El cation en el CaS

Please help balance _Ch7H16+_O2=_CO2+_H2O

iren2701 [21]

Answer:

C7H16 + 11 O2 → 7 CO2 + 8 H2O

This is an oxidation-reduction (redox) reaction:

7 C-16/7 - 44 e- → 7 CIV (oxidation)

22 O0 + 44 e- → 22 O-II (reduction)

C7H16 is a reducing agent, O2 is an oxidizing agent.

Reactants:

C7H16

O2

Names: Dioxygen source: wikidata, accessed: 2019-09-07, Oxygen source: ICSC, accessed: 2019-09-04source: wikidata, accessed: 2019-09-07, Oxygen (liquefied) source: ICSC, accessed: 2019-09-04

Appearance: Odourless compressed gas source: ICSC, accessed: 2019-09-04; Liquefied gas. colourless-to-blue extremely cold liquid source: ICSC, accessed: 2019-09-04

Products:

CO2

Names: Carbon dioxide source: wikipedia, accessed: 2019-09-27source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, {{plainlist| source: wikipedia, accessed: 2019-09-27, CO2 source: wikidata, accessed: 2019-09-02

Appearance: Colorless gas source: wikipedia, accessed: 2019-09-27; Odourless colourless compressed liquefied gas source: ICSC, accessed: 2019-09-04; Colorless, odorless gas. [Note: Shipped as a liquefied compressed gas. Solid form is utilized as dry ice.] source: NIOSH NPG, accessed: 2019-09-02

H2O – Water, oxidane source: wikipedia, accessed: 2019-09-27

Other names: Water (H2O) source: wikipedia, accessed: 2019-09-27, Hydrogen hydroxide (HH or HOH) source: wikipedia, accessed: 2019-09-27, Hydrogen oxide source: wikipedia, accessed: 2019-09-27

Appearance: White crystalline solid, almost colorless liquid with a hint of blue, colorless gas source: wikipedia, accessed: 2019-09-27

Search by reactants (C7H16, O2)

1 O2 + C7H16 → H2O + CO2

2 O2 + C7H16 → H2O + CO

3 O2 + C7H16 → H2O + CO2 + CO

4 O2 + C7H16 → H2 + CO2

Search by products (CO2, H2O)

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1 HCl + CaCO3 → H2O + CO2 + CaCl2

2 HCl + NaHCO3 → H2O + CO2 + NaCl

3 O2 + CH4 → H2O + CO2

4 NaHCO3 → H2O + CO2 + Na2CO3

5 H2SO4 + KMnO4 + (COOH)2 → H2O + CO2 + K2SO4 + MnSO4

6 O2 + C3H8 → H2O + CO2

7 H2SO4 + K2CO3 → H2O + CO2 + K2SO4

8 O2 + C2H6 → H2O + CO2

9 CH3COOH + NaHCO3 → H2O + CO2 + CH3COONa

10 HCl + Na2CO3 → H2O + CO2 + NaCl

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Search by reactants (C7H16, O2) and by products (CO2, H2O)

1 O2 + C7H16 → H2O + CO2

2 O2 + C7H16 → H2O + CO2 + CO

Explanation:

8 0

2 years ago

Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔ 2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2 [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ] ' ²

Substituting the new concentration terms ,

K ' = 1/2 [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ] ²

K ' = 1/4 [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ] ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' = [ NO ] ² [ Br₂ ] / [ NoBr ] ²

Hence ,

K' = K

5 0

3 years ago

Phosphorus can be stable with 12 electrons in its valence structure while nitrogen can never have more than 8 electrons in its v

dolphi86 [110]

Here we have explain that the maximum possible electrons present in nitrogen valence shell is 8 whereas in phosphorous 12 valence electrons are present.

Although both nitrogen (N) and phosphorous (P) belongs to the same series there are several properties which are different between both the element. The number of electrons present in nitrogen is seven which are present in the -s and -p orbitals. The electronic configuration of nitrogen is 1s²2s²2p³. In which the outermost electrons are the valence electrons i.e. 5 valence electrons are present. The maximum orbitals are possible under the principal quantum number 2 are -s and -p orbitals. Now the maximum capacity of the p orbital to contain 6 electrons, as it is half filled in nitrogen another 3 electrons can be incorporated. Thus the maximum number of electrons can be present in nitrogen is 10 among which 8 is the valence electrons.

On the other hand there are 15 electrons in phosphorous the electronic configuration is 1s²2s²2p⁶3s²3p³. Now the principal quantum number 3 can have three orbitals -s, -p and -d. So another 13 electrons can be incorporated (3 in -p orbital and 10 in -d orbital) among which upto 12 electrons can be its valence electrons.

8 0

2 years ago

How does electron repulsion in the outer shell cause anions to be larger than their atoms?

Alexeev081 [22]

The anion is also larger than the atom because of electron-electron repulsion. As more electrons are added to the outer shell, and even to higher principle energy levels, the repulsion bewteen the negatively charged particles grows, pushing the shells farther from the nucleus.

5 0

3 years ago

A gas mixture contains the following gases with the mole fractions indicated: nitrogen (0.21), oxygen (0.16), carbon dioxide (0.

alukav5142 [94]

Answer:

Mole fraction Ar = 0.31

Explanation:

Remember that the sum of the mole fractions in a mixture of gases = 1

Mole fraction = Moles from a gas / Total moles

Mole fraction N₂+ Mole fraction O₂+ Mole fraction SO₂+ Mole fraction CO₂ + Mole fraction Ar = 1

N₂ = 0.21

O₂= 0.16

CO₂ = 0.23

SO₂ = 0.09

Mole fraction Ar = 1 - 0.21 - 0.16 - 0.23 - 0.09

Mole fraction Ar = 0.31

4 0

3 years ago

El cation en el CaS​

El cation en el CaS