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bearhunter [10]
3 years ago
6

A burning match will burn more vigorously in pure oxygen than in air because _________ . Select one: a. oxygen is a catalyst for

combustion b. nitrogen is a reactant in combustion and its low concentration in pure oxygen catalyzes the combustion c. oxygen is a product of combustion d. nitrogen is a product of combustion and the system reaches equilibrium at a lower temperature e. oxygen is a reactant in combustion and pure oxygen increases the reactant concentration
Chemistry
1 answer:
Westkost [7]3 years ago
8 0

Answer:

e. oxygen is a reactant in combustion and pure oxygen increases the reactant concentration

Explanation:

The reaction of a burning match is combustion. In this combustion, the organic components of the match (such as cellulose, C₆H₁₀O₅) react with oxygen, producing water and carbon dioxide:

  • C₆H₁₀O₅(s) + 6O₂(g) → 5H₂O(g) + 6CO₂(g)

Seeing as oxygen is a reactant and not a catalyst nor product, and that nitrogen plays no part in the reaction, <em>the only correct answer is option e</em>.

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Question 1<br><br> Of the following which displays a positive energetic process?
TiliK225 [7]

We can describe a positive energetic process as any process which increases the internal energy of the system.

A positive energetic reaction or process is often referred to as being Endothermic. This means that the system which is performing the process absorbs energy. Some examples include:

  • Boiling an Egg
  • Roasting food over a fire (the food is the reference system)

etc

Therefore, we can confirm that a positive energetic process is one in which the system in question absorbs energy, thus increasing its internal energy.

<em>Since I could not locate the options online, I have provided a general explanation of the concept coupled with a few examples.</em>

<em />

<em />

To learn more visit:

brainly.com/question/4345448?referrer=searchResults

5 0
2 years ago
Which is the correct Lewis structure for carbon monoxide?
LuckyWell [14K]
<span>The Lewis structure for CO has 10 valence electrons. For the CO Lewis structure you'll need a triple bond between the Carbon and Oxygen atoms in order to satisfy the octets of each atom while still using the 10 valence electrons available for the CO molecule.</span>
7 0
3 years ago
Does the physical form of the material matter for mass-mole<br> and mole-mass calculations?
natulia [17]
Tin metal reacts with hydrogen fluoride to produce tin(II) fluoride and hydrogen gas according to the following balanced equation.

Sn(s)+2HF(g)→SnF2(s)+H2(g)
Sn(s)+2HF(g)→
SnF
2
(s)+
H
2
(g)

How many moles of hydrogen fluoride are required to react completely with 75.0 g of tin?

Step 1: List the known quantities and plan the problem.

Known

given: 75.0 g Sn
molar mass of Sn = 118.69 g/mol
1 mol Sn = 2 mol HF (mole ratio)
Unknown

mol HF
Use the molar mass of Sn to convert the grams of Sn to moles. Then use the mole ratio to convert from mol Sn to mol HF. This will be done in a single two-step calculation.

g Sn → mol Sn → mol HF

Step 2: Solve.

75.0 g Sn×1 mol Sn118.69 g Sn×2 mol HF1 mol Sn=1.26 mol HF
75.0 g Sn×
1
mol Sn
118.69
g Sn
×
2
mol HF
1
mol Sn
=1.26 mol HF

Step 3: Think about your result.

The mass of tin is less than one mole, but the 1:2 ratio means that more than one mole of HF is required for the reaction. The answer has three significant figures because the given mass has three significant figures.
4 0
3 years ago
Read 2 more answers
A chemical reaction produces formaldehyde, with a chemical formula of CH2O. Carbon is in group 4A, Oxygen is in group 6A, and Hy
steposvetlana [31]

Answer:

Explanation:

carbon in  group 4A needs 4 more electrons to be stable

hydrogen in group 1A has one electron needs 1 more t o be stale

oxygen in group 6A has 6 and needs 2 more to be stable

They all obtain this by sharing electrons  8 around C &O, 2 around H

H  : C:  H

       ::        

      :O:

5 0
2 years ago
Complete combustion of a 17.12mg sample of xylene In oxygen yielded 56.77mg
Veronika [31]

Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=

106.16×1000

17.12

=0.00016moles

Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO

2

=

44.01×1000

56.77

=0.0013

Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H

2

O==

18.02×1000

14.53

=0.0008

Moles ratios

\frac{0.0013}{0.0008}=1.625

0.0008

0.0013

=1.625

\frac{0.0008}{0.0008}=1

0.0008

0.0008

=1

Hence molecular fomula

The empirical formula is C 4H 5.

The molecular formula C8H10

8 0
2 years ago
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