All categories

1 year ago

3.-4.

Chemistry

1 answer:

creativ13 [48]1 year ago

3 0

The power for the first tank is 2kW and the second tank is 150 kw for the theoretical pumping power lifts a solution with a density of 1200 kg/m³ from a tank to a reservoir.

<h3>What is power?</h3>

Power is the amount of energy in kilowatt which is required to do some work that needs energy known as power it is the multiplication of pressure and the flow of liquid.

Power for the tank = pressure of water × flow of water

Substituting the value,

power = 1 × 2 = 2 kW = P1

power = 10 × 15 = 150 kW = P2

Therefore, a solution with a density of 1200 kg/m³ from a tank to a reservoir power for the first tank is 2kW and for the second tank is 150 kW for the theoretical pumping power lifts.

Learn more about power, here:

brainly.com/question/24091709

#SPJ1

You might be interested in

Families 1-2 and 13-18 are called the...

Zolol [24]

Answer:

C. representative elements.

Explanation:

Families 1 - 2 and 13 - 18 are called representative elements or often times, main group elements.

Elements on the periodic table belongs to four blocks which are s, p, d and f blocks.
Those in the s and p-blocks are the main group elements also called representative elements.
Their outer-shell are not completely filled with electrons but their inner shell is completely filled.

5 0

3 years ago

Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the

Olin [163]

Firstly we need to determine the partial pressure of O2:

$\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\ \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}$

We will now use the Henry's Law equation to determine the solubility of the gas:

$\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\ \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}$

Answer: Solubility is 2.7x10^-3 M

6 0

1 year ago

What is the pH of a 0.40 M solution of C6H5COOH?

melomori [17]

What are the answer choices?

4 0

3 years ago

50 POINTS! ANSWER FAST!

Korvikt [17]

<u>Given:</u>

The initial energy of the electron Einitial = 16.32 * 10⁻¹⁹ J

The energy released i.e the change in energy ΔE = 5.4 * 10⁻¹⁹ J

<u>To determine:</u>

The final energy state Efinal of the electron

<u>Explanation:</u>

Since energy is being released, this suggests that Efinal < Einitial

i.e. ΔE = Einitial - Efinal

Efinal = Einitial - ΔE = (16.32 - 5.4)*10⁻¹⁹ = 10.92 * 10⁻¹⁹ J

Ans: A)

The electron moved down to an energy level and has an energy of 10.92 * 10⁻¹⁹ J

6 0

3 years ago

Read 2 more answers

All of the following are features of karst topography EXCEPT sinkholes. Disappearing and emerging streams. Caverns. Continuous s

Minchanka [31]

D, continuous streams.

4 0

3 years ago