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:
Amino >Methoxy > Acetamido
Explanation:
Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.
The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.
The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.
Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.
Therefore, decreasing order will be as follows:
Amino >Methoxy > Acetamido
Answer:
328.1 K.
Explanation:
- To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT</em>.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in.
- If n is constant, and have two different values of (P, V and T):
<em>P₁V₁T₂ = P₂V₂T₁</em>
<em></em>
P₁ = 1.0 atm (standard P), V₁ = 72.1 L, T₁ = 25°C + 273 = 298 K (standard T).
P₂ = 93.6 kPa = 0.924 atm, V₂ = 85.9 L, T₂ = ??? K.
<em>T₂ = P₂V₂T₁/P₁V₁ = </em>(0.924 atm)(85.9 L)(298 K)/(1.0 atm)(72.1 L) <em>= 328.1 K.</em>
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Part (a) :
H₂(g) + I₂(s) → 2 HI(g)
From given table:
G HI = + 1.3 kJ/mol
G H₂ = 0
G I₂ = 0
ΔG = G(products) - G(reactants) = 2 (1.3) = 2.6 kJ/mol
Part (b):
MnO₂(s) + 2 CO(g) → Mn(s) + 2 CO₂(g)
G MnO₂ = - 465.2
G CO = -137.16
G CO₂ = - 394.39
G Mn = 0
ΔG = G(products) - G(reactants) = (1(0) + 2*-394.39) - (-465.2 + 2*-137.16) = - 49.3 kJ/mol
Part (c):
NH₄Cl(s) → NH₃(g) + HCl(g)
ΔG = ΔH - T ΔS
ΔG = (H(products) - H(reactants)) - 298 * (S(products) - S(reactants))
= (-92.31 - 45.94) - (-314.4) - (298 k) * (192.3 + 186.8 - 94.6) J/K
= 176.15 kJ - 84.78 kJ = 91.38 kJ
The theory of biological evolution known as Darwin's theory of evolution or Darwinism given by Charles Darwin an English naturalist. According to the theory, all the species of organisms develop and originate via the natural selection of minor, inherited changes, which enhances the tendency of an individual to survive, compete, and develop the tendency to give rise to new ones.
Thus, according to Darwin's theory of evolution in the given case of black and orange snakes, the black snakes will survive and reproduce, therefore, passing their traits to their offspring, while few orange snakes will remain in the population.
