The effluent flow in concentration and particulate mass flow will be 0.198m³/sec.
<h3>How to calculate the effluent flow?</h3>
It should be noted that the total inflow will be equal to the total outflow. Therefore,
0.2 + 0.048 = 0.05 + We
Collect like terms
Qe = 0.2 + 0.048 - 0.05
Qe = 0.198m³/sec
The concentration will be:
= (360 × 1000)/0.05
= 7200mg/L.
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Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
= 58.14g/mol
Moles=14.5g / 58.14g/mol
=0.249
Therefore there are approx 0.249 moles in a 14.5g sample of C4H10
C. a long chain of carbon atoms with hydrogen atoms attached to them.
Explanation:
Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
