Should hydrogen be placed in group one and why

Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose

Slav-nsk [51]

Answer:

The minimum mass of octane that could be left over is 43.0 grams

Explanation:

Step 1: Data given

Mass of octane = 73.0 grams

Mass of oxygen = 105.0 grams

Molar mass octane = 114.23 g/mol

Molar mass oxygen = 32.0 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate the number of moles

Moles = mass / molar mass

Moles octane = 73.0 grams / 114.23 g/mol

Moles octane = 0.639 moles

Moles O2 = 105.0 grams / 32.0 g/mol

Moles O2 = 3.28 moles

Step 4: Calculate the limiting reactant

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

O2 is the limiting reactant. It will completely be consumed. (3.28 moles). There will react 3.28 / 12.5 = 0.2624 moles. There will remain 0.639 - 0.2624 = 0.3766 moles octane

Step 5: Calculate mass octane remaining

Mass octane = moles * molar mass

Mass octane = 0.3766 moles * 114.23 g/mol

Mass octane = 43.0 grams

The minimum mass of octane that could be left over is 43.0 grams

3 0

3 years ago

En un vaso de precipitado de un litro se coloca exactamente 500 mL de agua destilada a temperatura ambiente y se realizan dos ex

Serga [27]

Answer:

1) La masa del agua a temperatura ambiente es de 500 gramos, 2) La masa del agua cuando se congela es de 500 gramos, 3) La masa de agua que queda después de la evaporación es de 400 gramos, 4) Se ha evaporado 100 gramos de agua.

Explanation:

1) ¿Cuál es la masa de agua a temperatura ambiente?

Podemos determinar la masa inicial del agua ( $m_{o}$ ), medido en gramos, al conocer su densidad ( $\rho_{w}$ ), medida en gramos por mililitro, y volumen inicial ocupado en el vaso de precipitado ( $V_{o}$ ), medido en mililitros, a partir de la siguiente expresión:

$m_{o} =\rho_{w}\cdot V_{o}$

Si sabemos que $\rho_{w} = 1\,\frac{g}{mL}$ y $V_{o} = 500\,mL$ , entonces:

$m_{o} = \left(1\,\frac{g}{mL} \right)\cdot (500\,mL)$

$m_{o} = 500\,g$

La masa del agua a temperatura ambiente es de 500 gramos.

2) ¿Cuál es la masa de agua cuando se congela?

Puesto que el proceso de congelación no implica transferencia de masa, la masa de agua se conserva al transformarse en hielo. Por tanto, la masa resultante es de 500 gramos.

3) ¿Cuál es la masa de agua que queda después de la evaporación?

Durante la evaporación una parte del agua es transferida al aire, entonces podemos calcular la masa final ( $m_{f}$ ), medido en gramos, de la sustancia al multiplicar el volumen final ( $V_{f}$ ), medido en mililitros, por la densidad del agua ( $\rho_{w}$ ), medida en gramos por mililitro,. Es decir,

$m_{f} =\rho_{w}\cdot V_{f}$

Si sabemos que $\rho_{w} = 1\,\frac{g}{mL}$ y $V_{f} = 400\,mL$ , entonces:

$m_{f} = \left(1\,\frac{g}{mL} \right)\cdot (400\,mL)$

$m_{f} = 400\,g$

La masa de agua que queda después de la evaporación es de 400 gramos.

4) ¿Qué masa de agua se evaporó?

Determinamos que la masa evaporada de agua ( $m_{v}$ ), medida en gramos, es igual a la diferencia entre las masas inicial y final, ambas medidas en gramos:

$m_{v} =m_{o}-m_{f}$

Si $m_{o} = 500\,g$ y $m_{f} = 400\,g$ , entonces tenemos que:

$m_{v} = 500\,g -400\,g$

$m_{v} = 100\,g$

Se ha evaporado 100 gramos de agua.

5 0

3 years ago

When the paramagnetic [co(cn)6] 4– ion is oxidized to [co(cn)6] 3– , the ion becomes diamagnetic. however, when the paramagnetic

S_A_V [24]

Answer:
Basically, paramagnetic and diamagnetic refer to the way a chemical species interacts with a magnetic field. More specifically, it refers to whether or not a chemical species has any unpaired electrons or not.
A diamagnetic species has no unpaired electrons, while a paramagnetic species has one or more unpaired electrons.
Now, I won't go into too much detail about crystal field theory in general, since I assume that you're familiar with it.
So, you're dealing with the hexafluorocobaltate(III) ion, [CoF6]3â’, and the hexacyanocobaltate(III) ion, [Co(CN)6]3â’.
You know that [CoF6]3â’ is paramagnetic and that [Co(CN)6]3â’ is diamagnetic, which means that you're going to have to determine why the former ion has unpaired electrons and the latter does not.
Both complex ions contain the cobalt(III) cation, Co3+, which has the following electron configuration
Co3+:1s22s22p63s23p63d6
For an isolated cobalt(III) cation, all these five 3d-orbitals are degenerate. The thing to remember now is that the position of the ligand on the spectrochemical series will determine how these d-orbtals will split.
More specifically, you can say that
a strong field ligand will produce a more significant splitting energy, Î” a weak field ligand will produce a less significant splitting energy, Î”
Now, the spectrochemical series looks like this
http://chemedu.pu.edu.tw/genchem/delement/9.htmhttp://chemedu.pu.edu.tw/genchem/delement/9.htm
Notice that the cyanide ion, CNâ’, is higher on the spectrochemical series than the fluoride ion, Fâ’. This means that the cyanide ion ligands will cause a more significant energy gap between the eg and t2g orbitals when compared with the fluoride ion ligands.
http://wps.prenhall.com/wps/media/objects/3313/3393071/blb2405.htmlhttp://wps.prenhall.com/wps/media...
In the case of the hexafluorocobaltate(III) ion, the splitting energy is smaller than the electron pairing energy, and so it is energetically favorable to promote two electrons from the t2g orbitals to the eg orbitals â†’ a high spin complex will be formed.
This will ensure that the hexafluorocobaltate(III) ion will have unpaired electrons, and thus be paramagnetic.
On the other hand, in the case of the hexacyanocobaltate(III) ion, the splitting energy is higher than the electron pairing energy, and so it is energetically favorable to pair up those four electrons in the t2g orbitals â†’ a low spin complex is formed.
Since it has no unpaired electrons, the hexacyanocobaltate(III) ion will be diamagnetic.

6 0

3 years ago

Could somebody please give me an example of some physical properties for any substance?

KIM [24]

Answer:

A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity.

Explanation:

6 0

3 years ago

¿Qué bacterias existen según su nutrición?

Maurinko [17]

Answer:

autotrofas estrictas son aquellas bacterias incapaces de crecer usando materia orgánica como fuente de carbono. Mixotrofas son aquellas bacterias con metabolismo energético litotrofo (obtienen energía de compuestos inorgánicos), pero requieren sustancias orgánicas como nutrientes para su metabolismo biosintético.

7 0

3 years ago