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Butoxors [25]
3 years ago
6

Which chemical property makes a substance suitable as a fuel?

Chemistry
2 answers:
inessss [21]3 years ago
4 0
<span> Ethanol's </span>chemical<span> formula is C2H5OH.</span>Properties<span>. Pure ethanol is a flammable, colorless liquid with a boiling point of 78.5° C. Its low m</span>
Goryan [66]3 years ago
4 0

High flammability is the answer :P

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Convert 1.75 g/mL to the unit g/m3. Answer Choices: (1.75 x 106 g/m3), (1.75 x 102 g/m3), (1.75 x 10-2 g/m3), (1.75 x 10-6 g/m3)
Marta_Voda [28]
 Answer : 1750000g/cm3
4 0
4 years ago
Mrs. Nogaki tested four black markers in rubbing alcohol and in water. Each filter paper had four evenly spaced dots. She found
ExtremeBDS [4]

Answer:

Lost pigment of marker when dipped in alcohol

Explanation:

dependent viable = output

so it's the output of what happens after the input.

  • she put the marker in the water which is the independent variable, that's the input
  • the output or the result of that decision is having lost pigment in the marker

8 0
3 years ago
A hydroxide group (OH) is a component of most...
qaws [65]

Answer: bases

Explanation:

5 0
3 years ago
What is the percentage composition of Na, O, and H in the compound NaOH?
lions [1.4K]

The percent composition of NaOH, also known as sodium hydroxide, is 57.48 percent sodium, 40 percent oxygen, and 2.52 percent hydrogen.

4 0
3 years ago
A student in chemistry 150-02 weighed out 55.5 g of octane (C8H18) and allowed it to react with oxygen, O2. The products formed
Anni [7]

Answer:

Explanation:

Given data:

Mass of octane = 55.5 g

Balanced chemical equation = ?

Mass of oxygen required to react  = ?

Mass of CO₂ for med = ?

Molecules of water produced = ?

Mass of octane required to produced 30.0 g of water = ?

Solution:

1)

Chemical equation:

2C₈H₁₈ + 25O₂     →  16CO₂ + 18H₂O

2)

Mass of oxygen required to react  = ?

Mass of octane = 55.0 g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with oxygen.

                        C₈H₁₈          :           O₂

                           2              :            25

                         0.48          :          25/2×0.48 = 6 mol

Mass of oxygen required:

Mass = number of moles × molar mass

Mass = 6 mol × 32 g/mol

Mass = 192 g

3)

Given data:

Mass of carbon dioxide produced = ?

Mass of octane = 55g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with CO₂.

                        C₈H₁₈          :           CO₂

                           2              :            16

                         0.48          :          16/2×0.48 = 3.84 mol

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 3.84 mol × 44 g/mol

Mass = 168.96 g

4)

Given data:

Molecules of water produced = ?

Mass of octane = 55g

Solution:

Number of moles of octane:

Number of moles = mass/ molar mass

Number of moles = 55.0 g/114.23 g/mol

Number of moles = 0.48 mol

Now we will compare the moles of octane with H₂O.

                        C₈H₁₈          :           H₂O

                           2              :            18

                         0.48          :          18/2×0.48 = 4.32 mol

Number  of molecules of water:

1 mol = 6.022× 10²³ molecules

4.32 mol × 6.022× 10²³ molecules/ 1 mol

26 × 10²³ molecules

5)

Given data:

Mass of octane required = ?

Mass of water produced = 30 g

Solution:

Number of moles of water.

Number of moles = mass/ molar mass

Number of moles = 30 g/ 18 gmol

Number  of moles = 1.67 mol

Now we will compare the moles of water and octane from balance chemical equation:

2C₈H₁₈ + 25O₂     →  16CO₂ + 18H₂O

                 

H₂O        :         C₈H₁₈

 18          :          2

 1.67       :       2/18×1.67 = 0.185 mol

Mass of octane:

Mass = number of moles ×molar mass

Mass = 0.185 × 114.23 g/mol

Mass = 21.13 g

5 0
3 years ago
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