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3 years ago

Which of the following can explain the daily change in sea level observed along a coast?

Chemistry

2 answers:

gavmur [86]3 years ago

8 0

Answer:

The gravitational pull of the moon on the water near the coast (option A) can explain the daily change in sea level observed along a coast.

Explanation:

The tides are rising and falling from sea level that occur several times a day. These periodic changes in the level of the sea are produced by the gravitational forces of attraction of the Sun and the Moon with respect to the Earth. But the force of attraction of the Moon is greater than that of the Sun, due to the proximity of the satellite.

High tide or high tide is the time when the sea reaches its maximum height. Low tide or low tide is the time when the sea reaches its minimum height.

The tide is noticed on the beaches because the coastline can go forward or backward many meters.

Finally, the gravitational pull of the moon on the water near the coast (option A) can explain the daily change in sea level observed along a coast.

sertanlavr [38]3 years ago

6 0

D.the concentration of salt in the water her a the coast

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Determine the empirical formula of a compound having the following percent composition by mass: K: 24.74%; Mn: 34.76%; O: 40.50%

Dovator [93]

Answer: The empirical formula of the compound is $KMnO_4$

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Let the mass of the compound be 100 g

Given values:

% of K = 24.7%

% of Mn = 34.76%

% of O = 40.50%

Mass of K = 24.7 g

Mass of Mn = 34.76 g

Mass of O = 40.50 g

To calculate the empirical formula of a compound, few steps need to be followed:

Step 1: Calculating the number of moles of each element

We know:

Molar mass of K = 39.10 g/mol

Molar mass of Mn = 54.94 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of K}=\frac{24.7g}{39.10g/mol}=0.632 mol$

$\text{Moles of Mn}=\frac{34.76g}{54.94g/mol}=0.633 mol$

$\text{Moles of O}=\frac{40.50g}{16g/mol}=2.53 mol$

Step 2: Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.632 moles

$\text{Mole fraction of K}=\frac{0.632}{0.632}=1$

$\text{Mole fraction of Mn}=\frac{0.633}{0.632}=1$

$\text{Mole fraction of O}=\frac{2.53}{0.632}=4$

Step 3: Writing the mole fraction as the subscripts of each of the element

The empirical formula of the compound becomes $K_1Mn_1O_4=KMnO_4$

Hence, the empirical formula of the compound is $KMnO_4$

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