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Rzqust [24]
3 years ago
15

What is the empirical formula of C6H18O3?

Chemistry
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

The answer to your question is C₂H₆O

Explanation:

Data

Molecular formula = C₆H₁₈O₃

Empirical formula = ?

Empirical formula is defined as the simplest ratio of the elements that form part of a molecule.

Process

To find the empirical formula find the greatest common factor of the subscripts.

                             6    18   3   2

                             3      9  3   3

                              1     3    1   3

                                     1

The GCF is 3, so factor 3 of the molecular formula

                         3 ( C₂H₆O)  

The result is the empirical formula C₂H₆O

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Bio-chemisty of protain​
sergejj [24]

Answer:

Protein biochemistry is the study of proteins. Protein biochemistry is a scientific field dedicated to the study of proteins, complex chains of amino acids which make up the building blocks of all living organisms.

Explanation:

I hope that helped

Copy and Pasted!

7 0
3 years ago
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
A chemical reaction that removes electrons from an atom is called a(n) ________ reaction.
ch4aika [34]
A chemical reaction that removes electrons from an atom is called "O<span>xidation".
The term came from late 18th century from French.
When the electrons are removed from an atom it increase its valence.</span>
3 0
3 years ago
When an aqueous solution containing hydrochloric acid is just neutralized with an aqueous solution containing sodium hydroxide,
Elza [17]

Answer:

c. salty.

Explanation:

From the question,

The reaction of an acid , i.e hydrochloric acid and a base i.e. , sodium hydroxide takes place , which is a type of neutralization reaction,

Where the acid and a base reacts to give salt along with water .

From the question,

The reaction is as follows -

2 HCl  +  NaOH  →  NaCl  +  H₂O  

Hence, the resulting product is a salt ,

Therefore , the resulting solution would taste like salt.

3 0
3 years ago
Calcium hydroxide is a strong base but is not very soluble ( Ksp = 5.02 X 10-6 ). What is the pH of a saturated solution of Ca(O
Misha Larkins [42]

Answer : The pH of a saturated solution is, 12.33

Explanation : Given,

K_{sp} = 5.02\times 10^{-6}

First we have to calculate the solubility of OH^- ion.

The balanced equilibrium reaction will be:

Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-

Let the solubility will be, 's'.

The concentration of Ca^{2+} ion = s

The concentration of OH^- ion = 2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][OH^-]^2

Let the solubility will be, 's'

K_{sp}=(s)\times (2s)^2

K_{sp}=(4s)^3

Now put the value of K_[sp} in this expression, we get the solubility.

5.02\times 10^{-6}=(4s)^3

s=1.079\times 10^{-2}M

The concentration of Ca^{2+} ion = s = 1.079\times 10^{-2}M

The concentration of OH^- ion = 2s = 2\times (1.079\times 10^{-2}M)=2.158\times 10^{-2}M

First we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (2.158\times 10^{-2})

pOH=1.67

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-1.67=12.33

Therefore, the pH of a saturated solution is, 12.33

7 0
3 years ago
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