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Kamila [148]
3 years ago
14

The standard entropy change of a reaction has a positive value. This reaction results in: Select the correct answer below: a dec

rease in entropy. an increase in entropy. no entropy change. neither an entropy increase nor decrease.
Chemistry
1 answer:
Ira Lisetskai [31]3 years ago
5 0

Explanation:

The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.

Positive entropy means the system has increased its degree of disorderness.

You might be interested in
What is the molarity of a solution of Cu2SO4 if 250.0 mL of solution contains 14.1 g of Cu2SO4?
balu736 [363]
Copper(I) compounds in aqueous solutions are unstable and disproportionate:
Cu₂SO₄ = Cu + CuSO₄
Necessary for dissolution of non-aqueous solvent.

M(Cu₂SO₄)=223.16 g/mol
m(Cu₂SO₄)=14.1 g
v=0.250 L

n(Cu₂SO₄)=m(Cu₂SO₄)/M(Cu₂SO₄)

c=n(Cu₂SO₄)/v=m(Cu₂SO₄)/(vM(Cu₂SO₄))

c=14.1/(0.250*223.16)=0.253 mol/L

0.253 M

5 0
3 years ago
Read 2 more answers
Describe two tests that you can run to determine if something is a substance or a mixture.
ryzh [129]

The difference between a substance and a mixture is that a substance is one of a kind, a material of the same composition throughout, on the contrary, a mixture is one or more different substances brought together and mixed together without changing the nature of each single substance.

One way to test it is to take two substances like sand and table salt. They should each be in granular form and in adequate amount to mix. Neither substance has changed after mixing the two. Even though it may not be easy or convenient to accomplish, each substance could be separated out from the mixture.

When it comes to two substances in lump form, it would not be a mixture when one lump is positioned next to the other lump because there are not enough pieces to combine.

However, there could be a mixture of three substances, like sand, table salt and graphite powder and there could be a mixture with four substances, etc., ad infinitum.

Mixtures are of solid substances in general. On the other hand, one starts referring to solutions when liquids are involved. Gases can be a mixture like for example, air is a mixture with nitrogen, oxygen, argon, etc.


4 0
2 years ago
A silver cube with an edge length of 2.38 cm and a gold cube with an edge length of 2.79 cm are both heated to 81.9 ∘C and place
never [62]

Answer:

The final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

Explanation:

<u>Given data;</u>

edge length of silver, a = 2.38 cm = 0.0238 m

edge length of gold, a = 2.79 cm = 0.0279 m

final temperature of silver, t = 81.9 ° C

final temperature of Gold, t = 81.9 ° C

initial temperature of water, t = 19.6 ° C

volume of water, v =  109.5 mL = 0.0001095 m³

<u>Known data:</u>

density gold 19300 kg/m³

density silver 10490 kg/m³

density water 1000 kg/m³

specific heat gold is 129 J/kgC

specific heat silver is 240 J/kgC

specific heat water is 4200 J/kgC

<u>Calculated data</u>

Apply Pythagoras theorem to determine the side of each cube;

Silver cube;

let L be the side of the silver cube

Taking the cross section of the cube (form a right angled triangle), the edge  length forms the <em>hypotenuse side</em>.

L² + L² = 0.0238²

2L² = 0.0238²

L² = 0.0238² / 2

L² = 0.00028322

L = √0.00028322

L = 0.0168

Volume of cube = L³

Volume of the silver cube = (0.0168)³ = 4.742 x 10⁻⁶ m³

Gold cube;

let L be the side of the gold cube

L² + L² = 0.0279²

2L² = 0.0279²

L² = 0.0279² / 2

L² = 0.0003892

L = √0.0003892

L = 0.0197

Volume of cube = L³

Volume of the silver cube = (0.0197)³ = 7.645 x 10⁻⁶ m³

Mass of silver cube;

density = mass / volume

mass = density x volume

mass of silver cube = 10490 (kg/m³) x 4.742 x 10⁻⁶ (m³) = 0.0497 kg

Mass of Gold cube

mass of gold cube = 19300 (kg/m³) x 7.645 x 10⁻⁶ (m³) = 0.148 kg

Mass of water

mass of water = 1000 (kg/m³) x 0.0001095 (m³) = 0.1095 kg

Let the heat gained by cold water be  Q₁

Let the heat lost  by silver cube = Q₂

Let the heat lost  by gold cube = Q₃

Let the final temperature of water = T

Q₁  = 0.1095 kg x 4200 J/kgC x (T–19.6)

Q₂ = 0.0497 kg x 240 J/kgC x (81.9–T)

Q₃ = 0.148 kg x 129 J/kgC x (81.9–T)

At thermal equilibrium;

Q₁ = Q₂ + Q₃

0.1095 x 4200  (T–19.6)  = 0.0497 x 240 x (81.9–T)  + 0.148 x 129 x (81.9–T)

459.9 (T–19.6) = 11.928 (81.9–T) + 19.092(81.9–T)

459.9T - 9014.04 = 976.9032 - 11.928T + 1563.6348 - 19.092T

459.9T + 11.928T + 19.092T = 9014.04  + 976.9032 + 1563.6348

490.92T = 11554.578

T = 11554.578 / 490.92

T = 23.54 ⁰C

Therefore, the final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

6 0
3 years ago
L avanzar hacia la derecha por el período 5, el tamaño atómico, la energía de ionización y los electrones de valencia: A. dismin
rusak2 [61]

Answer:

Al avanzar hacia la derecha por el período 5, el tamaño atómico, la energía de ionización y los electrones de valencia: A. disminuye, aumenta y aumentan, respectivamente.

Explanation:

El radio atómico representa la distancia que existe entre el núcleo y la capa de valencia, es decir la más externa. Por medio del radio atómico es posible determinar el tamaño del átomo.  En un período el tamaño atómico disminuye de izquierda a derecha pues en este sentido aumenta el  número atómico aumentando la carga nuclear mientras que el efecto pantalla y el número de  niveles permanecen constantes. En otras palabras, disminuye de izquierda a derecha debido a la atracción que ejerce el núcleo sobre los electrones de los orbitales más externos, disminuyendo así la distancia núcleo-electrón.

<u><em>Al avanzar hacia la derecha por el período 5, el tamaño atómico disminuye.</em></u>

La energía de ionización es la necesaria para remover un electrón a un átomo en estado  gaseoso. Mientras más lejos del núcleo esté el electrón, es más fácil removerlo porque se necesita  menos energía. Al aumentar el número atómico de los elementos de un  mismo período, se incrementa la atracción nuclear sobre el  electrón más externo, ya que disminuye el radio atómico y  aumenta la carga nuclear efectiva sobre él. Entonces en un período, al aumentar el número atómico, la energía de ionización aumenta.

<u><em>Entonces, al avanzar hacia la derecha por el período 5, la energía de ionización y los electrones de valencia aumenta.</em></u>

Los electrones de valencia  son los electrones que están en la última capa electrónica (llamados orbitales de valencia) y tienen una alta posibilidad de participar en una reacción química.

En cada período aparecen los elementos cuyo último nivel de su configuración electrónica coincide con el número del período, ordenados por orden creciente de número atómico. Por ejemplo, el período 3 incluye los elementos cuyos electrones más externos están en el nivel 3.

Los electrones de valencia aumentan en número a medida que se avanza en un período.

<u><em> Entonces, al avanzar hacia la derecha por el período 5, los electrones de valencia aumentan.</em></u>

4 0
3 years ago
You are examining the DNA sequences that code for the enzyme phosphofructokinase in skinks and Komodo dragons. You notice that t
jek_recluse [69]

Answer: D. Mutation in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.

Explanation: It was not likely to be that the coding sequences are replicated more often. The only possible explanation is that the mutations in coding is more likely to be deleterious to the organism than mutations because it is in a non coding sequence.

8 0
2 years ago
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