<u>Answer:</u> The concentration of nitrogen dioxide at equilibrium is 0.063 M
<u>Explanation:</u>
We are given:
Initial concentration of nitrogen dioxide = 0.0250 M
Initial concentration of dinitrogen tetraoxide = 0.0250 M
For the given chemical equation:
<u>Initial:</u> 0.025 0.025
<u>At eqllm:</u> 0.025-x 0.025+2x
The expression of 
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%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 nitrogen dioxide = (0.025 + 2x) = [0.025 + 2(0.019)] = 0.063 M
Hence, the concentration of nitrogen dioxide at equilibrium is 0.063 M
The answer is 360ms I hope this helps
<span>NaCl is one and that is the only one that I know is for sure.
i hope this help!!!!!!!!!!</span>
Answer:
After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.
Explanation:
The first order kinetics reaction is:
ln [A] = ln [A]₀ - kt
<em>Where [A] is concentration after t time, [A]₀ is intial concentration and k is reaction constant.</em>
To convert half-life to k you must use:
t(1/2) = ln 2 / K
221s = ln 2 / K
K = ln 2 / 221s
<h3>K = 3.1364x10⁻³s⁻¹</h3>
If [A] = 1/64, [A]₀ = 1:
ln [A] = ln [A]₀ - kt
ln (1/64) = ln 1 - 3.1364x10⁻³t
4.1588 = 3.1364x10⁻³s⁻¹t
1326s = t
<h3>After 1326s, the concentration of pyruvic acid fall to 1/64 of its initial concentration.</h3>
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