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mafiozo [28]
3 years ago
14

(a) Write the reaction for the overall oxidation of propane (C3H8) with oxygen to produce CO2 and water.

Chemistry
1 answer:
Pavlova-9 [17]3 years ago
4 0

Answer:

Oxidation is the loss of electrons, that is, addition of electronegetive elements, example is addition of oxygen. Also, removal of electropositive elements, example is removal of hydrogen.

Explanation: a) In the presence of excess oxygen, propane burns in air, which gives the following chemical equation:

C3H8 + 5O2⇒ 3CO2 + 4H2O +Heat

b) When insufficient oxygen or too much oxygen is present for complete combustion, the following equation is given:

2C3H8 + 9O2 ⇒ 4CO2 + 2CO + 8H2O + Heat

c) At the anode( negative terminal): O∧2- ⇒ O + e

Oxygen accepts electron.

d) At cathode ( positive terminal): H∧+ + e∧- ⇒ H

Hydrogen donates electron

d) Nernst equation for reversal potential is given as follows:

E= RT/zF  In{ion outside cell}/{ion inside cell}= 2.303 RT/zF In{ion outside cell}/{ion inside cell}

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Which part of the cell is affected by the movement of molecules through diffusion, osmosis, and active transport? A.Cell membran
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Answer:

membrana celular

Explanation:

El transporte celular es el intercambio de sustancias a través de la membrana plasmática, que es una membrana semipermeable.1​

El transporte es importante para la célula porque le permite expulsar de su interior los desechos del metabolismo, también el movimiento de sustancias que sintetiza como hormonas. Además, es la forma en que adquiere nutrientes mediante procesos de incorporación a la célula de nutrientes disueltos en el agua. Las vías de transporte a través de la membrana celular y los mecanismos básicos para las moléculas de pequeños tamaños son:

Índice

1 Transporte pasivo

1.1 Ósmosis

1.1.1 Ósmosis en una célula animal

1.1.2 Ósmosis en una célula vegetal

1.2 Difusión facilitada

2 Transporte activo

2.1 Transporte activo primario: Bomba de sodio y potasio o Bomba Na+/K+

2.2 Transporte activo secundario o cotransporte

3 Transporte en masa

3.1 Endocitosis

3.2 Exocitosis

4 Véase también

5 Referencias

6 Enlaces externos

6 0
2 years ago
What is the GFM of H2O2
vichka [17]

Answer:

Hydrogen peroxide

Explanation:

5 0
3 years ago
What is the [H₃O⁺] (hydronium concentration) of an aqueous solution that has a pH of 3.30?
alisha [4.7K]

Answer:

5.01 *10^(-4) M

Explanation:

pH = - log[H₃O⁺]

[H₃O⁺] = 10^(-pH)= 10^(-3.30) = 5.01 *10^(-4) M

4 0
3 years ago
Which best describes the effect of J. J. Thomson’s discovery?
worty [1.4K]

Answer:

A. The accepted model of the atom was changed.

Explanation:

Thomson described an atom that has the protons uniformly distributed within a circular atom with the electrons embedded . However, it cannot explain the emission of electrons from the surfaces of metals like photoelectric emission. it does not explain how the particle of the atom contribute significantly to the entire mass of an atom

7 0
3 years ago
A chemistry student weighs out 0.154 g of chloroacetic acid (HCH2CICO2) into a 250. mL volumetric flask and dilutes to the mark
erastovalidia [21]

Answer:

11.6 mL of 0.1400 M of NaOH is required to reach equivalence point.

Explanation:

Chloroacetic acid is an monoprotic acid.

Neutralization reaction: ClCH_{2}COOH+NaOH\rightleftharpoons ClCH_{2}COONa+H_{2}O

So, 1 mol of chloroacetic acid is neutralized by 1 mol of NaOH.

Molar mass of chloroacetic acid = 94.5 g/mol

So, 0.154 g of chloroacetic acid = \frac{0.154}{94.5} moles of chloroacetic acid

                                                     = 0.00163 moles of chloroacetic acid

Lets assume V mL of 0.1400 M of NaOH is required to reach equivalence point.

So, number of moles of NaOH needed to reach equivalence point

      = \frac{0.1400\times V}{1000} moles

So, \frac{0.1400\times V}{1000}=0.00163

or, V = 11.6

Hence, 11.6 mL of 0.1400 M of NaOH is required to reach equivalence point.

7 0
3 years ago
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