<u>Answer:</u> The concentration of original sulfuric acid solution is 1.62 M
<u>Explanation:</u>
Let the original concentration of sulfuric acid be 'x' M
To calculate the molarity of the diluted solution, we use the equation:

where,
are the molarity and volume of the concentrated sulfuric acid solution
are the molarity and volume of diluted sulfuric acid solution
We are given:

Putting values in above equation, we get:

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

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the concentration of original sulfuric acid solution is 1.62 M
Answer:
b)
Explanation:
It shows the natrual proggresion of a story and helps blend the narrtive together.
Answer:
Rb<K<Ga<As<Se<S
Explanation:
We must remember that first ionization energy decreases down the group and increases across the period.
First ionization energy decreases down the group because of the addition of more shells which increases the distance between the nucleus and the outermost electron. Hence, Rb has a lower ionization energy that K.
Across the period, increase in the size of the nuclear charge causes the pull of the nucleus on the outermost electrons to increase thereby increasing the ionization energy. Hence ionization energy increases across the period. For this reason, the ionization energy of Ga<As<Se as shown.
Answer:
12
Explanation:
pOH = -log(1.0 x 10^-2) = 2
pH = 14 - pOH
pH = 14 - 2 = 12
The empirical formula for the caproic acid, given the combustion analysis data is C₃H₆O
We'll begin bey obtaining the mass of carbon, hydrogen and oxygen in the compound. This is illustrated below:
How to determine the mass of C
- Mass of CO₂ = 9.78 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 9.78
Mass of C = 2.67 g
How to determine the mass of H
- Mass of H₂O = 20.99 g
- Molar mass of H₂O = 18 g/mol
- Molar of H = 2 × 1 = 2 g/mol
- Mass of H =?
Mass of H = (2 / 18) × 4
Mass of H = 0.44 g
How to determine the mass of O
- Mass of compound = 4.30 g
- Mass of C = 2.67 g
- Mass of H = 0.44 g
- Mass of O =?
Mass of O = (mass of compound) – (mass of C + mass of H)
Mass of O = 4.30 – (2.67 + 0.44)
Mass of O = 1.19 g
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as follow:
- C = 2.67 g
- H = 0.44 g
- O = 1.19 g
- Empirical formula =?
Divide by their molar mass
C = 2.67 / 12 = 0.2225
H = 0.44 / 1 = 0.44
O = 1.19 / 16 = 0.074
Divide by the smallest
C = 0.2225 / 0.074 = 3
H = 0.44 / 0.0744 = 6
O = 0.074 / 0.074 = 1
Thus, the empirical formula of the compound is C₃H₆O
Learn more about empirical formula:
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