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Margarita [4]
3 years ago
13

The _______________ system collects solid wastes—such as undigested food, intestinal bacteria, and old cells—and removes them fr

om the body, while the _____________ system filters the blood and collects wastes to remove from the body as urine.
Chemistry
1 answer:
zmey [24]3 years ago
6 0

Answer:

Excretory system, and Kidney.

Explanation:

The excretory system helps to regulate the chemical composition of the body fluids by removing metabolic waste such as undigested food, intestinal bacteria, and old cells and helping to retain the proper amount of water, nutrients, and salts in the body.

The main function of the kidney is to filter the blood, kidney help to remove waste from the body as urine, control fluid balance in the body, and keep the right level of electrolytes. Each kidney is bean-shaped in structure and 4-5 inches long in size.

You might be interested in
Jose and Sue were investigating the formation of precipitates in chemistry lab. They mixed a silver nitrate solution with a sodi
irinina [24]

In a double decomposition reaction, AgNo3 and NaCl combine to form NaNo3 and AgCl. The resulting AgCl precipitate is white as well. AgCl (silver chloride) with nitrate of sodium. It will be an insoluble solid, AgCl. The equation represents a double decomposition reaction. Therefore, option B is correct.

<h3>What is double decomposition reaction ?</h3>

A type of decomposition reaction known as a double decomposition reaction involves the exchange of positive and negative ions between two constituent reactants to create two new molecules.

When two compounds in a solution react to generate two new compounds through the exchange of radicals, this reaction is known as a double displacement reaction. Double decomposition reaction is another name for this kind of reaction.

Thus, option B is correct.

To learn more about double decomposition reaction, follow the link;

brainly.com/question/21489872

#SPJ1

7 0
1 year ago
Based on formal charges, draw the most preferred Lewis structure for the chlorate ion, ClO3−. To add lone pairs, click the butto
vova2212 [387]

Answer:

The structure is shown in the diagram.

Explanation:

Lewis structure : In order to draw Lewis structure we will calculate the total number of valence electrons in the molecule.

The valence electrons from Cl : 7

The valence electrons from O = 3 X 6 = 18

Charge negative so more electrons = 2

total electrons = 7 + 18 +2 = 27

Now we will distribute the electrons on each atom and in between atoms as shown in the diagram.

8 0
3 years ago
Read 2 more answers
If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was
Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

n_{acid}=n_{base}=n_{salt}

Whereas the moles of the salt are computed as shown below:

n_{salt}=0.021L*0.68mol/L=0.01428mol

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

[salt]=0.01428mol/0.0276L=0.517M

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

C_6H_5NH_3^++H_2O\rightleftharpoons  C_6H_5NH_2+H_3O^+

Whose equilibrium expression is:

Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

2.326x10^{-5}=\frac{x^2}{0.517M}

Whereas x is:

x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

pH=-log(3.47x10^{-3})\\\\pH=2.46

Regards!

6 0
3 years ago
In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidat
Fiesta28 [93]

Answer:

Part A: (1, 1, 4, 1, 1, 1)

Part B: (2, 6, 4, 2, 3, 8)

Explanation:

Redox reactions can be balanced using the half-reaction method. It has the following steps:

  1. We write both half-reactions (reduction and oxidation)
  2. We balance the masses using H⁺ and H₂O in acidic media or OH⁻ and H₂O in basic media.
  3. We add electrons to balance electrically the half-reaction
  4. We multiply the half-reaction by numbers to make sure the number of electrons gained and lost are the same.
  5. We add both half-reactions and take the numbers to the general equation.

<em>Acidic solution</em>

SO₄²⁻(aq) + Sn²⁺(aq) + X ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + Y

1.

Reduction: SO₄²⁻ ⇒ SO₃²⁻

Oxidation: Sn²⁺ ⇒ Sn⁴⁺

2.

2 H⁺ + SO₄²⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺

3.

2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O

Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻

4.

1 x [2 H⁺ + SO₄²⁻ + 2 e⁻ ⇒ SO₃²⁻ + H₂O]

1 x [Sn²⁺ ⇒ Sn⁴⁺ + 2 e⁻]

5.

2 H⁺ + SO₄²⁻ + 2 e⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺ + 2 e⁻

2 H⁺ + SO₄²⁻ + Sn²⁺ ⇄ SO₃²⁻ + H₂O + Sn⁴⁺

Taking this to the general equation:

SO₄²⁻(aq) + Sn²⁺(aq) + 2 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

Since H⁺ are spectator ions, they are not balanced automatically through this method and we have to balance them manually. In this case, we need to add 2 more H⁺ to the left.

SO₄²⁻(aq) + Sn²⁺(aq) + 4 H⁺(aq) ⇄ H₂SO₃(aq) + Sn⁴⁺(aq) + H₂O(l)

<em>Basic solution</em>

MnO₄⁻(aq) + F⁻(aq) + X ⇄ MnO₂(s) + F₂(aq) + Y

1.

Reduction: MnO₄⁻ ⇒ MnO₂

Oxidation: F⁻ ⇒ F₂

2.

2 H₂O + MnO₄⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂

3.

2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻

2 F⁻ ⇒ F₂ + 2 e⁻

4.

2 × (2 H₂O + MnO₄⁻ + 3 e⁻ ⇒ MnO₂ + 4 OH⁻)

3 × (2 F⁻ ⇒ F₂ + 2 e⁻)

5.

4 H₂O + 2 MnO₄⁻ + 6 e⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂ + 6 e⁻

4 H₂O + 2 MnO₄⁻ + 6 F⁻ ⇄ 2 MnO₂ + 8 OH⁻ + 3 F₂

Taking this to the general equation:

2 MnO₄⁻(aq) + 6 F⁻(aq) + 4 H₂O ⇄ 2 MnO₂(s) + 3 F₂(aq) + 8 OH⁻

This equation is balanced.

6 0
3 years ago
Determine the empirical and molecular formula for a compound containing 26.1
kobusy [5.1K]

determing that the compoumd formed is in 100%

therefore, 26.1 +4.3+69.6 = 100%

hence no need of adding O molecule

finding the numbr of moles of each element

<em>n</em><em>o</em><em>.</em><em> </em><em>o</em><em>f</em><em> </em><em>m</em><em>o</em><em>l</em><em>e</em><em>s</em><em> </em><em>o</em><em>f</em><em> </em><em>C</em><em>=</em><em> </em><em>g</em><em>i</em><em>v</em><em>e</em><em>n</em><em> </em><em>m</em><em>a</em><em>s</em><em>s</em><em>÷</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>m</em><em>a</em><em>s</em><em>s</em>

<em>=</em><em>2</em><em>6</em><em>.</em><em>1</em><em>÷</em><em>1</em><em>2</em>

<em>=</em><em><u>2</u></em><em><u>.</u></em><em><u>1</u></em><em><u> </u></em><em><u>m</u></em><em><u>o</u></em><em><u>l</u></em>

<em>n</em><em>o</em><em>.</em><em> </em><em>o</em><em>f</em><em> </em><em>m</em><em>o</em><em>l</em><em>e</em><em>s</em><em> </em><em>o</em><em>f</em><em> </em><em>H</em><em> </em><em>=</em><em> </em><em>g</em><em>i</em><em>v</em><em>e</em><em>n</em><em> </em><em>m</em><em>a</em><em>s</em><em>s</em><em>÷</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>m</em><em>a</em><em>s</em><em>s</em>

<em>=</em><em>4</em><em>.</em><em>3</em><em>÷</em><em>1</em>

<em>=</em><em><u>4</u></em><em><u>.</u></em><em><u>3</u></em><em><u> </u></em><em><u>m</u></em><em><u>o</u></em><em><u>l</u></em>

<em>n</em><em>o</em><em> </em><em>o</em><em>f</em><em> </em><em>m</em><em>o</em><em>l</em><em>e</em><em>s</em><em> </em><em>o</em><em>f</em><em> </em><em>O</em><em> </em><em>=</em><em>g</em><em>i</em><em>v</em><em>e</em><em>n</em><em> </em><em>m</em><em>a</em><em>s</em><em>s</em><em>÷</em><em>m</em><em>o</em><em>l</em><em>a</em><em>r</em><em> </em><em>m</em><em>a</em><em>s</em><em>s</em>

<em>=</em><em>6</em><em>9</em><em>.</em><em>6</em><em>÷</em><em>1</em><em>6</em>

<em><u>=</u></em><em><u>4</u></em><em><u>.</u></em><em><u>3</u></em>

the simplest whole nber of the element in the unknown compoumd is 1:2:2 of C:H:O

HENCE, the empirical formula is CH2O2

empirical formula mass of CH2O2 IS

=(12)+(2×1)+(16×2)

12+2+32

46 G/MOL

N= molar mass of the compoumd ÷ emporical mass of the compoumd

N= 138÷46

<em><u>N</u></em><em><u>=</u></em><em><u>3</u></em>

<em><u>m</u></em><em><u>o</u></em><em><u>l</u></em><em><u>e</u></em><em><u>c</u></em><em><u>u</u></em><em><u>l</u></em><em><u>a</u></em><em><u>r</u></em><em><u> </u></em><em><u>f</u></em><em><u>o</u></em><em><u>r</u></em><em><u>m</u></em><em><u>u</u></em><em><u>l</u></em><em><u>a</u></em><em><u> </u></em><em><u>=</u></em><em><u>N</u></em><em><u>×</u></em><em><u>e</u></em><em><u>m</u></em><em><u>p</u></em><em><u>i</u></em><em><u>r</u></em><em><u>i</u></em><em><u>c</u></em><em><u>a</u></em><em><u>l</u></em><em><u> </u></em><em><u>f</u></em><em><u>o</u></em><em><u>r</u></em><em><u>m</u></em><em><u>u</u></em><em><u>l</u></em><em><u>a</u></em>

<em><u>=</u></em><em><u>3</u></em><em><u> </u></em><em><u>×</u></em><em><u>C</u></em><em><u>H</u></em><em><u>2</u></em><em><u>O</u></em><em><u>2</u></em>

<em><u>C</u></em><em><u>3</u></em><em><u>H</u></em><em><u>6</u></em><em><u>O</u></em><em><u>6</u></em><em><u> </u></em><em><u>i</u></em><em><u>s</u></em><em><u> </u></em><em><u>t</u></em><em><u>h</u></em><em><u>e</u></em><em><u> </u></em><em><u>m</u></em><em><u>o</u></em><em><u>l</u></em><em><u>e</u></em><em><u>c</u></em><em><u>u</u></em><em><u>l</u></em><em><u>a</u></em><em><u>r</u></em><em><u> </u></em><em><u>f</u></em><em><u>o</u></em><em><u>r</u></em><em><u>m</u></em><em><u>u</u></em><em><u>l</u></em><em><u>a</u></em><em><u> </u></em>

<em><u>(</u></em><em><u>t</u></em><em><u>a</u></em><em><u>r</u></em><em><u>t</u></em><em><u>o</u></em><em><u>n</u></em><em><u>i</u></em><em><u>c</u></em><em><u> </u></em><em><u>a</u></em><em><u>c</u></em><em><u>i</u></em><em><u>d</u></em><em><u>)</u></em>

<em><u>h</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>i</u></em><em><u>t</u></em><em><u> </u></em><em><u>h</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u>e</u></em><em><u>d</u></em><em><u> </u></em><em><u>u</u></em><em><u>!</u></em>

5 0
3 years ago
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