<span> Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol
M(NaBr)= 102 g/mol
M(NaNO3) = 85 g/mol
1) 244g NaNO3 *1 mol NaNO3/85 g NaNO3 = 244/85 mol NaNO3
2)</span> Pb(NO3)2 + 2 NaBr ------> PbBr2 + 2 NaNO3
2 mol 2 mol<span>
x mol </span>244/85 mol
<span>
x=(2 mol*</span> 244/85 mol )/2 mol = 244/85 mol NaBr
<span>
3) </span> 244/85 mol NaBr*102g NaBr/1 mol = (244*102/85) g NaBr =292.8 g NaBr<span>
</span>
the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
0.00193 mol/L
Given that:
numbers of moles of H₂S = 0.59 moles
Volume = 3.0-L
Equilibrium constant = 9.30 × 10⁻⁸
The equation for the reaction is given as :
2H₂S ⇄ 2H₂(g) + S₂(g)
The initial concentration of H₂S =
The initial concentration of H₂S =
= 0.1966 mol/L
The ICE table is shown be as :
2H₂S ⇄ 2H₂(g) + S₂(g)
Initial 0.9166 0 0
Change -2 x +2 x + x
Equilibrium (0.9166 - 2x) 2x x
(since 2x < 0.1966 if solved through quadratic equation)
The equilibrium concentration for H₂(g) = 2x
∴
= 0.00193 mol/L
Thus, the equilibrium concentration of H₂(g) at 700°C = 0.00193 mol/L
To know more about equilibrium concentration
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The total pressure of the gaseous mixture has been 5.37 atm. Thus, option D is correct.
The partial pressure has been defined as the pressure exerted by each gas in the mixture.
According to the Dalton's law of partial pressure, the total pressure of gas has been the sum of the partial pressure of the gases in the mixture.
The given partial pressure of gases in the mixture has been:
- Partial pressure of Nitrogen,
- Partial pressure of Oxygen,
- Partial pressure of Argon,
- Partial pressure of Helium,
- Partial pressure of Hydrogen,
The total pressure of the gaseous mixture has been:
The total pressure of the gaseous mixture has been 5.37 atm. Thus, option D is correct.
For more information about partial pressure, refer to the link:
brainly.com/question/14623719
Answer:
Potassium chloride, KCl , is an ionic compound formed by the electrostatic force of attraction that holds the potassium cations and the chlorine anions together. Potassium, K , is located in group 1 of the Periodic Table.
Explanation:
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