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3 years ago

Which of the following reactions would result in the formation of a gaseous product?

Chemistry

2 answers:

emmainna [20.7K]3 years ago

8 0

The reaction that would result in formation of gaseous product is

KHCO3 + HBr ( third answer)

<em><u>explanation</u></em>

KHCO3 react with HBr to form KBr , CO2 and H2O according to the below equation.

KHCO3 + HBr → KBr + CO2 + H2O

CO2 ( carbon (IV) oxide) is the gas that is formed during reaction.

other reaction does not produce a a gas for example NaOH + HCl forms NaCl and H2O

Yanka [14]3 years ago

3 0

The third reaction will form a gaseous product, CO₂
KHCO₃ + HBr → KBr + H₂O + CO₂

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Write Balance chemical reaction for preparation of chlorine with or without application heat<br>

suter [353]

Answer:

2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)

Explanation:

Chlorine is a diatomic halogen gas known for its greenish-yellow colour. It has a pungent smell and is only moderately soluble in water.

It is a very reactive gas and is never found in free state in nature.

Chlorine can be prepared in the laboratory by oxidation of hydrochloric acid using KMnO4 as follows;

2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)

The set up does not need to be heated.

3 0

3 years ago

What does a student need to know about double bonds and triple bonds when predicting molecular geometry of molecules?

zhuklara [117]

This problem is asking for an explanation of what we need to know about double and triple bonds to successfully predict molecular geometries in molecules. At the end, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.

<h3>Molecular geometry:</h3>

In chemistry, molecules are not necessarily flat arrangements of atoms, yet they have specific bond angles, orientations and shapes, which define the molecular geometry. In such a way, we can use the VSEPR theory in order to know the molecular geometry of a molecule; however, we first need its Lewis structure or at least the number and type of bonds to do so.

Consider water and carbon dioxide; the former has two hydrogen to oxygen bonds (O-H) and 2 lone pairs because O has six valence electrons but just 2 are bonded to complete the octet, so 4 unpaired electrons lead to two lone pairs. On the other hand, the latter has two double bonds (C=O) and 0 lone pairs because carbon has four valence electrons and they are all bonded to complete the octet.

In such a way, one can see how the double bond affected the bonding in CO2 in contrast to the H2O; situation that also applies to triple bonds, because CO2 has a linear molecular geometry whereas water has a bent one (see attached picture)

Hence, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.

Learn more about molecular geometry: brainly.com/question/7558603

Learn more about the VSEPR theory: brainly.com/question/14225705

5 0

2 years ago

atomic size of element increasing with increasing in number of shell in the third period of periodic table name the smallest ele

Mars2501 [29]

Answer:

Argon (Ar) is smallest element in 3rd period in periodic table.

3 0

3 years ago

Explain how the development of the atomic theory is an example of science's self-correcting process of discovery in depth.

Stolb23 [73]

Answer:

En química y física, la teoría atómica es una teoría científica sobre la naturaleza de la materia que sostiene que está compuesta de unidades discretas llamadas átomos. Empezó como concepto filosófico en la Antigua Grecia y logró ampliar aceptación científica a principios del siglo XIX cuando los descubrimientos en el campo de la química demostraron que la materia realmente se comportaba como si estuviese hecha de átomos.

La palabra átomo proviene del adjetivo en griego antiguo átomos, que significa «indivisible». Los químicos del siglo XIX empezaron a utilizar el término en relación con el número creciente de elementos químicos irreducibles.1 Cerca del cambio al siguiente siglo, a través de varios experimentos con electromagnetismo y radiactividad, los físicos descubrieron que los "átomos indivisibles" eran de hecho un conglomerado de varias partículas subatómicas (principalmente, electrones, protones y neutrones), las que pueden existir separadas unas de otras. De hecho, en ciertos entornos extremos, como las estrellas de neutrones, la presión y la temperatura extremas impiden que los átomos puedan existir en absoluto.

Ya que se descubrió que los átomos podían dividirse, los físicos inventaron el término «partículas elementales» para describir las partes "indivisibles", aunque no indestructibles, de un átomo. El campo de ciencia que estudia las partículas subatómicas es la física de partículas y es en este campo donde los físicos esperan descubrir la auténtica naturaleza fundamental de la materia

5 0

3 years ago

An ideal gas at a given initial state expands to a fixed final volume. would the work be greater if the expansion occurs at cons

crimeas [40]

Answer:

Constant pressure

Explanation:

At constant pressure,

$w = -p\Delta V = -p(V_{f} - V_{i})$

At constant temperature,

$w = -RT \ln \left(\dfrac{V_{f}}{V_{i}} \right)$

1 mol of an ideal gas at STP has a volume of 22.71 L.

Let's compare the work done as it expands under each condition from an initial volume of 22.71 L.

Isobaric expansion

$w = -100p(V_{2} - 22.71}); \text{(1 bar$\cdot$L = 100 J)}$

A plot of -w vs V₂ gives a straight line (red) with a constant slope of 100 J/L as in the diagram below (Note that w is work done on the system, so -w is the work done by the system). \

Isothermal expansion

$w= -8.314 \times 273.15 \ln \left(\dfrac{V_{f}}{22.71} \right)\\\\= -2271 \left( \ln V_{f} -\ln22.71 \right)\\= -2271 \left(\ln V_{f} - 3.123 \right)\\= 7092 - 2271\ln V_{f}$

A plot of -w vs V₂ is a logarithmic curve. Its slope starts at 100 J/mol but decreases as the volume increases (the blue curve below).

Thus, the work done during an expansion at constant pressure is greater than if the system is at constant temperature.

4 0

3 years ago