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3 years ago

What returns the sand to the beach

Chemistry

2 answers:

r-ruslan [8.4K]3 years ago

5 0

Answer:

the waves

Explanation:

barxatty [35]3 years ago

3 0

Answer:

There is a constant flow of sand from the land into the ocean. ... Sand grains travel southward down the coast, while finer particles of sediment are carried and deposited further out to sea. Along the way, sand is washed ashore, temporarily resting on beaches, until it is re-suspended in the ocean by wave action or wind. !!!!(summarize this is copied)!!!!

Explanation:

You might be interested in

Write the resonance forms of ozone, O3, the component of the upper atmosphere that protects the Earth from ultraviolet radiation

aleksklad [387]

Answer:

The resonance forms of O3 are attached as an image.

Explanation:

A compound with different contributing structures comes together and forms a resonating or intermediate structure that best describes the properties of that compound.

The given compound is ozone, having the chemical formula

O3 = 6 electrons * 3 = 18 electrons

O → prefers to have a complete their octet

The bonding electrons and lone pair electrons are radical electrons that are present on the oxygen atoms tend to delocalize and results in various resonating forms of O3.

7 0

3 years ago

The molar concentration of sucrose in a can of soda is 0.375 M. How much sucrose would be found in a 2 liter bottle of soda of t

antoniya [11.8K]

Answer:

C) 0.75 mol.

Explanation:

Molarity is defined as the no. of moles of solute dissolved in 1.0 L of the solution.

So, The molar concentration of sucrose in a can of soda is 0.375 M means that: every 1.0 L of sucrose contains 0.375 mol of sucrose.

∴ 2 liter bottle of soda of the same concentration contains (2 * 0.375 mol = 0.75 mol) of sucrose.

<em>Thus, the right choice is: C) 0.75 mol.</em>

5 0

3 years ago

Perform the following operation and express the answer in scientific notation 8.6500x10^3+6.5500x10^5

Dafna1 [17]

Taking into account the scientific notation, the result of the sum is 6.6365×10⁵.

<h3>Scientific notation</h3>

First, remember that scientific notation is a quick way to represent a number using powers of base ten.

The numbers are written as a product:

a×10ⁿ

where:

a is a real number greater than or equal to 1 and less than 10, to which a decimal point is added after the first digit if it is a non-integer number.
n is an integer, which is called an exponent or an order of magnitude. Represents the number of times the comma is shifted. It is always an integer, positive if it is shifted to the left, negative if it is shifted to the right.

<h3>Sum in scientific notation</h3>

You want to add two numbers in scientific notation. It should be noted that when the numbers to be added do not have the same base 10 exponent, the base 10 power with the highest exponent must be found.

In this case, the highest exponent is 5.

Then all the values are expressed as a function of the base 10 exponent with the highest exponent. In this case: 8.6500×10³= 0.086500×10⁵

Taking the quantities to the same exponent, all you have to do is add what was previously called the number "a". In this case:

0.086500×10⁵ + 6.5500×10⁵= (0.086500+ 6.5500)×10⁵= <u><em>6.6365×10⁵</em></u>

Finally, the result of the sum is 6.6365×10⁵.

Learn more about scientific notation:

brainly.com/question/11403716

brainly.com/question/853571

#SPJ1

8 0

2 years ago

Learn how changes in binding free energy affect binding and the ratio of unbound and bound molecules.

Delvig [45]

C)[D]/[ED] = 5.20

D)[D]/[ED] = 5.20

E)[D']_T = 1.495* 10 ^-7 M

F)[D'] / [ED'] = 0.0579

Explanation:

E = 250 nM =2.5* 10 ^-7 mol/L , T=298.15 K

Dissociation constant of K_D = 1.3 μM (1.3 *10 ^-6 M)

E + D ⇄ ED → K_a = [ED] / [D][E] (association constant)

ED ⇄ E + D → K_D = [E][D] / [ED] (dissociation constant)

[E] =2.5*10^-7 mol/L

K_D = 1.3* 10^-6 M

K_D = [E][D] / [ED] → [D]/[ED] = K_D / [E]

= [D]/[ED] = 1.3* 10 ^-6 / 2.5 *10^-7

= 13/25 * 10

=130/25 = 5.20

[D]/[ED] = 5.20

ΔG =RTln Kd

ΔG_2 for E and D = 1.987 * 298.15 * ln 1.3*10^-6

ΔG_2 592.454 * [ln 1.3 +ln 10^-6]

ΔG_1 = 592.424 [0.2623 - 13.8155]

ΔG_2 = -592.424 * 13.553

ΔG_1 = -8184.633 cal/ mol

ΔG_1 = -8184.633 * 4.18 J/mol = -34244.508 J?mol

ΔG_1 = -34.245 KL/mol

so, ΔG_2 = ΔG_1 - 10.5 KJ/mol

ΔG_2 = -34.245 - 10.5

ΔG_2 = -44.745KJ / mol

ΔG_2 =RT ln K_D

-44.745 *10^3

=8.314 *298.15 lnK_D

lnK_D' = - 44745 / 2478.81 g

ln K_D' = -18.051

K_D' = -18.051

K_D' = e^-18.051

[D]/[ED] = 5.20

[E] = 2.5* 10 ^-7 mol/ L = a

K_D' = [E][D] / [ED'] E +D' → ED'

K_D' = a/2(x-(a/2) / (a/2)

KD' = x - a/2

=2.447 *10^-8 = (2.5/2) * 10^-7

x=2.447 * 10^-8 + 1.25 * 10^-7

x = 2.447 *10^-8 + 1.25 * 10 ^-7

x= 10^-7 [1.25 + 0.2447]

x = 1.4947 * 10^-7

[D']_T = 1.495* 10 ^-7 M

K_D' = [E][D'] / [ED']

[D'] / [ED'] = KD' / [E]

[D'] / [ED'] = 1.447 *10^-8 / 2.5* 10^-7

[D'] / [ED'] = 0.5788 * 10^-1

[D'] / [ED'] = 0.0579

5 0

3 years ago

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass %

11Alexandr11 [23.1K]

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

<em>Note: The question is stated more clearly below:</em>

<em>Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.</em>

<em>What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?</em>

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

6 0

3 years ago