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Bogdan [553]
3 years ago
9

Please help. I am very stuck on this portfolio and I have no idea what to do.

Chemistry
1 answer:
DaniilM [7]3 years ago
5 0
Same I don’t know it
You might be interested in
Hey! I was wondering how to get the solution to this question!
zzz [600]

Answer:

6.22 × 10⁻⁵

Explanation:

Step 1: Write the dissociation reaction

HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺

Step 2: Calculate the concentration of H⁺

The pH of the solution is 2.78.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M

Step 3: Calculate the molar concentration of the benzoic acid

We will use the following expression.

Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution

Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M

Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid

We will use the following expression.

Ka = [H⁺]²/Ca

Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵

3 0
3 years ago
A student wants to remove the salt from a mixture of sand and salt in order to get only pure sand. He adds water to the mixture.
raketka [301]

Answer:

Here is one way: Add water to the mixture. Only the sugar dissolves. This is a physical change.

Explanation:

The sugar would dissolve in water. You could then pour off the solution and wash the remaining sand with a bit more water. Heat the water to evaporate it from the sugar, and the two are separated.

5 0
3 years ago
Read 2 more answers
Draw a bond-line structure for CH3CH2O(CH2)2CH(CH3)2.<br><br> Include Lone Pairs in your answer.
mash [69]

Answer:

See explanation and image attached

Explanation:

A bond line structure refers to any structure of a covalent molecule wherein the covalent bonds present in the molecule are represented with a single line for each level of bond order.

The bond-line structure of CH3CH2O(CH2)2CH(CH3)2 has been shown in the image attached. We know that oxygen has a lone pair of electrons and this has been clearly shown also in the image attached.

7 0
3 years ago
Determine the equilibrium constant for the acid-base reaction between ethanol and hydrobromic acid? Acid pKa Hydrobromic Acid −5
siniylev [52]

Answer:

10^{-3.4

Explanation:

Let us first take a look at the image below;

In the acid - base  reaction; we can see the transfer of electrons that takes place;

We can also see that the reaction goes in the direction which converts the stronger acid and the stronger base to the weaker acid and the weaker base.

The stronger acid is shown with the one with more negative P_{Ka} Value.

∴ The equilibrium constant for the acid-base reaction is expressed as:

K_{eq}= \frac{K_a of reactant acid}{K_a of product acid}

      = \frac{10^{-pK} of reactant acid}{10^{-pk} of product acid}

From  P_{Ka} Value (shown in the image below), it is clear and vivid that hydrobromic acid is a stronger acid than the ethyloxonium ion, therefore the equilibrium lies to the right.

From the chemical equation (shown in the attached image); the equilibrium constant for the acid-base reaction can be expressed as:

K_{eq}=\frac{10^{-pK} of hydrobromic acid}{10^{-pk} of Ethyloxonium acid}

K_{eq}=\frac{10^{-5.8} of hydrobromic acid}{10^{-2.4} of Ethyloxonium acid}

       = 10^{-3.4}

6 0
3 years ago
Question 5 (1 point)
Marianna [84]

Answer:

P=12.16 atm

Explanation:

Using the formula of ideal gas law:

PV = nRT

P= nRT/V

 n= number of moles

 R= Avogadro constant = 0.0821

 T= Temperature in K => ºC + 273.15 K

P= (1.50 moles)(0.0821)( 296.15 K)/ 3.00L

P= 12.15

7 0
3 years ago
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