All categories

4 years ago

The water that penetrates into the ground is stored in

Chemistry

2 answers:

miss Akunina [59]4 years ago

6 0

geysers,
here you go
I hope this helps and you pass you ? or test

Studentka2010 [4]4 years ago

4 0

Answer:

Aquifers and groundwater

Explanation:

Also known as "Groundwater", these are underground layers of water-bearing permeable rocks, unconsolidated materials or rock fractures.

Grandwater can also be extracted using a water well.

Hydrogeology is the study of water flow in aquifers and the characterization of aquifers.

You might be interested in

“Fine sediment buries the organism, and pressure eventually squeezes out the liquid and gas parts of the remains, leaving only a

wlad13 [49]

Decomposition I believe. hope this helps

4 0

3 years ago

How many moles of Boron (B) are in 5.03 x 1024 B atoms?

goldfiish [28.3K]

Hey there!:

Number of moles = ( number of atoms / 6.023*10²³ atoms )

given number of atoms = 5.03*10²⁴

Therefore:

Number of moles B = 5.03*10²⁴ / 6.023*10²³

Number of moles B = 8.35 moles

Hope that helps!

3 0

3 years ago

How many grams of hydrogen reacts with 32 g of oxygen to produce 34 g of hydrogen peroxide

nordsb [41]

Answer:

2 grams.

Explanation:

H2 + O2 ---> H2O2

Using molar masses:

2*1 g hydrogen reacts with 2*16 g oxygen.

so 2g H2 reacts with 32 g O2.

8 0

3 years ago

What is an example of ionic compound? A.co2 B.Kcl C.H2o D.Co

julia-pushkina [17]

An iconic compound is the highest occupied energy level of an atom so your answer would be....... B

3 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers