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Crazy boy [7]
3 years ago
14

Consider this system at equilibrium. A(aq)<--->B(aq)deltaH=+550 kJ/mol What can be said about Q and K immediately afteran

increase in temperature? A) Q > K because Q increased B) Q > K because K decreased C) Q < K because Q decreased D) Q< K because K increased E) Q = K because neither charged How will the system respond to a temperature increase? A) shift left B) shift right C) no change
Chemistry
1 answer:
Nataliya [291]3 years ago
3 0

Answer:

Q < K because K increased

The system will SHIFT TO THE RIGHT

Explanation:

First, the reaction is endothermic. That is, a reaction in which heat is absorbed from the surroundings. ΔH is positive for endothermic reactions.

At the addition of temperature, the system absorbed more heat causing more products to be formed, shifting the equilibrium to the right.

The equilibrium constant k will therefore increase as the concentration of product increases

K = [B] / [A].

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The two reactions illustrated in the diagrams below often occur when a foreign substance enters the body.
zheka24 [161]

The specific proteins produced by cell B in response to the foreign substance are antibodies.

<h3>What are antibodies?</h3>

Antibodies are specific proteins produced by the immune cells of the body in response to a foreign substance called antigen which produces the immune response.

Antibodies are released by immune cells such as B cells.

The antibodies bind to antigen and tag them for destruction by phagocytes.

Therefore, cell B will produce antibodies.

Learn more about antibodies at: brainly.com/question/15382995

7 0
2 years ago
Which base is strong but never concentrated
GREYUIT [131]
I am pretty sure that it is magnesium hydroxide. pls brainliest 
3 0
3 years ago
An unknown gas is found to diffuse through a porous membrane 4.11 times more slowly than h2 what is the molecular weight of the
kirill [66]
To answer this question, you need to know <span>Graham's Law of Effusion/Diffusion formula. In this formula, the rate of diffusion/effusion would be influenced by the mass. As the molecule has bigger mass, the rate should be slower because it will be harder to pass the membrane. The calculation should be:</span>
<span>Rate 1 / Rate 2 = √[M2/M1] 
</span>4.11/1= √[M2/2] 
M2=33.78 g/mol

5 0
3 years ago
Read 2 more answers
I need help identifying the type of reactions. Can anyone help?
Eddi Din [679]

Answer:

Reaction 5: Decomposition reaction.

Reaction 6: Single replacement reaction

Reaction 7: Combination reaction.

Reaction 8: Combustion reaction.

Explanation:

<u><em>Reaction 5:</em></u>    2KClO₃ → 2KCl + 3O₂.

  • It is a decomposition reaction.
  • A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more elements or new compounds.
  • In this reaction: potassium chlorate decomposes into two single components (potassium chloride and oxygen).
  • So, it is a decomposition reaction.

<u><em>Reaction 6:</em></u> Zn + 2HCl → H₂ + ZnCl₂.

  • It is a single replacement reaction.
  • A single-replacement reaction, a single-displacement reaction, is a reaction by which one (or more) element(s) replaces an/other element(s) in a compound.
  • It is most often occur if element is more reactive than the other, thus giving a more stable product.
  • In this reaction, zinc metal (more active) displaces the hydrogen to form hydrogen gas and zinc chloride, a salt. Zinc reacts quickly with the acid to form bubbles of hydrogen.

<u><em>Reaction 7:</em></u> N₂O₅ + H₂O → 2HNO₃.

  • It is a combination "synthesis" reaction.
  • A synthesis reaction has two or more reactants and only one product.
  • In this reaction, dinitrogen pentoxide reacts with water to produce nitric acid.
  • So, it is considered as a synthetic "combination" reaction.

<u><em>Reaction 8:</em></u> 2C₂H₆ + 7O₂ →  4CO₂ + 6H₂O.

  • It is a combustion reaction.
  • A combustion reaction is a reaction where hydrocarbon alkane is completely burned in oxygen to produce water and carbon dioxide.
  • In this reaction 1.0 mole of ethane is burned to give 4.0 moles of carbon dioxide and 6.0 moles of water.
  • So, it is considered as a combustion reaction.
7 0
3 years ago
Read 2 more answers
A 13.5g sample of gold is heated, then placed in a calorimeter containing 60g of water. The temperature of water increases from
sweet-ann [11.9K]

Answer:

T_i~=163.1 ºC

Explanation:

We have to start with the variables of the problem:

Mass of water = 60 g

Mass of gold = 13.5 g

Initial temperature of water= 19 ºC

Final temperature of water= 20 ºC

<u>Initial temperature of gold= Unknow</u>

Final temperature of gold= 20 ºC

Specific heat of gold = 0.13J/gºC

Specific heat of water = 4.186 J/g°C

Now if we remember the <u>heat equation</u>:

Q_H_2_O=m_H_2_O*Cp_H_2_O*deltaT

Q_A_u=m_A_u*Cp_A_u*deltaT

We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:

m_H_2_O*Cp_H_2_O*deltaT=~-~m_A_u*Cp_A_u*deltaT

Now we can <u>put the values into the equation</u>:

60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C=-(13.5~g*0.13~J/g{\circ}C*(20-T_i)~{\circ}C)

Now we can <u>solve for the initial temperature of gold</u>, so:

T_i~=(\frac{60~g*4.186~J/g{\circ}C*(20-19)~{\circ}C}{13.5~g*0.13~J/g{\circ}C})+20

T_i~=163.1 ºC

I hope it helps!

5 0
3 years ago
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