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The chemical system will shift to the left in order to produce more N₂ , H₂.
The missing data in the question is
The equilibrium constant (K) of the reaction below is K = 6.0 x 10⁻² with initial concentrations as follows: H₂ = 1.0 x 10⁻² M, N₂ = 4.0 M, and NH₃ = 1.0 x 10⁻⁴M.
<h3>What is Chemical Equilibrium?</h3>When the rate of forward reaction is equal to the rate of the backward reaction then the equation is said to be in Chemical Equilibrium.
N₂ + 3H₂ --> 2NH₃
The reaction quotient, Q, has the same algebraic expressions but uses the actual concentrations of reactants.
[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M
Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
Q = (5.6 x 10⁻³ )² / (1.0 x 10⁻²) ³(4.0 )
Q = 7.84
As Q > K,
The chemical system will shift to the left in order to produce more N₂ , H₂
To know more about chemical equilibrium
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Before addition of HCl,
conc. of CH3COOH = 0.450 M
conc. of CH3COONa = 0.550 M
After addition of 0.08 M HCl, following reaction occurs in system:
HCl + CH3COONa ↔ CH3COOH + NaCl
Thus, in reaction system conc. of CH3COOH will increase to 0.53 M (0.08M + 0.450M)
And, conc to CH3COONa will reduce to 0.47 M (0.550M - 0.08M)
Now, conc. of H+ ions = ka![\frac{[acid]}{[conjugated base]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Bacid%5D%7D%7B%5Bconjugated%20base%5D%7D)
where ka = dissociation constant for acid = 10^-5 for Ch3COOH
∴ conc. of H+ ions =
= 1.1277 x 10^-5
Now, pH = -log [H+] = -log (1.1227 x 10^-5) = 4.94
conc. of CH3COOH = 0.450 M
conc. of CH3COONa = 0.550 M
After addition of 0.08 M HCl, following reaction occurs in system:
HCl + CH3COONa ↔ CH3COOH + NaCl
Thus, in reaction system conc. of CH3COOH will increase to 0.53 M (0.08M + 0.450M)
And, conc to CH3COONa will reduce to 0.47 M (0.550M - 0.08M)
Now, conc. of H+ ions = ka
where ka = dissociation constant for acid = 10^-5 for Ch3COOH
∴ conc. of H+ ions =
= 1.1277 x 10^-5
Now, pH = -log [H+] = -log (1.1227 x 10^-5) = 4.94
Answer:
The mass of CO in 12,000 Liters of air is .
Explanation:
The ppm is the amount of solute (in milligrams) present in one Liter of a solvent. It is also known as parts-per million.
To calculate the ppm of oxygen in sea water, we use the equation:
The ppm concentration of CO = 10 ppm
Mass of CO = m
Volume of air = 12,000 L
The mass of CO in 12,000 Liters of air is .