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Delvig [45]
2 years ago
10

Which of the following acids (listed with pKa values) and their conjugate base would form a buffer with a pH of 8.10?

Chemistry
1 answer:
Amiraneli [1.4K]2 years ago
7 0

Answer:

(C) HClO, pKa = 7.54

Explanation:

A buffer is a solution that can resist abrupt changes in pH when acids or bases are added. It is formed by two components:

  • A weak acid and its conjugate base.
  • A weak base and its conjugate acid.

In this case, acid and base are defined according to Bronsted-Löwry theory, which states that acids are substances that <em>release H⁺</em> and bases are substances that <em>accept H⁺. </em>Therefore, when an acid loses an H⁺ transforms into its conjugated base. For example, HF/F⁻ is a conjugate acid-base pair.

In buffers, when an acid is added, it reacts with the base to diminish its amount:

F⁻ + H⁺ ⇄ HF

Also in buffers, when a base is added, it reacts with the acid to diminish its amount:

HF + OH⁻ = F⁻ + H₂O

The optimum pH range of work of a buffer system (known as buffer range) is between 1 unit less and 1 unit more of pH than its pKa.

So, the buffer formed by HClO/ClO⁻ works optimally in the pH range 6.54-8.54. Since pH = 8.10 is in that interval, this would be the optimal choice.

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What is the iupac name of the following compound? <br><br> 2,2-dimethyl-4-ethylheptane
Lady bird [3.3K]

Answer:

The IUPAC name of the compound has already been given which is 2,2-dimethyl-4-ethylheptane.

Explanation:

The IUPAC (International Union of Pure and Applied Chemistry) is an authority in chemistry that provides a guideline and standardized methods in the naming of compounds formed from the periodic table.

In order the give an IUPAC name to a compound, certain steps needs to be followed, these includes:

--> Identify the functional group in the compound as this will form the suffix. For example if the functional group is an alkane the suffix will be -ane.

--> Identify the longest carbon chain (it may not be a straight chain) that contains the functional group. This forms the prefix. Example: if the longest carbon chain is 7 carbon atoms then the prefix will be hept-

--> All the carbons of the longest chain should be numbered

--> Identify branched groups on the chain and name them according to the number of carbon atoms. They usually end with -yl.

--> Finally, combine the elements of the name is a single word.

The structural formula of the IUPAC compound can be found in the attached file for a better understanding. The branched groups are circled.

4 0
3 years ago
I need help please my teaching is going to go crazy if I didn’t do it
victus00 [196]

Answer:

No One is going to answer all this because it hard to type it all in her. its easy to search it up

Explanation:

4 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
Which of the following is true?
Umnica [9.8K]

Answer:

The answer should be D

Explanation:

because turning 1 mole of propane to grams means its still one mole of propane just in a different unit.

8 0
2 years ago
A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera
Grace [21]

Answer: The initial temperature of the iron was 515^0C

Explanation:

heat_{absorbed}=heat_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]         .................(1)

where,

q = heat absorbed or released

m_1 = mass of iron = 360 g

m_2 = mass of water = 750 g

T_{final} = final temperature = 46.7^0C

T_1 = temperature of iron = ?

T_2 = temperature of water = 22.5^oC

c_1 = specific heat of iron = 0.450J/g^0C

c_2 = specific heat of water= 4.184J/g^0C

Now put all the given values in equation (1), we get

-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]

T_i=515^0C

Therefore, the initial temperature of the iron was 515^0C

4 0
3 years ago
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