Answer:
Explanation:
Entropy is measure of disorder so as we lower the temperature of gas , its entropy decreases .
Hence at - 200°C entropy of nitrogen will be less than that at - 190°C .
At freezing point ,
entropy of fusion = latent heat / freezing temperature
= .71 kJ / ( 273 - 210 )
= 710 / 63 J mol⁻¹ K⁻¹ .
= 11.27 J mol⁻¹ K⁻¹ .
entropy of fusion = 11.27 J mol⁻¹ K⁻¹ .
According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
= 7.55 - 1.95 = 5.6
Answer:
127°C
Explanation:
This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.
We convert our value to K → -73°C + 273 = 200 K
The moles are the same, and the volume is also the same:
P₁ / T₁ = P₂ / T₂
But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂
P₁ / 200K = 2P₁ / T₂
1 /2OOK = (2P₁ / T₂) / P₁
See how's P₁ term is cancelled.
200K⁻¹ = 2/ T₂
T₂ = 2 / 200K⁻¹ → 400K
We convert the T° to C → 400 K - 273 = 127°C
Density, Volume and Mass
3. A metal weighing 7.101 g is placed in a graduated cylinder containing 33.0 mL of water. The water
level rose to the 37.4 mL mark.
a) Calculate the density of the metal (in g/mL).
b) If you were to do this with an equal mass of aluminum (d = 2.7 g/mL), how high would the water rise?
Answer:
2.2 % and 0 %
Explanation:
The equation we will be using to solve this question is:
N/N₀ = e⁻λ t
where N₀ : Number of paricles at t= 0
N= Number of particles after time t
λ= Radioactive decay constant
e= Euler´s constant
We are not given λ , but it can be determined from the half life with the equation:
λ = 0.693 / t 1/2 where t 1/2 is the half-life
Substituting our values:
λ = 0.693 / 55 s = 0.0126/s
a) For t = 5 min = 300 s
N / N₀ = e^-(0.0126/s x 300 s) = e^-3.8 = 0.022 = 2.2 %
b) For t = 1 hr = 3600 s
N / N₀ = e^-(0.0126/s x 3600 s) = 2.9 x 10 ⁻²⁰ = 0 % (For all practical purposes)
