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gizmo_the_mogwai [7]
2 years ago
8

A 2.87 g sample of carbon reacts with hydrogen to form 3.41 g of car fuel. What is the empirical formula for the car fuel?

Chemistry
1 answer:
olga_2 [115]2 years ago
5 0

CH₇ is the empirical formula of the car fuel.

Explanation:

To find the empirical formula we use the following algorithm.

First divide each mass the the molar weight of each element:

for carbon 2.87 / 12 = 0.239

for hydrogen 3.41 / 2 = 1.705

And now divide each quantity by the lowest number which is 0.239:

for carbon 0.239 / 0.239 = 1

for hydrogen 1.705 / 0.239 = 7.13 ≈ 7

The empirical formula of the car fuel is CH₇.

I have to tell you that in reality this formula is wrong because is not possible to exist. However the algorithm for finding the empirical formula is right, the problem may reside in the amounts of carbon and hydrogen given.

Learn more about:

empirical formula

brainly.com/question/5297213

#learnwithBrainly

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A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s
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11.31g NaClO₂

Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>

You can change the concentration of the substance if you write the moles of the substances:

[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>

Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =

<h3>11.31g NaClO₂</h3>
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3 years ago
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