Answer:
Q = 14.578 m³/s
Explanation:
Given
We use the Manning Equation as follows
Q = (1/n)*A*(∛R²)*(√S)
where
- Q = volumetric water flow rate passing through the stretch of channel (m³/s for S.I.)
-
A = cross-sectional area of flow perpendicular to the flow direction, (m² for S.I.)
-
S = bottom slope of channel, m/m (dimensionless) = 2.5% = 0.025
-
n = Manning roughness coefficient (empirical constant), dimensionless = 0.023
-
R = hydraulic radius = A/P (m for S.I.) where
:
-
A = cross-sectional area of flow as defined above,
-
P = wetted perimeter of cross-sectional flow area (m for S.I.)
we get A as follows
A = (B*h)/2
where
B = 5 m (the top width of the flowing channel)
h = (B/2)*(m) = (5 m/2)*(1/2) = 1.25 m (the deep)
A = (5 m*1.25 m/2) = 3.125 m²
then we find P
P = 2*√((B/2)²+h²) ⇒ P = 2*√((2.5 m)²+(1.25 m)²) = 5.59 m
⇒ R = A/P ⇒ R = 3.125 m²/5.59 m = 0.559 m
Substituting values into the Manning equation gives:
Q = (1/0.023)*(3.125 m²)*(∛(0.559 m)²)*(√0.025)
⇒ Q = 14.578 m³/s
Answer:
58.44 g/mol The Molarity of this concentration is 0.154 molar
Explanation:
the molar mass of NaCl is 58.44 g/mol,
0.9 % is the same thing as 0.9g of NaCl , so this means that 100 ml's of physiological saline contains 0.9 g of NaCl. One liter of physiological saline must contain 9 g of NaCl. We can determine the molarity of a physiological saline solution by dividing 9 g by 58 g... since we have 9 g of NaCl in a liter of physiological saline, but we have 58 grams of NaCl in a mole of NaCl. When we divide 9 g by 58 g, we find that physiological saline contains 0.154 moles of NaCl per liter. That means that physiological saline (0.9% NaCl) has a molarity of 0.154 molar. We can either express this as 0.154 M or 154 millimolar (154 mM).
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Answer:
See explaination
Explanation:
Kindly check attachment for the step by step solution of the given problem.
Answer:
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