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3 years ago

WHAT IS RAIN WATER HARVESTING

Chemistry

2 answers:

NeTakaya3 years ago

7 0

Explanation:

Rainwater harvesting is collecting the run-off from a structure or other impervious surface in order to store it for later use. Traditionally, this involves harvesting the rain from a roof. The rain will collect in gutters that channel the water into downspouts and then into some sort of storage vessel.

Svetlanka [38]3 years ago

7 0

Answer:

Rainwater harvesting is the accumulation and storage of rainwater for reuse on-site, rather than allowing it to run off. Rainwater can be collected from rivers or roofs, and in many places, the water collected is redirected to a deep pit (well, shaft, or borehole), a reservoir with percolation, or collected from dew or fog with nets or other tools. Its uses include water for gardens, livestock, irrigation, domestic use with proper treatment, indoor heating for houses, etc. The harvested water can also be used as drinking water, longer-term storage, and for other purposes such as groundwater recharge.

Rainwater harvesting is one of the simplest and oldest methods of self-supply of water for households usually financed by the user.

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What is the maximum number of electrons that can be contained in the first level

Fantom [35]

Your answer is:

The first level (or shell) can hold up to 2 electrons.

<h3>Hope this helps! C:</h3>

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3 years ago

if .40 L if water is added to the volume of cup 3, what would be the new molarity of a 2 M solution of kool-aid

stiks02 [169]

Answer:

0.7692 M ≅ 0.77 M.

Explanation:

It is known that the no. of millimoles of a solution before dilution is equal to the no. of millimoles of the solution after the dilution.

It can be expressed as:

(MV) before dilution = (MV) after dilution.

M before dilution = 2.0 M, V before dilution = 0.25 L.

M after dilution = ??? M, V after dilution = 0.25 L + 0.40 L = 0.65 L.

∴ M after dilution = (MV) before dilution/(V) after dilution = (2.0 M)(0.25 L)/(0.65 L) = 0.7692 M ≅ 0.77 M.

8 0

3 years ago

A researcher studying the nutritional value of a new candy places a 4.70g sample of the candy inside a bomb calorimeter and comb

Luba_88 [7]

Answer:

Calories of candy = 5.1 kcal /g

Explanation:

Given data:

Mass of candy = 4.70 g

Change in temperature = 2.46 °C

Specific heat capacity of calorimeter = 40.50 Kj/°C

Calories present per gram = ?

Solution:

Formula

Cal of candy = specific heat capacity × change in temperature / mass

C = c . ΔT/m

C = 40.50 Kj/°C × 2.46 °C / 4.70 g

C = 99.63 kj /4.70 g

C = 21.2 kj/ g

C = 21.2 kj/ g × 0.239 kcal/kj

C = 5.1 kcal /g

3 0

3 years ago

A child is male because he has one X chromosome and one Y chromosome where did these chromosomes come from

Deffense [45]

it would come from your mom and dad

3 0

3 years ago

Calculate the amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C.

qwelly [4]

Answer:

The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

Between heat and temperature there is a direct proportional relationship. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

c= 4.184 $\frac{J}{g*C}$
m= 32 g
ΔT= Tfinal - Tinitial= 22°C - 8°C= 14°C

Replacing:

Q= 32 g* 4.184 $\frac{J}{g*C}$ *14 °C

Solving:

Q= 1,874.432 J

The amount of heat required to raise the temperature of a 32g sample of water from 8°C to 22°C is 1,874.432 J

7 0

3 years ago