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Ulleksa [173]
3 years ago
9

The process where a substance reacts with oxygen

Chemistry
1 answer:
daser333 [38]3 years ago
3 0

Answer:

The process where substance react with oxygen is called combustion.

Explanation:

When substance react with oxygen combustion is occur. The substance which burned is called fuel and in this process large amount of heat is released to the surrounding. It is exothermic process.

For example:

4Li + O₂    →      2Li₂O

2Mg + O₂    →    2MgO

  S +  O₂      →      SO₂  

The product which is formed as a result of combustion reaction are called oxides.

In given examples we can see that lithium, magnesium and sulfur react with oxygen and product formed is oxides of respective elements such as lithium oxide ( Li₂O), magnesium oxide (MgO) and sulfur oxide ( SO₂ ).

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Write the correct ionic formula when given two elements that bond ionically.
Tanzania [10]
Answer:
            Ionic bond is formed by the opposite ions attraction between the 2 atoms in an ionically bonded compound. The two ions i.e. Cation and Anions are formed by oxidation and reduction reactions respectively. General Ionic formula is as follow,

                                             Mⁿ⁺ + Nⁿ⁻   →  MN
where;
           Mⁿ⁺   =  Cation

           Nⁿ⁻   =  Anion

           MN  =  Salt

Explanation:
                   Ionic bond is the electrostatic forces of attraction between positively charged cations and negatively charged Anions. These forces are very stronger resulting in increasing several physical properties of Ionic compounds like melting point and boiling point e.t.c.

Example:

Sodium Chloride:
                           NaCl is formed by Na⁺ cation and Cl⁻ anion as follow,

Oxidation of Na;

                                      2 Na  →  2 Na⁺  +  2 e⁻
Reduction of Cl₂;

                                    Cl₂  +  2 e⁻  →  2 Cl⁻

Crystal Lattice formation is as follow,

                                  Na⁺  +  Cl⁻   →  NaCl
4 0
3 years ago
Give explanations for the large drops in melting point from C to Si and from Ge to Sn.
____ [38]

As we move down the group, the metallic bond becomes more stable and the formation of forming covalent bond decreases down the group due to the large size of elements.

Covalent and metallic bonding leads to higher melting points. Due to a decrease in attractive forces from carbon to lead there is a drop in melting point.

Carbon forms large covalent molecules than silicon and hence has a higher melting point than silicon.

Similarly, Ge also forms a large number of covalent bonds and has a smaller size as compared to that of Sn. Hence melting point decreases from Ge to Sn.

The order will be C>Si>Ge>Pb>Sn.

To learn more about the covalent bond, visit: brainly.com/question/10777799

#SPJ4

3 0
1 year ago
Which highway run through comlumbus
Alexeev081 [22]

Answer:

Highway 22 ,GA 31820

Explanation:

3 0
3 years ago
Chemist and physicist marie curie became famous for her pioneering research on radioactivity. What are the two chemical elements
wolverine [178]

Answer:

The two elements are POLONIUM and RADIUM.

Explanation:

Maria Curie is a French physicist and chemist, though she was of a Polish naturals. She was the first woman to receive a Noble Price which she earned for conducting leading and head way research on radioactivity. She discovered the theory of radioactivity; also the techniques isolating radioactive isotopes. These helped her and her husband discover Polium and Radium.

4 0
3 years ago
Read 2 more answers
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
3 years ago
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