<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M
<u>Explanation:</u>
Molarity is calculated by using the equation:

Moles of HI = 0.550 moles
Volume of container = 2.00 L

For the given chemical equation:

<u>Initial:</u> 0.275
<u>At eqllm:</u> 0.275-2x x x
The expression of
for above equation follows:
![K_c=\frac{[H_2][I_2]}{[HI]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
We are given:

Putting values in above expression, we get:

Neglecting the negative value of 'x' because concentration cannot be negative
So, equilibrium concentration of hydrogen gas = x = 0.0275 M
Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M
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>
Answer:
c = 0.13 j/ g.°C
Explanation:
Given data:
Mass of mercury = 29.5 g
Initial temperature = 32°C
Final temperature = 161°C
Heat absorbed = 499.2 j
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Q = m.c. ΔT
ΔT = T2 - T1
ΔT = 161°C - 32°C
ΔT = 129 °C
Q = m.c. ΔT
c = Q / m. ΔT
c = 499.2 j / 29.5 g. 129 °C
c = 499.2 j / 3805.5 g. °C
c = 0.13 j/ g.°C
it is A Acid
Explanation : Any substance that yields a hydrogen ion when placed in a water solution is called an acid. According to Arrhenius theory of Acids and Bases, Acids are those substances which on dissociation in solution generates ions. Whereas a Base is a substance that dissociates in the solution to produce ions.
Answer:
1. 2.510kJ
2. Q = 1.5 kJ
Explanation:
Hello there!
In this case, according to the given information for this calorimetry problem, we can proceed as follows:
1. Here, we consider the following equivalence statement for converting from calories to joules and from joules to kilojoules:

Then, we perform the conversion as follows:

2. Here, we use the general heat equation:

And we plug in the given mass, specific heat and initial and final temperature to obtain:

Regards!