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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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The rate constant for the first-order decomposition of n2o is 3.40 s-1. what is the half-life of the decomposition?
stiks02 [169]

Hey There:

First order  half life equation :

T 1/2  = ln ( 2 ) /K

T 1/2 = 0.693 / 3.40

T 1/2 = 0.204 s

Answer B

3 0
3 years ago
Read 2 more answers
If iron metal reacts with an aqueous solution of silver nitrate and zinc reacts with an aqueous solution of iron sulfate, rank t
Lelechka [254]

Answer:

yes!you are right a cloudy formation will be formed when they will react.its because if nitrogen.

5 0
4 years ago
When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for
Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

3 0
3 years ago
Help please #7 and #8
ira [324]
7 A lead
B Gold
Csilver

8 Afeathrr
B water
C silver
6 0
3 years ago
How many grams of iron(II) chloride are needed to produce 44.3 g iron(II) phosphate in the presence of excess sodium phosphate?
zalisa [80]

Answer:

47.2 g

Explanation:

Let's consider the following double displacement reaction.

3 FeCl₂ + 2 Na₃PO₄ → Fe₃(PO₄)₂ + 6 NaCl

The molar mass of Fe₃(PO₄)₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:

44.3 g × (1 mol / 357.48 g) = 0.124 mol

The molar ratio of Fe₃(PO₄)₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:

3 × 0.124 mol = 0.372 mol

The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:

0.372 mol × (126.75 g/mol) = 47.2 g

5 0
3 years ago
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