Interpret the electron configuration

1. Consider the following: h2so4 2 naoh → na2so4 2 h2o if a 24. 0 ml sample of h2so4 (aq) was neutralized by 20. 0 ml of 0. 125

jasenka [17]

The molarity of the acid sample H₂SO₄ is 0.052M .

<h3>What is Molarity ?</h3>

Molarity (M) is the amount of a substance in a certain volume of solution.

Molarity is defined as the moles of a solute per liters of a solution.

Molarity is also known as the molar concentration of a solution

Now to determine the molarity of the acid sample

V( H₂SO₄) = 24.0 mL in liters = 24.0 / 1000 = 0.024 L

M(H₂SO₄) = ?

V(NaOH) = 20.0 mL = 20.0 / 1000 = 0.02 L

M(NaOH) = 0.125 M

Number of moles NaOH :

n = M x V

n = 0.125 x 0.02

n = 0.0025 moles of NaOH

H₂SO₄(aq) + 2 NaOH(aq) = Na₂SO₄(aq) + 2 H₂O(l)

1 mole H₂SO₄ ---------- 2 mole NaOH

? mole H₂SO₄ ---------- 0.0025 moles NaOH

moles = 0.0025 * 1 / 2

= 0.00125 moles of H₂SO₄

M(H₂SO₄) = n / V

M = 0.00125 / 0.024

= 0.052 M

Therefore the molarity of the acid sample H₂SO₄ is 0.052M .

To know more about molarity

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In a solution, the substance that changes phase is the _____, while the substance that does not change phase is the____.

iren [92.7K]

Phase change is the solute
the one that doesn't phase change is the solvent.
A

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3 years ago

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How does a lithium cation compare to a lithium atom?

ExtremeBDS [4]

The correct answer is gonna be C) A lithium cation ion is smaller

A lithium cation has lost its valence electrons, which causes the remaining electrons to be pulled in stronger by the positive charge in the nucleus. As they get closer to the nucleus, the overall size of the atom is decreased.

6 0

3 years ago

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7 f Find the volume in dm3 and in mole of 0.505m of NaoH required to react with 40ml of 0.505m

Anna [14]

The volume of NaOH required is 0.08 dm³

To solve this question, we'll begin by writing the balanced equation for the reaction between H₂SO₄ and NaOH. This is illustrated below:

H₂SO₄ + 2NaOH —> Na₂SO₄ + 2H₂O

From the balanced equation above,

Mole ratio of the acid, H₂SO₄ $(n_{A}) = 1$

Mole ratio of the base, NaOH $(n_{B}) = 2$

Next, we shall determine the volume of NaOH required to react with H₂SO₄. This can be obtained as follow:

Molarity of the base, NaOH $(M_{B}) = 0.505 M$

Volume of the acid, H₂SO₄ $(V_{A}) = 40 mL$

Molarity of the acid, H₂SO₄ $(M_{A}) = 0.505 M$

<h3>Volume of the base, NaOH $(V_{B})$ =? </h3>

$\frac{M_{A} * V_{A}}{M_{B} * V_{B}} = \frac{n_{A}}{n_{B}}\\\\\frac{0.505 * 40}{0.505 *V_{B}} = \frac{1}{2}\\\\\frac{20.2}{0.505 *V_{B}} = \frac{1}{2}$