Answer:
A. 0.90 L.
Explanation:
- NaOH solution will react with H₂SO₄ according to the balanced reaction:
<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.</em>
<em>1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.</em>
- For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.
<em>∴ (MV) NaOH = (xMV) H₂SO₄.</em>
x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.
<em>∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH</em> = 2(0.6 L)(3.0 M)/(4.0 M) = <em>0.90 L.</em>
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
<h3>2.75 = pKa</h3>
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
<h3>1.78x10⁻³ = Ka</h3>
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
Answer:
The advantages described below
Explanation:
Advantages of a balanced chemical equation versus word equation:
- easier to read: chemical equations typically only take one line and they include all the relevant information needed. They are short-hand notations for what we describe in words.
- balanced chemical equations show molar ratio in which reactants react and the molar ratio of the products. Those are coefficients in front of the species. This is typically not included in a word equation, for example, hydrochloric acid reacts with potassium hydroxide. The latter statement doesn't describe the molar ratio and stoichiometry.
- includes relevant information, such as catalysts, temperature and pressure above the arrow in the equation. We wouldn't have this in a word equation most of the time.
- shows the stoichiometry of each compound itself, e. g. if we state 'ammonia', we don't know what atoms it consists of as opposed to
![NH_3]](https://tex.z-dn.net/?f=NH_3%5D)
. - includes states of matter: aqueous, liquid, gas, solid. This would often be included in a word equation, however.
Answer:
CS₂
Explanation:
To find the empirical formula we need to determine first the percentage of each atom in the molecule. Then, we need to find the moles and, as empirical formula is the simplest whole-number ratio of atoms we can solve the empirical formula:
<em>%C:</em>
0.33gC / 2.12g * 100 = 15.6%
<em>%S:</em>
1.53g S / 1.82g * 100 = 84.1%
In a basis of 100, the moles of each atom are:
<em>C:</em>
15.6g C * (1mol / 12.01g) = 1.30 moles
<em>S:</em>
84.1g S * (1mol / 32.065g) = 2.62 moles
The ratio of Sulphur-Carbon is:
2.62mol / 1.30mol = 2
That means empirical formula is:
<h3>CS₂</h3>
