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kozerog [31]
2 years ago
15

Compare your circulatory system to a busy street in a city or town.

Chemistry
1 answer:
White raven [17]2 years ago
7 0

City individuals -different cells. Group of individuals - Organ systems. City Road System- Circulatory system.

<u>Explanation:</u>

Any city has population. The individuals of the population can be considered as the cells that are present in a human body. As a city cannot be formed with individuals, the human body cannot survive without cells. These individuals will belong to different profession and they go to an organisation for doing various work.

These can be compared with the organ system of human body. These organ systems will be performing different functions. The network of any rod in a city can be compared with the circulatory system of human body. As it helps individuals to reach their home of any destination, the presence of arteries and veins can be compared with the highways and capillaries can be associated with side ways of the road system.

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How do you solve for density
vodka [1.7K]

The Density Calculator uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). The calculator can use any two of the values to calculate the third. Density is defined as mass per unit volume.

5 0
2 years ago
Read 2 more answers
Sulfur,Iodine,.Carbon,Hydrogen put these lowest to highest
natima [27]

Answer:

The order is Hydrogen, Carbon, Sulfur, Iodine

Explanation:

According to the periodic table. The order is Hydrogen, Carbon, Sulfur, Iodine. You can search up an image of a periodic table, it will help you.

7 0
3 years ago
If 56.8 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 554 mg sample of na2so4 (forming baso4), what is
IrinaVladis [17]
<span>0.0687 m The balanced equation is BaCl2 + Na2SO4 ==> BaSO4 + 2 NaCl Looking at the equation, it indicates that there's a 1 to 1 ratio of BaCl2 and Na2SO4 in the reaction. So the number of moles of each will be equal. Now calculate the number of moles of Na2SO4 we had. Start by looking up atomic weights. Atomic weight sodium = 22.989769 Atomic weight sulfur = 32.065 Atomic weight oxygen = 15.999 Molar mass Na2SO4 = 2 * 22.989769 + 32.065 + 4 * 15.999 = 142.040538 g/mol Moles Na2SO4 = 0.554 g / 142.040538 g/mol = 0.003900295 mol Molarity is defined as moles per liter, so let's do the division. 0.003900295 mol / 0.0568 l = 0.068667165 mol/l = 0.068667165 m Rounding to 3 significant figures gives 0.0687 m</span>
3 0
3 years ago
¿ ppr que no podemos automedicarnos ,cuando pensamos q tenemos el virus del CIVD-19?
GREYUIT [131]

Answer:

Porque como tu no eres profesional en medicina, no sabes nada sobre las dosis de medicamentos ni de los efectos que pueden tener en tu cuerpo si se administran incorrectamente.

Explanation:

Es logico que solo vayas al medico para que te den la cura allí.

3 0
3 years ago
When balancing redox reactions under basic conditions in aqueous solution, the first step is to:________.
natta225 [31]

Answer:

When balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.

Explanation:

Oxidation-reduction reactions or redox reactions are those in which an electron transfer occurs between the reagents. An electron transfer implies that there is a change in the number of oxidation between the reagents and the products.

The gain of electrons is called reduction and the loss of electrons oxidation. That is to say, there is oxidation whenever an atom or group of atoms loses electrons (or increases its positive charges) and in the reduction an atom or group of atoms gains electrons, increasing its negative charges or decreasing the positive ones.

The oxidation and reduction half-reactions, in a basic medium, adjust the oxygens and hydrogens as follows:

In the member of the half-reaction that presents excess oxygen, you add as many water molecules as there are too many oxygen. Then, in the opposite member, the necessary hydroxyl ions are added to fully adjust the half-reaction. Normally, twice as many hydroxyl ions, OH-, are required as water molecules have previously been added.

In short, you first adjust the oxygens with OH-, then you adjust the H with H₂O, and finally you adjust the charge with e-

So, <u><em>when balancing redox reactions under basic conditions in aqueous solution, the first step is to balance oxygen.</em></u>

3 0
3 years ago
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