Answer:
Kc for this equilibrium is 2.30*10⁻⁶
Explanation:
Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.
Being:
aA + bB ⇔ cC + dD
the equilibrium constant Kc is defined as:
![Kc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%2A%5BD%5D%5E%7Bd%7D%20%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.
In this case, being:
2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)
the equilibrium constant Kc is:
![Kc=\frac{[N_{2} ]*[H_{2} ]^{3} }{[NH_{3} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BN_%7B2%7D%20%5D%2A%5BH_%7B2%7D%20%5D%5E%7B3%7D%20%20%7D%7B%5BNH_%7B3%7D%20%5D%5E%7B2%7D%20%7D)
Being:
- [N₂]= 0.0551 M
- [H₂]= 0.0183 M
- [NH₃]= 0.383 M
and replacing:

you get:
Kc= 2.30*10⁻⁶
<u><em>Kc for this equilibrium is 2.30*10⁻⁶</em></u>
Answer:
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Explanation:

Moles of hydrochloric acid = n
Volume of hydrochloric acid solution = 200.0 mL = 0.200 L
Molarity of the hydrochloric acid = 0.089 M
of HCL

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.
Then 0.0178 moles of HCl wil be neutralized by :
of sodium bicarbonate
Mass of 0.0178 moles of sodium bicarbonate:
0.0178 mol × 72 g/mol = 1.4952 g
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Answer:
the answer is distillation
CH4 is an emprirical formula as it shows the simplest ratio of the numbers of different atoms present in the molecule. The empirical formula for CH4 is also the same as the molecular formula.
The other compunds can be simplified so they are not the empirical formula of compounds.
Hope this helps :).