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SpyIntel [72]
2 years ago
6

i want to make a presentation on mixtures Clearly describes the process through illustrations or other visuals. Describe termino

logy relevant to the process Provide at least 3-4 daily life applications of the process. Presentation (layout, headings/subheadings,verbal explanation)
Chemistry
1 answer:
Furkat [3]2 years ago
4 0

Answer:

Mixture

Explanation:

Definition:

              An impure substance that contain two or more pure substances that retains their individual chemical characteristics.

Examples:

1.  Smoke and fog(smog)

2.  Dirt and water(mud)

3.  Sand , water and gravel(cement)

4.  Water and salt (sea water)

5.  Petroleum, hydrocarbons and fuel addictives(Gasoline)

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What is the oxidation state of Hydrogen in H20<br><br> +2<br> +1<br> -2<br> -1
xeze [42]
<h3>What is the oxidation number of oxygen in H2O?</h3>

Oxygen almost always has an oxidation number of -2, except in peroxides (H 2 O 2) where it is -1 and in compounds with fluorine (OF 2) where it is +2. Hydrogen has an oxidation number of +1 when combined with non-metals, but it has an oxidation number of -1 when combined with metals.

<h3><em>Sure hoep this helps you :)</em></h3>
6 0
2 years ago
What is Gallium Sulfide(Ga2S3).<br> Ionic or Covalent?
mrs_skeptik [129]
The answer is Covalent







6 0
2 years ago
When you turn on the heat in the car, heat is being transferred by (1 point)
OLEGan [10]
A. Radiation 

Air in a car is heated up with the radiator and then blown out with an air pump.
5 0
3 years ago
Write an equilibrium expression for each chemical equation involving one or more solid or liquid reactants or products.
Alex_Xolod [135]

Answer:

a.

Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}

b.

Keq=[O_2]^3

c.

Keq=\frac{[H_3O^+][F^-]}{[HF]}

d.

Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}

Explanation:

Hello there!

In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:

a.

Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}

b.

Keq=[O_2]^3

c.

Keq=\frac{[H_3O^+][F^-]}{[HF]}

d.

Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}

Regards!

5 0
2 years ago
Dihydrogen dioxide decomposes into water and oxygen gas. Calculate the amounts requested if 1.34 moles of dihydrogen dioxide rea
a_sh-v [17]

Answers:

<em><u>a. Moles of oxygen formed:</u></em><u> 0.670 mol</u>

<em><u>b. Moles of water formed:</u></em><u> 1.34 mol</u>

<em><u>c. Mass of water formed:</u></em><u> 24.1 g</u>

<em><u>d. Mass of oxygen formed:</u></em><u> 21.4 g</u>

<u></u>

Explanation:

Dihdyrogen dioxide is the chemical name for a compound made of two hydrogen atoms and two oxide atoms, i.e. H₂O₂, which is also known as hydrogen peroxide or oxygenated water.

The decomposition reaction of dihydrogen dioxide into water and oxygen gas is represented by the balanced chemical equation:

2H_2O_2(l)\rightarrow 2H_2O(l)+O_2(g)

The mole ratios derived from that balanced chemical equation are:

  • 2 mol H₂O₂ : 2 mol H₂O : 1 mol O₂

<em><u>a. Moles of oxygen formed</u></em>

  • Set the proportion using the theoretical mole ratio of H₂O₂ to O₂ and the amount of moles of dyhydrogen dioxide that react:

2\text{ mol }H_2O_2/1\text{ mol }O_2=1.34\text{ mol }H_2O_2/x

When you solve for x, you get:

  • x = 1.34 mol H₂O₂ × 1 mol O₂ / 2 mol H₂O₂ = 0.670 mol O₂

<u />

<em><u>b. Moles of water formed</u></em>

  • Set the proportion using the theoretical mole ratio of H₂O₂ to H₂O and the amount of moles of dyhydrogen dioxide that react:

2\text{ mol }H_2O_2/2\text{ mol }H_2O=1.34\text{ mol }H_2O_2/x

When you solve for x, you get:

  • x = 1.34 mol H₂O₂ × 2 mol H₂O / 2 mol H₂O₂ = 1.34 mol H₂O

<em><u>c. Mass of water formed</u></em>

Using the number of moles of water calculated in the part b., you calculate the mass of water formed, in grams, using the molar mass of water:

  • Molar mass of water = 18.015 g/mol

  • Number of moles = mass in grams / molar mass

       ⇒ mass in grams = number of moles × molar mass

       ⇒ mass in grams = 1.34 mol × 18.015 g/mol = 24.1 g

<em><u>d. Mass of oxygen formed</u></em>

Using the number of moles of oxygen determined in the part a., you calculate the mass in grams using the molar mass of O₂.

  • Molar mass of O₂ = 32.00 g/mol
  • mass = molar mass × number of moles
  • mass = 32.00 g/mol × 0.670 mol = 21.4 g.
5 0
3 years ago
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