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Setler79 [48]
3 years ago
13

Hich of these reactions is a combustion reaction?

Chemistry
2 answers:
4vir4ik [10]3 years ago
8 0
For combustion reaction we need oxygen and the results most of the time produce heat which mean first one is correct.
tiny-mole [99]3 years ago
6 0

Answer : The correct option is, burning wood.

Explanation :

Combustion reaction : It is a type of chemical reaction in which a hydrocarbon molecule react with oxygen to give or produce carbon dioxide and water as a product.

Decomposition reaction : It is a type of chemical reaction in which the larger molecule decomposes into two or more smaller molecules.

Oxidation reaction : It is a type of chemical reaction in which the substance oxidizes by gain of oxygen or react with oxygen.

From this we conclude that, the burning of wood is an example of combustion reaction because wood is a type of hydrocarbon that react or burn with oxygen to give carbon dioxide and water as a product.

Hence, the correct option is, burning wood.

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What is the result of having two batteries on an electric motor
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Answer:

Well bro lemme tell u,

by connecting batteries, you can increase the voltage, amperage, or both.

Explanation:

8 0
3 years ago
1. What is the theory that is accepted as a true<br> called? I
Harlamova29_29 [7]

Answer: Scientific Theory

Explanation:

A scientific theory is a broad explanation for events that is widely accepted as true. To become a theory, a hypothesis must be tested over and over again, and it must be supported by a great deal of evidence. People commonly use the word theory to describe a guess about how or why something happens.

5 0
3 years ago
For each of the reactions, calculate the mass (in grams) of the product formed when 15.12g of the underlined reactant completely
Ludmilka [50]

Answer:

Part A : amount of product (KCl) =  28.88 g

Part B :  amount of product (KBr) =  46.13 g

Part C : amount of product (Cr₂O₃) =  17.3 g

Part D: amount of product (SrO) =  35.76 g

Explanation:

Part A:

Data Given:

Reaction :

                      2K(s) + Cl₂(g) --------> 2KCl

Amount of underline Reactant  (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is chlorine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

                                2K(s) + Cl₂(g) --------> 2KCl

                                2mol    1mol                 2mol

it shows that

2 mole of K give 2 mole of  KCl

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KCl = (39 + 35.5)

Molar mass of KCl = 74.5 g/mol

So the look again to reaction in terms of grams

                                      2K(s)     +    Cl₂(g) --------> 2KCl

                            2mole (39 g/mol)                      2mole (74.5 g/mol)

                                      78 g                                  149 g

Apply the Unity formula

                          78 g of Potassium ≅ 149 g of KCl

Then

                        15.12 g of Potassium ≅ how many g of Product (KCl)

By doing cross multiplication

               X g of Product (KCl) = 149 g of KCl  x 15.12 g of K /  78 g of K

              X g of Product (KCl) = 149 g of KCl  x 15.12 g of K /  78 g of K

              X g of Product (KCl) = 28.88 g

So the amount of product (KCl) =  28.88 g

_________________________________________

Part B:

Data Given:

Reaction :

                      2K(s) + Br₂(g) --------> 2KBr

Amount of underline Reactant  (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is Bromine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

                                2K(s) + Br₂(g) --------> 2KBr

                                2mol    1mol                 2mol

it shows that

2 mole of K give 2 mole of  KBr

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KBr = (39 + 80)

Molar mass of KBr =  119 g/mol

So, look again to reaction in terms of grams

                                      2K(s)     +    Br₂(g) --------> 2KBr

                            2mole (39 g/mol)                      2mole (119 g/mol)

                                      78 g                                  238 g

Apply the Unity formula

                          78 g of Potassium ≅  238 g of KBr

Then

                        15.12 g of Potassium ≅ how many g of Product (KBr)

By doing cross multiplication

               X g of Product (KBr) = 238 g of KBr  x 15.12 g of K /  78 g of K

              X g of Product (KBr) = 238 g of KBr  x 15.12 g of K /  78 g of K

              X g of Product (KBr) = 46.13 g

So the amount of product (KBr) =  46.13 g

__________________________________________

Part C:

Data Given:

Reaction :

                      4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

Amount of underline Reactant  (Cr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Chromium (Cr) is 15.12g and other reactant that is Oxygen is more than enough so the Cr is limiting reagent.

So, amount of product depend on the amount of Chromium (Cr)

Now Look at Given Reaction:

                                  4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

                                  4mol      3mol                 2mol

it shows that

4 mole of Cr give 2 mole of  Cr₂O₃

if we represent mole in grams

Then

Molar mass of Cr = 52 g/mol

Molar mass of 2Cr₂O₃ = 2 [2 (52) + 3(16) ] = 2 (104+ 48)

Molar mass of  2Cr₂O₃ =  304 g/mol

So, look again to reaction in terms of grams

                                     4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

                                 4 mol (52 g/mol)              2 mole (304 g/mol)

                                      208 g                                  608 g

Apply the Unity formula

                          208 g of Chromium ≅  608 g of Cr₂O₃

Then

                        15.12 g of Chromium ≅ how many g of Product (Cr₂O₃)

By doing cross multiplication

        X g of Product (Cr₂O₃) = 238 g of Cr₂O₃  x 15.12 g of Cr /  208 g of Cr

        X g of Product (Cr₂O₃) = 238 g of Cr₂O₃  x 15.12 g of Cr /  208 g of Cr

        X g of Product (Cr₂O₃) = 17.3 g

So the amount of product (Cr₂O₃) =  17.3 g

________________________________________

Part D:

Data Given:

Reaction :

                      2Sr(s) + O₂(g) --------> 2SrO(s)

Amount of underline Reactant  (Sr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Strontium (Sr) is 15.12g and other reactant that is Oxygen is more than enough so the Sr is limiting reagent.

So, amount of product depend on the amount of Strontium (Sr)

Now Look at Given Reaction:

                                    2Sr(s) + O₂(g) --------> 2SrO(s)

                                    2mol      1mol              2mol

it shows that

2 mole of Sr give 2 mole of SrO

if we represent mole in grams

Then

Molar mass of Sr = 87.6 g/mol

Molar mass of 2SrO = 2 [87.6 + 16] = 2 (103.6)

Molar mass of 2SrO =  207.2 g/mol

So, look again to reaction in terms of grams

                                      2Sr(s) + O₂(g) --------> 2SrO(s)

                                 2 mol ( 87.6 g/mol)         2 mole (207.2 g/mol)

                                      175.2 g                                  414.4 g

Apply the Unity formula

                          175.2 g of Strontium ≅  414.4 g of SrO

Then

                        15.12 g of Strontium ≅ how many g of Product (SrO)

By doing cross multiplication

       X g of Product (SrO) = 414.4 g of SrO  x 15.12 g of Sr /  175.2 g of Sr

        X g of Product (SrO) = 414.4 g of SrO  x 15.12 g of Sr /  175.2 g of Sr

        X g of Product (SrO) = 35.76 g

So the amount of product (SrO) =  35.76 g

8 0
3 years ago
Which statements best describe half-lives of radioactive isotopes? Check all that apply.
Igoryamba

Answer:

The half-life varies depending on the isotope.

Half-lives range from fractions of a second to billions of years.

The half-life of a particular isotope is constant.

3 0
3 years ago
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