Answer:
Q = 3,534.4 lbm/s = 212,062 lbm/min
Explanation:
Mass flowrate of discharge or leakage mass flowrate (Q) is given as
Q = AC₀√(2ρgP)
A = Cross sectional Area of leakage = (πD²/4) = (π×0.7²)/4
A = 0.385 ft²
C₀ = discharge coefficient = 0.98 (For maximum discharge flow rate, the flow is turbulent with discharge coefficient within 1% of 0.98)
ρ = density of butane at 76°F = 35.771 lbm/ft³
g = acceleration due to gravity = 32.2 lbm.ft/lbf.s²
P = Gauge Pressure in the tank = (absolute pressure) - (external pressure) = 19 - 1 = 18 atm = 38091.9 lbf/ft²
Q = AC₀√(2ρgP)
Q = (0.385)(0.98)√(2×35.771×32.2×38091.9)
Q = 3,534.4 lbm/s = 212,062 lbm/min
Hope this Helps!!!
Answer:
D. Ni²⁺
Explanation:
We know at once that the answer cannot be A or C, because Ni and Cu are already in their lowest oxidation states.
The correct answer must be either B or D.
An electrolytic cell is the opposite of a galvanic cell. In the former, the reaction proceeds spontaneously. In the latter, you must force the reaction to occur.
One strategy to solve this problem is:
- Look up the standard reduction potentials for the half reaction·
- Figure out the spontaneous direction.
- Write the equation in the reverse direction.
1. Standard reduction potentials
E°/V
Cu²⁺ + 2e⁻ ⟶ Cu; 0.3419
Ni²⁺ + 2e⁻ ⟶ Ni; -0.257
2. Galvanic Cell
We reverse the direction of the more negative half cell and add.
<u>E°/V
</u>
Ni ⟶ Ni²⁺ + 2e⁻; 0.257
<u>Cu²⁺ + 2e⁻ ⟶ Cu; </u> 0.3419
Ni + Cu²⁺ ⟶ Cu + Ni²⁺; 0.599
This is the spontaneous direction.
Cu²⁺ is reduced to Cu.
3. Electrochemical cell
<u>E°/V</u>
Ni²⁺ + 2e⁻ ⟶ Ni; -0.257
<u>Cu ⟶ Cu²⁺ + 2e⁻; </u> <u>-0.3419</u>
Cu + Ni²⁺ ⟶ Ni + Cu²⁺; -0.599
This is the non-spontaneous direction.
Ni²⁺ is reduced to Ni in the electrolytic cell.
The ratio of buffer C₂H₃O₂ /HC₂H₃O₂ must you use are1:0.199 or 10:2
the ratio of buffer C₂H₃O₂ /HC₂H₃O₂ can be calculate using the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:
pH = pKa + log ([base]/[acid]
Where,
[base] = concentration of C₂H₃O₂in molarity or moles
[acid] = concentration of HC₂H₃O₂ in molarity or moles
For the sake of easy calculation, allow us to assume that:
[base] =1
[acid] = x
Therefore using equation 1,
5.44 = 4.74 + log (1 / x)
log [base / acid] = 0.7
1 / x = 5.0118
x = 0.199
The required ratio of buffer C₂H₃O₂ /HC₂H₃O₂ is 1:0.199 or 10:2
learn more about buffer ratio here brainly.com/question/4342532
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Hello!
Data:
P (pressure) = 1 atm
V (volume) = 18.5 L
T (temperature) = 300 K
n (number of mols) = ? (in mol)
R (Gas constant) = 0.082 (atm*L/mol*K)
Apply the data to the Clapeyron equation (ideal gas equation), see:
Note:
If the feedback is to be considered, the closest r
esponse is 0.751 mol Nacl
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I hope this helps. =)
