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Degger
3 months ago
6

Suppose you perform a calorimeter experiment to determine the molar heat of neutralization of an unknown acid, H A HA, with sodi

um hydroxide, N a O H NaOH. You mix 37.2 mL of 0.50 M H A HA with 56.8 mL of 0.75 M N a O H NaOH and calculate the heat of reaction as -1.6 kJ. What is the molar heat of neutralization (in kJ/mol) for the unknown acid
Chemistry
1 answer:
Alekssandra [3K]3 months ago
5 0

Answer:

-86.02 kJ/ mole

Explanation:

The amount of acid utilized in moles = Molarity × Volume (L) =

= 0.50 (0.0372 L)

= 0.0186 moles

The energy released = -1.6 kJ

Thus, 0.0186 moles of HA neutralization results in: -1.6 kJ

The molar heat of neutralization for one mole of the unidentified acid is = -1.6/0.0186

= -86.02 kJ/ mole

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How many molecules are in 13.5g of sulfur dioxide, so2?
alisha [2963]
Answer: The number of sulfur dioxide molecules present is 1.27·10²³.
Calculating: m(SO₂) equals 13.5 g.
Using the formula n(SO₂) = m(SO₂) ÷ M(SO₂).
This gives n(SO₂) = 13.5 g ÷ 64 g/mol.
Resulting in n(SO₂) = 0.21 mol.
Subsequently, N(SO₂) = n(SO₂) ·Na.
Therefore, N(SO₂) = 0.21 mol · 6.022·10²³ 1/mol.
Ultimately, N(SO₂) equals 1.27·10²³.
Where n represents amount of substance.
M refers to molar mass.
Na is Avogadro's number.
5 0
3 months ago
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