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Vinil7 [7]
2 years ago
13

Predict what must happen when a beaker containing a solution of lithium chloride is heated. check all that apply. view available

hint(s) check all that apply. energy will be transferred into the system. the δh3 value will increase. the δhsoln value will decrease. the reaction will become endothermic.
Chemistry
2 answers:
weqwewe [10]2 years ago
6 0

Answer:  energy will be transferred into the system and the reaction will become endothermic.

Explanation:   When the solution of Lithium chloride is heated that means the system is solution of lithium chloride and by heating, we are providing the sufficient amount of energy to the system. Thus in other words, energy will be transferred into the system.

Moreover, the system now will represent an example of endothermic reaction as the energy is being provided to the system and the process of dissolution is basically an endothermic reaction in which the heat is being transferred to the system.

slava [35]2 years ago
5 0

The correct answer is that the energy will be transferred into the system and as well as the solubility of lithium chloride will likely increase. It Is because LiCl is soluble when is dispersed in hot water that is why the solubility of lithium chloride will increase and that the heating solution will likely transfer the heat energy in the system in which the energy will be transferred in the system is correct.

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At an elevated temperature, Kp=4.2 x 10^-9 for the reaction 2HBr (g)---> +H2(g) + Br2 (g). If the initial partial pressures o
Damm [24]

Answer : The partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

Explanation :

The partial pressure of HBr = 1.0\times 10^{-2}atm

The partial pressure of H_2 = 2.0\times 10^{-4}atm

The partial pressure of Br_2 = 2.0\times 10^{-4}atm

K_p=4.2\times 10^{-9}

The balanced equilibrium reaction is,

                                2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)

Initial pressure    1.0×10⁻²       2.0×10⁻⁴      2.0×10⁻⁴

At eqm.            (1.0×10⁻²-2p)   (2.0×10⁻⁴+p)  (2.0×10⁻⁴+p)

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{H_2})(p_{Br_2})}{(p_{HBr})^2}

Now put all the values in this expression, we get :

4.2\times 10^{-9}=\frac{(2.0\times 10^{-4}+p)(2.0\times 10^{-4}+p)}{(1.0\times 10^{-2}-2p)^2}

p=-1.99\times 10^{-4}

The partial pressure of H_2 at equilibrium = (2.0×10⁻⁴+(-1.99×10⁻⁴) )= 1.0 × 10⁻⁶

Therefore, the partial pressure of H_2 at equilibrium is, 1.0 × 10⁻⁶

4 0
3 years ago
How many moles of ba(oh)2 are present in 275 ml of 0.200 m ba(oh)2?
DENIUS [597]
You have to put your attention to the unit of concentration. It is expressed in terms of molarity, which is represented in M. It is the number of moles solute per liter solution. So, you simply have to multiply the molarity with the volume in liters.

Volume = 275 mL * 1 L/1000 mL = 0.275 L
<em>Moles Ba(OH)₂ = (0.200 M)(0.275 L) = 0.055 mol</em>
6 0
3 years ago
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