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maks197457 [2]
3 years ago
10

A student is using a coffee-cup calorimeter to determine the enthalpy change of the endothermic reaction of two aqueous solution

s. After both solutions are added to the cup, the student neglects to put the lid on the cup. This would cause the magnitude of the calculated ΔH° value to be: the answer is: too small, since the solution will absorb heat from the room. But why? Wouldn't depend on if the reaction releases or absorbs heat. Wouldn't it be too large because heat escapes the cup? I'm so confused
Chemistry
1 answer:
Ugo [173]3 years ago
6 0

Answer:

Explanation:

In all calorimetric experiment , the calorimeter must be isolated from the surrounding . Otherwise the heat change in the experiment can not be determined with precision .

The reaction is endothermic . Hence, there is lowering of temperature due to absorption of heat in the reaction equal to ΔH°. The value of ΔH° can be calculated by measuring fall in the temperature of the content . The fall in the temperature will be less when heat is allowed to come from the surrounding . Less fall of temperature will result in less ΔH° to be calculated .

Hence in the given experiment , if the student neglects to put lid on the cup , the experiment will give less value of ΔH°.

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A certain second-order reaction (B→products) has a rate constant of 1.30×10−3 M−1⋅s−1 at 27 ∘C and an initial half-life of 224 s
soldier1979 [14.2K]

Answer:

       \large\boxed{\large\boxed{0.291M}}

Explanation:

By definition one <em>half-life</em> is the time to reduce the initial concentration to half.

For a <em>second order reaction </em>the rate law equations are:

              \dfrac{d[B]}{dt}=-k[B]^2

             \dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt

The <em>half-life</em> equation is:

            t_{1/2}=\dfrac{1}{k[A]_0}

Thus, substitute the<em> rate constant</em>  1.30\times 10^{-3}M^{-1}\cdot s^{-1} and the <em>half-life </em>time <em>224s</em> to find [A]₀:

           224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}

           [A]_o=0.291M

7 0
3 years ago
In a solution, when the concentrations of a weak acid and its conjugate base are equal, ________.
NARA [144]

In a solution, when the concentrations of a weak acid and its conjugate base are equal, the -log of the concentration of H+ and the -log of the Ka are equal.

<h3>What is Henderson-Hasselbalch equation ?</h3>
  • This formula is frequently used to determine the pH of buffer solutions. A weak acid and its conjugate base make up buffer solutions.
  • To find the pH of the buffer solution, we obtain the pKa of the weak acid and add it to the log of the concentration of the conjugate base divided by the concentration of weak acid.
  • So whenever the concentration of the weak acid is equal to the concentration of the conjugate base, the pH of the buffer solution is equal to the pKa of the weak acid.
<h3 />

To know more about buffer solutions you can refer to the link below:

brainly.com/question/10695579

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6 0
2 years ago
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What was the name of the disease that killed nearly half the people of Europe?
Licemer1 [7]
The answer is the plague, or the black death
7 0
3 years ago
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Algunas drogas bloquean los químicos secretados por los axones. ¿Cómo podrían afectar estas drogas al impulso nervioso? ¿Qué pod
olya-2409 [2.1K]

Answer:

Algunas drogas intervienen en el impulso nervioso inhibiendo la liberacion de  neurotransmisores, por ende estas neuronas no generaran efectos postsinapticos.

Podria pasar que algunos procesos internos estaran inhibidos o fomentando indirectamente la aparicion de otros, un ejemplo claro de esto es la xerestomia generada por las drogas betabloqueantes.

Explanation:

Es asi que los procesos internos se veran desregularizados, las drogas que inhiben la liberacion de los neurotransmisores son reguladores por lo general de la tension arterial o antiarritmicos cardiacos, es por eso que su consumo es fundamental y vale el riesgo beneficio a la hora de evaluar la inhibicion de las respuestas postsinapticas.

Los neurotransmisores mas conocidos son la noradrenalina y adrenalia, le membrana presinaptica es la nerviosa y la postsinaptica puede ser muscularo o nerviosa.

4 0
3 years ago
Which of the following is not a conjugate acid-base pair? A) NH4+/NH3 B) H30-OH OC) H2SO3/HSO3 D) C2H302-/HC2H302
raketka [301]

Answer : The option (d) is not a conjugate acid-base pair.

Explanation :

According to the Bronsted Lowry concept, Bronsted Lowry-acid is a substance that donates one or more hydrogen ion in a reaction and Bronsted Lowry-base is a substance that accepts one or more hydrogen ion in a reaction.

Or we can say that, conjugate acid is proton donor and conjugate base is proton acceptor.

(a) The equilibrium reaction will be,

NH_4^++OH^-\rightleftharpoons NH_3+H_2O

In this reaction, NH_4^+/NH_3 are act as a conjugate acid-base.

(b) The equilibrium reaction will be,

H_3O^++OH^-\rightleftharpoons H_2O+H_2O

In this reaction, H_3O^+/OH^- are act as a conjugate acid-base.

(c) The equilibrium reaction will be,

H_2SO_3+OH^-\rightleftharpoons HSO_3^-+H_2O

In this reaction, H_2SO_3/HSO_3^- are act as a conjugate acid-base.

(d) The equilibrium reaction will be,

C_2H_3O_2^-+H_2O\rightleftharpoons HC_2H_3O_2+OH^-

In this reaction, C_2H_3O_2^-/HC_2H_3O_2 are act as a conjugate base-acid.

Hence, from this we conclude that, the option (d) is not a conjugate acid-base pair but it is a act as conjugate base-acid.

6 0
3 years ago
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