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Mamont248 [21]
2 years ago
10

A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 26.6

mL of the base is added. The concentration of acetic acid in the sample was ________ M.
Chemistry
1 answer:
joja [24]2 years ago
4 0

<u>Answer:</u> The concentration of CH_3COOH comes out to be 0.16 M.

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is CH_3COOH

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=?M\\V_1=25mL\\n_2=1\\M_2=0.175M\\V_2=26.6mL

Putting values in above equation, we get:

1\times M_1\times 25=1\times 0.175\times 26.6\\\\M_1=0.1862M

Hence, the concentration of CH_3COOH comes out to be 0.1862 M.

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Explanation:

It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.

             \frac{1 g}{1000 mg} or \frac{1000 mg}{1 g}

So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.

4 0
3 years ago
How do you solve this ??
kotegsom [21]
Answer : Option A) 2.00 eV

Explanation : The conversion of J to eV is done with the following formula;

E_{eV} = E_{J} X (6.241 X 10^{18})

Here, we have the value of particle in terms of Joules which is 3.2 X 10^{-19}

So, on substituting we get,
E_{eV} = 3.2 X 10^{-19}  X  (6.241 X 10^{18} )


E_{eV} = 1.99 eV so, it can be rounded off to 2.00 eV.
3 0
3 years ago
The concept of empiricism states that all rationally accepted knowledge is determined from experience. Francis bacon was one of
lawyer [7]

the answer is option D "people emphasized obtaining knowledge through scientific experiments" (on plato)

6 0
2 years ago
Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc
Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is C_5H_7N

(b) The empirical formula for the given compound is C_4H_5N_2O

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = \frac{6.17}{1.23}=5.01\approx 5

For Hydrogen  = \frac{8.70}{1.23}=7.07\approx 7

For Nitrogen = \frac{1.23}{1.23}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is C_5H_7N_1=C_5H_7N

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = \frac{4.12}{1.03}=4

For Hydrogen  = \frac{5.19}{1.03}=5.03\approx 5

For Nitrogen = \frac{2.06}{1.03}=2

For Nitrogen = \frac{1.03}{1.03}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is C_4H_5N_2O_1=C_4H_5N_2O

6 0
3 years ago
Please help I'm not good at science​
Citrus2011 [14]

Answer:

i'm not able to see images but if you type it all i can answer:)

Explanation:

8 0
3 years ago
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