Answer:
λ = 1×10²⁶m
Explanation:
Given data:
Wavelength of radiation = ?
Frequency of radiation = 3×10⁻¹⁸Hz
Solution:
Formula:
c = f × λ
c = speed of wave = 3×10⁸ m/s
by putting values,
3×10⁸ m/s = 3×10⁻¹⁸Hz × λ
λ = 3×10⁸ m/s / 3×10⁻¹⁸s⁻¹
λ = 1×10²⁶m
2C4H10 + 13O2 = 8CO2 + 10H2O
1. (2.06g C4H10)/(58.12 g/mol C4H10) = 0.035mol C4H10
2. (0.035molC4H10)(10 mol H2O/2mol C4H10) = 0.177mol H2O
3. (0.177mol H2O)(18.01g/mol H2O) = 3.19g H2O
Answer: option D) energy was absorbed and entropy increased.
Explanation:
1) Given balanced equation:
2H₂O (l) + 571.6 kJ → 2 H₂ (g) + O₂(g).
2) Being the energy placed on the side of the reactants means that the energy is used (consumed or absorbed). This is an endothermic reaction.
So, the first part is that energy was absorbed.
3) As for the entropy, it is a measure of the disorder or radomness of the system.
Since, two molecules of liquid water were transformed into three molecules of gas, i.e. more molecules and more kinetic energy, therefore the new state has a greater degree of radomness, is more disordered, and you conclude that the entropy increased.
With that, you have shown that the right option is D) energy was absorbed and increased.
Answer:
-1 Coulomb meter = -2.997 × 10²⁹ Debye
Explanation:
Given:
Coulomb meter = -1 CM
Find:
In debye
Computation:
We know that,
1 Coulomb meter = 299,792,458,178,090,000,000,000,000,000 Debye
So,
-1 Coulomb meter = -299,792,458,178,090,000,000,000,000,000 Debye
-1 Coulomb meter = -2.997 × 10²⁹ Debye
Answer:
a) 88.48%
b) 0.05625 mol
Explanation:
2CH₃CH₂OH(l) → CH₃CH₂OCH₂CH₃(l) + H₂O(g) Reaction 1
CH₃CH₂OH(l) → CH₂═CH₂(g) + H₂O(g) Reaction 2
a) CH₃CH₂OH = 46.0684 g/mol
CH₃CH₂OCH₂CH₃ = 74.12 g/mol
1 mol CH₃CH₂OH ______ 46.0684 g
x ______ 50.0 g
x = 1.085 mol CH₃CH₂OH
1 mol CH₃CH₂OCH₂CH₃ ______ 74.12 g g
y ______ 35.9 g
y = 0.48 mol CH₃CH₂OCH₂CH₃
100% yield _____ 0.5425 mol CH₃CH₂OCH₂CH₃
w _____ 0.48 mol CH₃CH₂OCH₂CH₃
w = 88.48%
b) Only 0.96 mol of ethanol reacted to form diethyl ether. This means that 0.125 mol of ethanol did not react. 45% of 0.125 mol reacted to form ethylene. Therefore, 0.05625 mol of ethanol reacted by the side reaction (reaction 2). Since 1 mol of ethanol leads to 1 mol of ethylene, 0.05625 mol of ethanol produces 0.05625 mol of ethylene.