Freezing Point of Sea water would be lower than that of Pure Water. It is because of salinity of the water
Hope this helps!
Answer:
P_2 =0.51 atm
Explanation:
Given that:
Volume (V1) = 2.50 L
Temperature (T1) = 298 K
Volume (V2) = 4.50 L
at standard temperature and pressure;
Pressure (P1) = 1 atm
Temperature (T2) = 273 K
Pressure P2 = ??
Using combined gas law:




Answer:
11.9 g of nitrogen monoxide
Explanation:
We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:
Mass of NH₃ = 6.75 g
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mole of NH₃ =?
Mole = mass /molar mass
Mole of NH₃ = 6.75 / 17
Mole of NH₃ = 0.397 mole
Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:
4NH₃ + 5O₂ —> 4NO + 6H₂O
From the balanced equation above,
4 moles of NH₃ reacted to produce 4 moles of NO.
Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.
Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:
Mole of NO = 0.397 mole
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO =?
Mass = mole × molar mass
Mass of NO = 0.397 × 30
Mass of NO = 11.9 g
Thus, the mass of NO produced is 11.9 g
Answer:
The system is not in equilibrium and will evolve left to right to reach equilibrium.
Explanation:
The reaction quotient Qc is defined for a generic reaction:
aA + bB → cC + dD
![Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where the concentrations are not those of equilibrium, but other given concentrations
Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:
![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%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where the concentrations are those of equilibrium.
This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
Comparing Qc with Kc allows to find out the status and evolution of the system:
- If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
- If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
- If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.
In this case:
![Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BSo_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BSO_%7B2%7D%20%5D%5E%7B2%7D%2A%20%5BO_%7B2%7D%5D%20%7D)

Q=100,000
100,000 < 4,300,000 (4.3*10⁶)
Q < Kc
<u><em>
The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>
Group 2 elements are the alkaline earth metals