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Jet001 [13]
2 years ago
13

¿Cómo se escribe la fórmula molecular (orden de los iones)?

Chemistry
1 answer:
cluponka [151]2 years ago
6 0

Answer:

A la izquierda el catión y a la derecha el anión.

Explanation:

¡Hola!

En este caso, y basado en las normas IUPAC para la escritura de las fórmulas moleculares, es necesario primero escribir el catión a la izquerda, seguido del anión a la derecha, tal y como se muestra en los siguientes ejemplos, recordando que el catión es el ion cargado positivamente y el anión, negativamente:

K^+Cl^-\\\\Ag_2^+(SO_4)^{2-}

Los cuales son cloruro de potasio y sulfato de plata respectivamente. También es necesario tener en cuenta que los metales tienden a ser cationes por su capacidad de perder electrones, mientras que los no metales a ganarlos y por ende resultar como aniones.

¡Un gusto ayudarte!

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What is the charge of a cation?
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Answer:

A cation has more protons than electrons, consequently giving it a net positive charge. For a cation to form, one or more electrons must be lost, typically pulled away by atoms with a stronger affinity for them.

Were i found my answer: Cation vs Anion: Definition, Chart and the Periodic Table

Explanation:

5 0
2 years ago
Which of the following methods can be used to identify the suspected use of heat lamps in a house without entering the house?
LuckyWell [14K]

Answer:

b I see some of ask you to do it

7 0
2 years ago
Propane gas, C3H8, is sometimes used as a fuel. In order to measure its energy output as a fuel a 1.860 g sample was combined wi
lisov135 [29]

Answer:

The heat of the reaction is 105.308 kJ/mol.

Explanation:

Let the heat released during reaction be q.

Heat gained by water: Q

Mass of water ,m= 1kg = 1000 g

Heat capacity of water ,c= 4.184 J/g°C

Change in temperature = ΔT = 26.061°C - 25.000°C=1.061 °C

Q=mcΔT

Heat gained by bomb calorimeter =Q'

Heat capacity of bomb calorimeter ,C= 4.643 J/g°C

Change in temperature = ΔT'= ΔT= 26.061°C - 25.000°C=1.061 °C

Q'=CΔT'=CΔT

Total heat released during reaction is equal to total heat gained by water and bomb calorimeter.

q= -(Q+Q')

q = -mcΔT - CΔT=-ΔT(mc+C)

q=-1.061^oC(1000 g\times 4.184J/g^oC+4.643 J/^oC )=-4,444.15J=-4.444 kJ

Moles of propane =\frac{1.860 g}{44 g/mol}=0.0422 mol

0.0422 moles of propane on reaction with oxygen releases 4.444 kJ of heat.

The heat of the reaction will be:

\frac{4.444 kJ}{0.0422 mol}=105.308 kJ/mol

3 0
3 years ago
The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu
MAVERICK [17]

Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation-     k=Ae^{\frac{-E_{a}}{RT}}    , where k is rate constant, A is pre-exponential factor, E_{a} is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

                                   \frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}

Here \frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005} , T = 298 K , R = 8.314 J/(mol.K) and E_{a}^{ala-pro}=60kJ/mol

So, \frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}

   \Rightarrow E_{a}^{phe-pro}=65705J/mol=66kJ/mol (rounded off to two significant digit)

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7 0
2 years ago
Which solution has the higher boiling point?
dedylja [7]

NaCl has higher boiling point than C₂H₆O₂ in H₂O.

<h3>What is molality?</h3>
  • The number of moles of solute in a solution equal to 1 kg or 1000 g of solvent is referred to as its molality.
  • The definition of molarity, on the other hand, is based on a certain volume of solution.
  • Mol/kg is a typical molality measurement unit in chemistry.
<h3>Calculation of boiling point:</h3>

When the non-volatile solute is dissolved in a solvent, the boiling point rises along with the molality (concentration) of the solute.

Given,

Mass of  C₂H₆O₂ (solute) = 15.0 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of C₂H₆O₂ = 62 g/mol

Thus, the moles of solute = 15.0 g x 1.0 mol solute/62 g/mol

= 0.2419 mol

Therefore, molality(m) of C₂H₆O₂ (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2419/0.50

= 0.4838 m

Similarly,

Given ,

Mass of  NaCl (solute) = 15 g

Mass of solvent (H₂O) = 0.50 kg

Molar mass of NaCl (solute) = 58.44 g/mol

Thus, the moles of NaCl (solute) = 15.0 g x 1.0 mol solute/58.44

= 0.2566 mol

Therefore, molality(m) of NaCl (solute) = Amount of solute (mol)/ Mass of solvent

= 0.2566 mol/0.50 kg  

= 0.51 m

Hence, the molality of NaCl (solute) is more than the molality of C₂H₆O₂(solute).

So, with an increase in the solute's concentration (molality), the boiling point rises.

Therefore, NaCl has higher boiling point than C₂H₆O₂ in H₂O.

Learn more about boiling point here:

brainly.com/question/24168079

#SPJ4

7 0
1 year ago
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