Answer: Q=5.46 L/s
COP=2.58
Explanation:
Given that
Cp = 4.18 kJ/(kg.C
density = 1 kg/L
Heat rejected Qr= 570 kJ/min
Power in put W= 2.65 KW
From first law of thermodynamics
U = W+ q
q = Heat absorbed
U = internal energy
W = workdone
U = 570 kJ/min = 9.5 KW
9.5 = 2.65 + q
q = 6.85 KW
COP = q/W
COP = 6.58 / 2.65
COP=2.58
Lets take volume flow rate is Q
So mass flow rate of water m = ρ Q
q = m Cp ΔT
6.85 = 1 x Q x 4.18 ( 23-5)
Q=0.091 L/min
Q=5.46 L/s
D. Decreasing its temperature
Explanation:
Decreasing the temperature of the carbon dioxide gas to be dissolved in the carbonated drink will most likely increase the solubility of the gas in the drink.
Temperature has considerable effects on the solubility of gases in liquids.
- Dissolution involves the surrounding of ions by water molecules, in this case, the carbon dioxide gas is to be surrounded by the liquid beverage medium.
- Increasing pressure increases the rate at which gases are soluble. At high pressure, the gases are brought more in contact with the liquid medium.
- Decreasing temperature aids gas solubility.
- If the temperature of gases are increased, they will not want to stay in solution as they gain a high amount of kinetic energy.
- Therefore, it will increase their randomness and the urge to leave the solution.
- Decrease in temperature and increase in pressure makes gas solubility to be fast.
Learn more:
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Answer:
when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
Explanation:
When you turn down the volume of the television, you are actually reducing the intensity of the sound wave, which is directly proportional to the amplitude of the sound. Amplitude is height of the sound wave.
Therefore, when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
Explanation:
1) N₂ + O₂ → 2 NO
Kc = [NO]² / ([N₂] [O₂])
Set up an ICE table:
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_{2}&0.114&-x&0.114-x\\O_{2}&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CN_%7B2%7D%260.114%26-x%260.114-x%5C%5CO_%7B2%7D%260.114%26-x%260.114-x%5C%5CNO%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
Plug into the equilibrium equation and solve for x.
1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))
1.00×10⁻⁵ = (2x)² / (0.114 − x)²
√(1.00×10⁻⁵) = 2x / (0.114 − x)
0.00316 = 2x / (0.114 − x)
0.00361 − 0.00316x = 2x
0.00361 = 2.00316x
x = 0.00018
The volume is 1.00 L, so the concentrations at equilibrium are:
[N₂] = 0.114 − x = 0.11382
[O₂] = 0.114 − x = 0.11382
[NO] = 2x = 0.00036
2(a) Cl₂ → 2 Cl
Kc = [Cl]² / [Cl₂]
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_{2}&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CCl_%7B2%7D%262.0%26-x%262.0-x%5C%5CCl%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
1.2×10⁻⁷ = (2x)² / (2 − x)
1.2×10⁻⁷ (2 − x) = 4x²
2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²
2.4×10⁻⁷ ≈ 4x²
x² ≈ 6×10⁻⁸
x ≈ 0.000245
2x ≈ 0.00049
2(b) F₂ → 2 F
Kc = [F]² / [F₂]
![\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_{2}&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26Initial%26Change%26Equilibrium%5C%5CF_%7B2%7D%262.0%26-x%262.0-x%5C%5CF%260%26%2B2x%262x%5Cend%7Barray%7D%5Cright%5D)
1.2×10⁻⁴ = (2x)² / (2 − x)
1.2×10⁻⁴ (2 − x) = 4x²
2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²
2.4×10⁻⁴ ≈ 4x²
x² ≈ 6×10⁻⁵
x ≈ 0.00775
2x ≈ 0.0155
F₂ dissociates more, so Cl₂ is more stable at 1000 K.
