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Art [367]
3 years ago
9

A common fuel additive that is composed of C, H, and O enhanced the performance of gasoline began being phased out in 1999 becau

se of its contamination of drinking water. When 12.1 g of the compound are burned, 30.2 g of CO2and 14.8 g of H2O are formed. What is the empirical formula of the compound?
Chemistry
1 answer:
spayn [35]3 years ago
6 0

Answer:

C₅ H₁₂ O

Explanation:

44 g of CO₂ contains 12 g of C

30.2 g of CO₂ will contain 12 x 30.2 / 44 = 8.236 g of C .

18 g of H₂O contains 2 g of hydrogen

14.8 g of H₂0 will contain 1.644 g of  H .

total compound = 12.1 out of which 8.236 g is C and 1.644 g is H , rest will be O

gram of O = 2.22

moles of C, O, H in the given compound =  8.236 / 12 , 2.22 / 16 , 1.644 / 1

= .6863 , .13875 , 1.644

ratio of their moles = 4.946 : 1 : 11.84

rounding off to digits

ratio = 5 : 1 : 12

empirical formula = C₅ H₁₂ O

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a sample of 25.0g of an unknown metal is added to 25.0ml of water in a graduated cylinder and the final volume is 28.5ml what is
leva [86]

Answer:

<h2>The answer is 7.14 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass of metal = 25 g

volume = final volume of water - initial volume of water

volume = 28.5 - 25 = 3.5 mL

It's density is

density =  \frac{25}{3.5}  \\  = 7.142857...

We have the final answer as

<h3>7.14 g/mL</h3>

Hope this helps you

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3 years ago
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3 years ago
What is a synthetic reaction?​
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4 0
3 years ago
Read 2 more answers
What is the molarity of a solution that is prepared by adding 3 moles of CaSO4 in 1.5 L of water?
blondinia [14]

Answer:

[CaSO_4] = 2 mol/L

Explanation:

[CaSO_4] = \frac{3 mol}{1.5 L} = 2 mol/L

6 0
2 years ago
A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba 2 . When the concentration of F - exceeds __________ M,
Stella [2.4K]

Answer:

When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

Explanation:

Ksp of BaF₂ is:

BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.

As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to  ksp just when:

1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

1.7x10⁻⁶ = [0.0144M] [F⁻]²

1.18x10⁻⁴ = [F⁻]²

0.0109M = [F⁻]

That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

5 0
3 years ago
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