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mel-nik [20]
3 years ago
13

Fluorine is a reactive nonmetal in the Halogen group. What other elements below are also likely to be reactive like fluorine?

Chemistry
1 answer:
Svetradugi [14.3K]3 years ago
7 0

Answer:okay

Explanation:

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The empirical formula of hydrazine, molecular mass 32 grams, is NH2. What is the molecular formula of hydrazine?
fgiga [73]

Answer:

C) I think..

Explanation:

7 0
2 years ago
In what ways are solid solid mixture categorised​
AlladinOne [14]

Mixtures can be classified as homogeneous or heterogeneous . Mixtures are composed of substances that are not chemically combined.

Homogeneous mixtures are solutions. The components of a solution are evenly distributed throughout, so that every part of the solution is the same. The components that make up a solution include one or more solutes dissolved in a solvent. Solutes can be solids, liquids, or gases, and solvents can also be solids, liquids or gases.

Brass is an example of a solid/solid solution, saline solution is an example of a solid/liquid solution, diluted ethanol is an example of a liquid/liquid solution. There are many examples of solutions. The components of a solution can be separated by physical means, such as distillation, evaporation, and chromatography, among others.

7 0
3 years ago
For the following reaction, 25.4 grams of sulfur dioxide are allowed to react with 11.6 grams of water . sulfur dioxide(g) water
Gekata [30.6K]

Answer:

The maximum mass of sulfurous acid that can be formed is 32.54 grams.

Explanation:

The balanced reaction is:

SO₂ + H₂O → H₂SO₃

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • SO₂: 1 mole
  • H₂O: 1 mole
  • H₂SO3: 1 mole

Being the molar masses of each compound:

  • SO₂: 64 g/mole
  • H₂O: 18 g/mole
  • H₂SO₃: 82 g/mole

Then, by stoichiometry the following quantities of mass participate in the reaction:

  • SO₂: 1 mole* 64 g/mole= 64 g
  • H₂O: 1 mole* 18 g/mole= 18 g
  • H₂SO₃: 1 mole* 82 g/mole=82 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: If 18 grams of water react with 64 grams of sulfur dioxide, how much mass of sulfur dioxide does 11.6 grams of water react with?

mass of sulfur dioxide=\frac{11.6 grams of water*64 grams of sulfur dioxide}{18 grams of water}

mass of sulfur dioxide= 41.24 grams

But 41.24 grams of sulfur dioxide are not available, 25.4 grams are available. Since you have less mass than you need to react with 11.6 grams of water, sulfur dioxide will be the limiting reagent.

Then you can apply the following rule of three: if by reaction stoichiometry 64 grams of sulfur dioxide produce 82 grams of sulfurous acid, 25.4 grams of sulfur dioxide, how much mass of sulfurous acid will it produce?

mass of sulfurous acid=\frac{25.4 grams of sulfur doixide*82 grams of sulfurous acid}{64 grams of sulfur doixide}

mass of sulfurous acid= 32.54 grams

<u><em>The maximum mass of sulfurous acid that can be formed is 32.54 grams.</em></u>

<u><em></em></u>

7 0
3 years ago
Add one or more curved arrows to show the movement of electrons in the reaction. To draw the arrows, select More in the drawing
DIA [1.3K]

Answer:

H3C - O - O - CH -HCOCH2.

Explanation:

Before chemical reaction can occur, some steps has to be followed in order to give the final product (s), this steps can be shown by drawing the chemical compound and showing how bonds are being broken or formed. Showing this in a stepwise manner is known as reaction mechanism.

From the question above, it is given that the products after heating the reactant gives H3C—0. +.0—CH HCO CH2.

The starting material can be deduce as;

H3C - O - O - CH -HCOCH2.

Heat causes something to divide or melt. Kindly check the attached picture which shows how the deduced starting compound splits to give H3C—0. and .0—CH HCO CH2.

3 0
3 years ago
I need help with my chemistry​
german

Answer:

Ok I will help you in your chemistry

3 0
3 years ago
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