Answer:
1. Gases can be easily liquefied into very small volumes and stored in liquid form Eg in LPGA cylinders and used in homes.
2. Balloons can be easily filled with air.
Answer:
The other signal that would indicate the presence of a C= C bond appears close to 3100 
.
Explanation:
Bands that appear above 3000 are often unsaturation diagnoses suggest. The band at 3000-
3100 is characteristics for C-H stretching frequencies and normally is overlaps with the ones for alkanes because it is a band of weak intensity.
Answer:
1. ΔE = 0 J
2. ΔH = 0 J
3. q = 3.2 × 10³ J
4. w = -3.2 × 10³ J
Explanation:
The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.
The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.
P₁ × V₁ = P₂ × V₂
3.2 atm × 20.0 L = 1.6 atm × V₂
V₂ = 40 L
The work (w) can be calculated using the following expression.
w = - P . ΔV
where,
P is the external pressure for which the process happened
ΔV is the change in the volume
w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J
The change in the internal energy is:
ΔE = q + w
0 = q + w
q = - w = 3.2 × 10³ J
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:
![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
Answer:
The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.
Explanation:
The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:
H2O (s) ⇒ H2O (l)
The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:
Entropy change (ΔSsys) = ΔSproduct - ΔSreactant
= (69.9 - 47.89) J mol/K
= 22.0 J mol/K
Therefore, the value of entropy change is positive.
Now the value of entropy for surrounding ΔSsurr will be,
ΔSsurr = -ΔHfusion/T
= -6012 j/mol/273
= -22.0 J/molK
Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.
