol
ΔG = ΔGº + RTLnQ → -20.5KJ/mol = - 6.7 KJ/mol + 2.5KJ/mol* LnQ
→ 5.5 = LnQ → Q= 245 =2.5 * 10^2
Answer is: a lower freezing point has solution of K₂SO₄.
Change in freezing
point from pure solvent to solution: ΔT =i · Kf · b.<span>
Kf - molal freezing-point depression constant for water is 1.86°C/m.
b - molality, moles of solute per
kilogram of solvent.
i - </span>Van't
Hoff factor.<span>
b(K</span>₂SO₄<span>) = 0.35 m.
</span>b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.
It glow, so light energy go out of the system, exotermic
Answer:
Explanation:
As per Boltzman equation, <em>kinetic energy (KE)</em> is in direct relation to the <em>temperature</em>, measured in absolute scale Kelvin.
Then, <em>the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be </em><em>3 times</em><em> such temperature.</em>
So, you must just convert the given temperature, 32°F, to kelvin scale.
You can do that in two stages.
- First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80
- Second, convert 0°C to kelvin:
T (K) = T(°C) + 273.15 K= 273.15 K
Then, <u>3 times</u> gives you: 3 × 273.15 K = 819.45 K
Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.
Answer:
D. 91.98K
Explanation:
The General Gas Law equation is given by,
From the question,
the initial pressure,
the initial volume,
the final temperature,
the final pressure,
the final volume,
Making
the subject of the expression, we obtain
By substitution,
Hence the initial temperature was 91.98 K
