Answer:
The flocculation basin often has a number of compartments with decreasing mixing speeds as the water advances through the basin. ... This compartmentalized chamber allows increasingly larger floes to form without being broken apart by the mixing blades.
A = change in velocity / time
so change in velocity = 13-7 = 6
so 6/1 =6:
acceleration = 6m/s^2
Answer:
The answer to the question above is
The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C = 29.36 kJ
Explanation:
The given variables are
ΔHfus = 7.27 kJ/mol
Cliq = 2.16 J/g°C
Cgas = 1.29 J/g°C
Csol = 1.65 J/g°C
Tmelting = -95.0°C.
Initial temperature = -154.0°C
Final temperature = -42.0°C?
Mass of acetone = 87.1 g
Molar mass of acetone = 58.08 g/mol
Solution
Heat required to raise the temperature of solid acetone from -154 °C to -95 °C or 59 °C is given by
H = mCsolT = 87.1 g* 1.65 J/g°C* 59 °C = 8479.185 J
Heat required to melt the acetone at -95 °C = ΔHfus*number of moles =
But number of moles = mass÷(molar mass) = 87.1÷58.08 = 1.5
Heat required to melt the acetone at -95 °C =1.5 moles*7.27 kJ/mol = 10.905 kJ
The heat required to raise the temperature to -42 degrees is
H = m*Cliq*T = 87.1 g* 2.16 J/g°C * 53 °C = 9971.21 J
Total heat = 9971.21 J + 10.905 kJ + 8479.185 J = 29355.393 J = 29.36 kJ
The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C is 29.36 kJ
Answer:
- Partial pressure He = 276 torr
- Partial pressure Ar = 457 torr
- Total pressure = 733 torr
Explanation:
Assuming temperature remains constant, we can use Boyle's law to solve this problem: P₁V₁=P₂V₂.
Once the two flasks are connected and the stopock opened, the total volume is:
Now we use Boyle's law <em>twice</em>, to <u>calculate the new pressure of </u><em><u>each</u></em><u> gas</u>:
- He ⇒ 752 torr * 275 mL = P₂He * 750 mL
P₂He = 276 torr
- Ar ⇒ 722 torr * 475 mL = P₂Ar * 750 mL
P₂Ar = 457 torr
Finally we <u>calculate the total pressure</u>, adding the partial pressures:
- Total pressure = P₂He + P₂Ar = 733 torr
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