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Andre45 [30]
2 years ago
9

34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an H2SO4 solution. What is the concentration of the original sulfu

ric acid solution?
Chemistry
1 answer:
Igoryamba2 years ago
5 0

0.1045M is the concentration of the original sulfuric acid solution

Explanation:

Titration is done to know the volume or concentration of unknown electrolyte.

Data given:

volume of acid = 50 ml

molarity of acid =?

volume of base NaOH = 34.62 ml

molarity of the base = 0.1510

For titration the formula used is,

Macid x Vacid = Mbase  x V base

Putting the values in above equation:

M acid x 50 = 34.62 x 0.1510

Macid = \frac{5.22762}{50}

           = 0.1045 M is the molarity of sulphuric acid solution used in neutralization of 0.1510 M base solution.

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 <span>Calculating average atomic mass is exactly like calculating a weighted average. Perform the following calculation: 

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One of the other answers given is correct, though the explanation is lacking a bit. Two of the answers can be eliminated immediately: 35.0 amu and 37.0 amu cannot be the average. If the mixture of isotopes was 50% and 50%, then 36.0 amu would be correct; however, the mixture is 75/25. This leaves only one possible answer choice.</span>
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A 23.2 g sample of an organic compound containing carbon, hydrogen and oxygen was burned in excess oxygen and yielded 52.8 g of
allochka39001 [22]

Answer:

The answer to your question is   C₃H₆O

Explanation:

Data

mass of sample = 23.2 g

mass of carbon dioxide = 52.8 g

mass of water = 21.6 g

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

                       44 g of CO₂ ---------------  12 g of C

                        52.8 g          ---------------  x

                        x = (52.8 x 12)/44

                        x = 633.6/44

                        x = 14.4 g of C

                        12 g of C ------------------  1 mol

                        14.4 g of C ---------------   x

                         x = (14.4 x 1)/(12)

                         x = 1.2 moles of C

2.- Calculate the grams and moles of Hydrogen

                         18 g of H₂O ---------------  2 g of H

                         21.6 g of H₂O -------------  x

                          x = (21.6 x 2) / 18

                         x = 2.4 g of H

                         1 g of H -------------------- 1 mol of H

                         2.4 g of H -----------------  x

                          x = (2.4 x 1)/1

                          x = 2.4 moles of H

3.- Calculate the grams and moles of Oxygen

Mass of Oxygen = 23.2 - 14.4 - 2.4

                           = 6.4 g

                         16 g of O ----------------  1 mol

                          6.4 g of O --------------  x

                          x = (6.4 x 1)/16

                          x = 0.4 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon = 1.2 / 0.4 = 3

Hydrogen = 2.4/ 0.4 = 6

Oxygen = 0.4 / 0.4 = 1

5.- Write the empirical formula

                                C₃H₆O

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