Which of these zones of the ocean is most hospitable to life?

Which of the following is an oxidation-reduction reaction?

Hatshy [7]

B. The chemical reaction that represents oxidation - reduction reaction is CaCO3(s) → CaO(s) + CO2(g).

<h3>Oxidation reduction reaction</h3>

In oxidation reduction reaction, the oxidizing agent is reduced while reducing agent is oxidized.

<h3>In the chemical equation below;</h3>

CaCO3(s) → CaO(s) + CO2(g)

Calcium (ca) is oxidized calcium (II) ion while the carbon (IV) oxide is reduced carbon (IV) oxide.

Thus, the chemical reaction that represents oxidation - reduction reaction is CaCO3(s) → CaO(s) + CO2(g).

Learn more about oxidation reduction reaction here: brainly.com/question/4222605

#SPJ1

8 0

2 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

$\mathbf{K_{entrance}}$ <em> </em>≈ $\mathbf{K_{exit}}$ ≈ 1.0, $\mathbf{K_{elbow}}$ <em> </em> ≈ 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 ≈<u> 2.598 m/s</u>

(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 ≈ <u>0.025</u>

(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 <u>227:1000</u>

(e) The power required <em>P</em> 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 ≈ <u>1,216.67 W</u>

Learn more about flow in pipes here:

brainly.com/question/7246532

6 0

2 years ago

The cell cycle can be divided into two phases interphase and mitosis (cell division), Mitosis is further subdivided into prophas

stealth61 [152]

Well the answer is C

7 0

3 years ago

Read 2 more answers

What is the difference between radiant energy,sound energy and electric energy?

aleksandr82 [10.1K]

Answer:

Radiant Energy - is a combination of heat and light energy. Light energy, like sound energy, travels out in all directions in waves. ... Electrical Energy - Energy produced by electrons moving through a substance is known as electrical energy. We mostly see electric energy in batteries and from the outlets in our homes.

Explanation:

6 0

3 years ago

The three naturally occurring isotopes of potassium are 39K, 38.963707u; 40K, 39.963999u; and 41K.The percent natural abundances

sweet-ann [11.9K]

Answer:

The isotopic mass of 41K is 40.9574 amu

Explanation:

Step 1: Data given

The isotopes are:

39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%

40K with an isotopic mass of 39.963999u

41K wit natural abundance of 6.7302 %

Average atomic mass =39.098 amu

Step 2: Calculate natural abundance of 40 K

100 % - 93.2581 % - 6.7302 %

100 % = 0.0117 %

Step 3: Calculate isotopic mass of 41K

39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302

39.098 = 36.33681 + 0.0046758 + X * 2.067302

X = 40.9574 amu

The isotopic mass of 41K is 40.9574 amu

8 0

3 years ago