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3 years ago

7

Liquid octane reacts with oxygen gas to form carbon dioxide gas and water vapor.express your answer as a chemical equation. iden

tify all of the phases in your answer.

1 answer:

FrozenT [24]3 years ago

5 0

<span>C8H18 + 12.5O2 â†’ 8CO2 + 9H2O</span>

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Which property is dependent only on the atomic structure of the compound? *

shusha [124]

I think its mass of volume

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2 years ago

II. Binding Forces A. Write a brief, one or two sentence, description of each binding force listed below. 1. London dispersion f

Leya [2.2K]

Answer:

All description is given in explanation.

Explanation:

Van der Waals forces:

It is the general term used to describe the attraction or repulsion between the molecules. Vander waals force consist of two types of forces:

1. London dispersion forces

2. Dipole-dipole forces

1. London dispersion forces:

These are the weakest intermolecular forces. These are the temporary because when the electrons of atoms come close together they create temporary dipole, one end of an atom where the electronic density is high is create negative pole while the other becomes positive . These forces are also called induce dipole- induce dipole interaction.

2. Dipole-dipole forces:

These are attractive forces , present between the molecules that are permanently polar. They are present between the positive end of one polar molecules and the negative end of the other polar molecule.

Hydrogen bonding:

It is the electrostatic attraction present between the atoms which are chemically bonded. The one atom is hydrogen while the other electronegative atoms are oxygen, nitrogen or flourine. This is weaker than covalent and ionic bond.

Ionic bond or electrostatic attraction:

It is the electrostatic attraction present between the oppositely charged ions. This is formed when an atom loses its electron and create positive charge and other atom accept its electron and create negative charge.

Hydrophobic interaction:

It is the interaction between the water and hydrophobic material. The hydrophobic materials are long chain carbon containing compound. These or insoluble in water.

Covalent bond:

These compounds are formed by the sharing of electrons between the atoms of same elements are between the different element's atoms. The covalent bond is less stronger than ionic bond so require less energy to break as compared to the energy require to break the ionic bond.

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3 years ago

Assignment<br>Explain why it is not advisable to wear<br>black silky cloth in the sun

Crazy boy [7]

Answer:

It is not water absorbant and also the colour black, due to it's dark nature makes the person wearing the cloth feel hot. Unlike cotton, silk is not at all water absorbant as it does not absord any sweat produced in the wearers body

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3 years ago

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

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3 years ago

Which of these is an example of food engineering?

Andrej [43]

Answer:

a.labeling food with an ingredient list to inform consumers

Explanation:

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3 years ago

Read 2 more answers