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svetoff [14.1K]
4 years ago
8

How many carbon atoms are there in the empirical formula corresponding to C10H22O2?

Chemistry
1 answer:
podryga [215]4 years ago
6 0

Empirical formula of the compound contains 5 carbon atoms.

Explanation:

We have the molecular formula of the compound C₁₀H₂₂O₂.

To find the empirical formula we divide each number of atoms by the lowest number which is 2:

for carbon (C) 10 / 2 = 5

for hydrogen (H) 22 / 2 = 11

for oxygen (O) 2 / 2 = 1

The empirical formula is C₅H₁₁O and contains 5 carbon atoms.

Learn more about:

empirical formula

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Which of these is an organism bacteria lung virus heart?
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The following equation describes how sodium and chlorine react to produce sodium chloride: 2Na + Cl2 -> 2NaCl Is the equation
DedPeter [7]

Answer:

The equation shows balance.

Explanation:

You can easily count the number of elements on each side.

On the left side of the equation, you have 2 moles of Na and 2 moles of Cl.

On the right side of the equation, you also have 2 moles of Na and 2 moles of Cl because the two elements formed a compound meaning that whatever number is in front of them, both elements will receive the same number.

5 0
4 years ago
assuming that each human being has 60 trillion body cells in that the earths population is 6 billion calculate the total number
sesenic [268]

3.6 x 10²³

Explanation:

Given parameters:

  Number of cell per body = 60 trillion body cells

   Population of the earth = 6 billion

Unknown:

Total number of living human body cells = ?

Solution:

  To solve this problem, simply find the product:

 Total number of living human body cells = number of cell per body x population of the earth

          1 trillion = 1 x 10¹²

         60 trillion = 6 x 10¹³

         1 billion = 1 x 10⁹

         6 billion = 6 x 10⁹

Total number of living human body cells = 6 x 10¹³ x  6 x 10⁹

                                                                      = 6 x 6 x 10⁽¹³⁺⁹⁾

                                                                       = 36 x 10²² or 3.6 x 10²³

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7 0
4 years ago
27.8 mL solution of 0.797 M HCHO2 with 0.928 M NaOH. What is the pH for the solution at the equivalence point in the titration?
kati45 [8]

Answer:

8.69 is the pH at the equivalence point

Explanation:

Formic acid, HCHO₂, reacts with NaOH as follows:

HCHO₂ + NaOH → NaCHO₂ + H₂O.

At the equivalence point you will have in the reaction just NaCHO₂ and H₂O. The concentration of NaCHO₂ will be:

<em />

<em>Moles: </em>0.0278L * 0.797mol/L = 0.02216moles

To reach the equivalence point it is necessary to add:

0.02216mol * (1L / 0.928mol) = 0.0239L

Total volume in the equivalence point:

0.0278L + 0.0239L = 0.0517L

Concentration: 0.02216moles / 0.0517L = 0.429M

The equilibrium of NaCHO₂, CHO₂⁻, in water is:

CHO₂⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HCHO₂(aq)

Where Kb, 5.56x10⁻¹¹ is defined as:

5.56x10⁻¹¹ = [OH⁻] [HCHO₂] / [CHO₂⁻]

In the equilibrium, it is produced X OH⁻ and HCHO₂, and as concentration of NaCHO₂ is 0.429M:

5.56x10⁻¹¹ = [X] [X] / [0.429M]

2.383x10⁻¹¹ = X²

4.88x10⁻⁶ = X = [OH⁻]

As pOH = -log [OH⁻]

pOH = 5.31

And pH = 14 - pH

pH = 8.69 is the pH at the equivalence point

3 0
3 years ago
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