Answer:
1.68 × 10²³ Molecules
Explanation:
As we know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of particles (molecules) contained by 0.280 moles of Br₂, we will use following relation,
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Solving for Number of Molecules,
Number of Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹
Putting values,
Number of Molecules = 0.280 mol × 6.022 × 10²³ Molecules.mol⁻¹
Number of Molecules = 1.68 × 10²³ Molecules
Hence,
There are 1.68 × 10²³ Molecules present in 0.280 moles of Br₂.
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Answer:
I believe it’s the mitochondria
Explanation:
The mass, in grams, of the sample of methanol (CH₃OH) is 64 grams.
<h3>How we calculate mass from moles?</h3>
Mass of any substance can be calculated by using moles as:
n = W/M, where
W = required mass
M = molar mass
In the question that:
Moles of methanol = 2mole
Molar mass of methanol = 32g/mole
On putting these values in the above equation, we get
W = n × M
W = 2mole × 32g/mole = 64g
Hence, 64 grams is the mass of the sample.
To know more about moles, visit the below link:
brainly.com/question/15374113
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.
