The figure shows a nutrition label. How much energy does this food contain?

Given the chemical equation: n2 + 3 h2 → 2 nh3. On a molecular level, what do the coefficients mean? 1 atom of nitrogen reacts w

Over [174]

The answer is, "1 molecule of nitrogen reacts with 3 molecules of hydrogen to give 2 molecules of ammonia."

Considering the reaction:

N₂ + 3 H₂ ------> 2 NH₃.

As this can be seen from the reaction that, "1 molecule of nitrogen reacts with 3 molecules of hydrogen to give 2 molecules of ammonia."

So the answer is that "1 molecule of nitrogen reacts with 3 molecules of hydrogen to give 2 molecules of ammonia."

4 0

4 years ago

A pump with an 80% efficiency drives water up between two reservoirs through a piping system of total length L = 15 and circular

Sunny_sXe [5.5K]

The losses in the pipe increases the power requirement of the pipe to

maintain a given flowrate.

Responses (approximate value);

(a) 2.598 m/s

(b) 181,058.58

(c) 0.025

(d) 227:1000

(e) 1,216.67 W

<h3>Which methods can be used to calculate the pressure head in the pipe?</h3>

The given parameters are;

Pump efficiency, η = 80%

Length of the pipe, L = 15 m

Cross-sectional diameter, d = 7 cm

Reservoir temperature, T = 20°C = 293.15 K

$\mathbf{K_{entrance}}$ ≈ $\mathbf{K_{exit}}$ ≈ 1.0, $\mathbf{K_{elbow}}$ ≈ 0.4

Volumetric flow rate, Q = 10 Liters/s = 0.01 m³/s

Surface roughness, ∈ = 0.15 mm

(a) The cross sectional area of the pipe, A = π·r²

Where;

$r = \mathbf{\dfrac{d}{2}}$

Which gives;

$r = \dfrac{0.07 \, cm}{2} = \mathbf{0.035 \, cm}$

$Average \ water \ velocity, \ v =\mathbf{ \dfrac{Q}{A}}$

Therefore;

$v = \dfrac{0.01}{ \pi \times 0.035^2} \approx 2.598$

The average velocity of the water, v ≈ 2.598 m/s

(b) The viscosity of water at 20°C is 0.001003 kg/(m·s) given as follows;

Density of water at 20°C, ρ = 998.23 kg/m³

Reynolds' number, Re, is found as follows;

$Re = \mathbf{\dfrac{\rho \cdot V \cdot D}{\mu}}$

Which gives;

$Re = \dfrac{998.58 \times 2.598 \times 0.07 }{0.001003} \approx \underline{181,058.58}$

(c) The friction factor is given by the following formula;

$\dfrac{1}{\sqrt{f} } = \mathbf{-2.0 \cdot log \left(\dfrac{\epsilon/D}{3.7} + \dfrac{5.74}{Re^{0.9}} } \right)}$

Which gives;

f ≈ 0.025

(d) Friction head loss is given as follows;

$h_f = \mathbf{f \times \dfrac{L}{D} \times \dfrac{V^2}{2 \cdot g}}$

Which gives;

$h_f = 0.025 \times \dfrac{15}{0.07} \times \dfrac{2.598^2}{2 \times 9.81} \approx \mathbf{1.84}$

$Other \ head \ losses, \ h_l= \sum K \cdot \dfrac{V^2}{2}$

Which gives;

$h_l=(1 + 1+0.4) \times \dfrac{ 2.598^2}{2} \approx \mathbf{8.0995}$

Ratio between friction head loss and other head loss is therefore;

$\dfrac{h_f}{h_l} \approx \dfrac{1.84}{8.0995} \approx \underline{0.227}$

The ratio between friction head loss and other head loss is approximately 227:1000

(e) The power required P is found as follows;

$P= \mathbf{ \dfrac{\rho \cdot g \cdot Q \cdot H}{\eta}}$

Which gives;

$P= \dfrac{998.23 \times 9.81 \times 0.01 \times (1.84 + 8.0995)}{0.8} \approx \mathbf{ 1216.67}$

The power required to drive the pump, P ≈ 1,216.67 W

Learn more about flow in pipes here:

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How many ethyne molecules are contained in 84.3 grams of ethyne (C2H2)?

ololo11 [35]

(~26grams/mole) and Avogadros # (6.022x10^23) 84.3grams x 1mole/26grams x 6.022x10^23 molecules/mole = 1.95x10^24 molecules of C2H2

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In the evolution of extremely large stars, they begin as supergiants or supernovas, if the star nucleus is large enough it will

fenix001 [56]

Black hole, or a singularity contained within an event horizon through which no light can escape.

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HURRY!! I NEEE HELP!!!

Ludmilka [50]

Answer:

NEEE

Explanation:

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