Answer:
The balanced chemical equation: NH₃ + 2 HF → NH₄⁺ + HF₂⁻
Explanation:
According to the Brønsted–Lowry acid–base theory, the acid- base reaction is a type of chemical reaction between the acid and base to give a conjugate acid and a conjugate base.
In this reaction, a Brønsted–Lowry acid loses a proton to form a conjugate base. Whereas, a Brønsted–Lowry base accepts a proton to form a conjugate acid.
Acid + Base ⇌ Conjugate Base + Conjugate Acid
The acid dissociation constant (Kₐ) <em>signifies the acidic strength of a chemical species.</em>
∵ pKₐ = - log Kₐ
Thus for a strong acid, Kₐ value is large and pKₐ value is small.
pKₐ (HF) = 3.2 → strong acid
pKₐ (NH₃) = 38 → weak acid
<u>The chemical reaction involved in the dissolution process:</u>
NH₃ + 2 HF → NH₄⁺ + HF₂⁻
In this acid-base reaction, the acid HF reacts with NH₃ base to give the conjugate base HF₂⁻ and conjugate acid NH₄⁺.
<u>HF (acid) donates a proton to form the conjugate base, HF₂⁻ ion. NH₃ (base) accepts a proton to form the conjugate acid. </u>
Im pretty sure the answer is 4 but not 100 percent
Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
Answer:
0.11mole
Explanation:
Let us assume that the condition is at standard temperature and pressure(STP);
Given parameters:
Volume of water = 2.45L
Unknown:
Number of moles found in this volume of water = ?
Solution;
At STP;
Number of moles = 
Input the parameters and solve;
Number of moles of water = 
= 0.11mole
The number of moles of water found is 0.11mole
