Answer:
202 g/mol
Explanation:
Let's consider the neutralization between a generic monoprotic acid and KOH.
HA + KOH → KA + H₂O
The moles of KOH that reacted are:
0.0164 L × 0.08133 mol/L = 1.33 × 10⁻³ mol
The molar ratio of HA to KOH is 1:1. Then, the moles of HA that reacted are 1.33 × 10⁻³ moles.
1.33 × 10⁻³ moles of HA have a mass of 0.2688 g. The molar mass of the acid is:
0.2688 g/1.33 × 10⁻³ mol = 202 g/mol
∆H = m x s x ∆T, where m is the mass of the reactants, s is the specific heat of the product, and ∆T is the change in temperature from the reaction.
Answer:
2.12 moles of gas were added.
Explanation:
We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:
Where in this case:
We <u>input the data</u>:
- 6.13 L * n₂ = 11.3 L * 2.51 mol
As <em>4.63 moles is the final number of moles</em>, the number of moles added is:
Hello.
The answer is C.Amine
When an amine is combined (reacted) with a carboxyl group, an AMIDE + water is formed, and if you carry on heating under a vacuum, an imidazoline is formed.
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