Answer:
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Explanation:
i can help if you get the pic to work
The given sentence is part of a longer question.
I found this question with the same sentence. So, I will help you using this question:
For the reaction N2O4<span>(g) ⇄ 2NO</span>2(g), a reaction mixture at a certain temperature initially contains both N2O4 and NO2 in their standard states (meaning they are gases with a pressure of 1 atm<span>). If </span>Kp = 0.15, which statement is true of the reaction mixture before
any reaction occurs?
(a) Q = K<span>; The reaction </span>is at equilibrium.
(b) Q < K<span>;
The reaction </span>will proceed to
the right.
(c) Q > K<span>; The reaction </span>will proceed to the left.
The answer is the option (c) Q > K<span>; The reaction will proceed to the </span>left,
since Qp<span> = </span>1<span>, and 1 > 0.15.</span>
Explanation:
Kp is the equilibrium constant in term of the partial pressures of the gases.
Q is the reaction quotient. It is a measure of the progress of a chemical reaction.
The reaction quotient has the same form of the equilibrium constant but using the concentrations or partial pressures at any moment.
At equilibrium both Kp and Q are equal. Q = Kp
If Q < Kp then the reaction will go to the right (forward reaction) trying to reach the equilibrium,
If Q > Kp then the reaction will go to the left (reverse reaction) trying to reach the equilibrium.
Here, the state is that both pressures are 1 atm, so Q = (1)^2 / 1 = 1.
Since, Q = 1 and Kp = 0.15, Q > Kp and the reaction will proceed to the left.
Answer:
34,6g of (NH₄)₂SO₄
Explanation:
The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:
ΔT = kb×m
Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.
For the problem:
ΔT = 109,7°C-108,3°C = 1,4°C
kb = 1.07 °C kg/mol
Solving:
m = 1,31 mol/kg
As mass of X = 600g = 0,600kg:
1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:
0,785 moles of ions×
= 0,262 moles of (NH₄)₂SO₄
As molar mass of (NH₄)₂SO₄ is 132,14g/mol:
0,262 moles of (NH₄)₂SO₄×
= <em>34,6g of (NH₄)₂SO₄</em>
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I hope it helps!
Answer:
The pressure contribution from the heavy particles is 17.5 atm
Explanation:
According to Dalton's law of partial pressures, if there is a mixture of gases which do not react chemically together, then the total pressure exerted by the mixture is the sum of the partial pressures of the individual gases that make up the mixture.
In the simulation:
the pressure of the 50 light particles alone was determined to be 5.9 atm, the pressure of the 150 heavy particles alone was measured to be 17.5 atm,
the total pressure of the mixture of 150 heavy and 50 light particles was measured to be 23.4 atm
Total pressure = partial pressure of Heavy particles + partial pressure of light particles
23.4 atm = partial pressure of Heavy particles + 5.9 atm
Partial pressure of Heavy particles = (23.4 - 5.9) atm
Partial pressure of Heavy particles = 17.5 atm
Therefore, the pressure contribution from the heavy particles is 17.5 atm
The two main factors the temperature of seawater are density and the salinity of the water.
