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4 years ago

The enthalpy change for the complete burning of one mole of a substance

Chemistry

1 answer:

Jet001 [13]4 years ago

7 0

Answer:

combustion

Explanation:

The enthalpy change for the complete burning of one mole of a substance

is the enthalpy of __combustion_____ .

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Define or explain the following terms. a) Equilibrium constant expression - b) Equilibrium constant - c) Absorbance - d) Beer's

tankabanditka [31]

Explanation:

(a) Equilibrium constant expression is ratio of concentrations of products over reactants each raised to their power of stoichiometric coefficients.

For example consider an equilibrium which is:

$aA+bB\rightleftharpoons cC+dD$

The expression is:

$K_c=\frac {[C]^c[D]^d}{[A]^a[B]^b}$

(b) Equilibrium constant of the chemical reaction is value of the reaction quotient of the reaction at the stage of chemical equilibrium which is a state that is approached by the dynamic chemical system at which the composition of the reactant and the product has no measurable tendency towards the change.

(c) Absorbance is the measure of capacity of the substance to absorb the light of a specific wavelength. Absorbance is equal to the logarithm of reciprocal of transmittance.

(d) The Beer's law relates attenuation of the light to properties of material through which light is travelling.

The expression for the law is:

A = ε × l× c

Where,

A is the absorbance

ε is molar absorptivity coefficient

l is the path length

c is the concentration.

3 0

3 years ago

How many mole of CO2 are in 58.9g of the compound

vivado [14]

Convert grams to moles and use Avogadro's number 6.022x10tho the 23rd power. Hope this helps

8 0

3 years ago

What steps would you take in a lab to prepare 1000 mL of a 2 M NaCl solution

amm1812

Answer:

The answer to your question is below

Explanation:

Data

Volume = 1000 ml

Concentration = 2M

molecule = NaCl

Process

1.- Calculate the number of moles of NaCl

Molarity = moles/Volume

-Solve for volume

moles = Molarity x Volume (liters)

-Substitution

moles = 2 x 1

-Result

moles = 2

2.- Determine the molar mass of NaCl

NaCl = 23 + 35.5 = 58.5 g

3.- Calculate the mass of NaCl to prepare the solution

58.5 g ----------------- 1mol

x ----------------- 2 moles

x = (2 x 58.5) / 1

x = 117g

4.- Weight 117 g of NaCl, place them in a volumetric flask (1 l), and add enough water to prepare the solution.

7 0

4 years ago

400 liters of a certain gas is collected at STP. What will the volume be at 273 C and 190 torr pressure?

Firlakuza [10]

Answer:

2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP? STP is a common abbreviation for "standard temperature and pressure." You have to recognize that five values are given in the problem and the sixth is an x. Also ... 273 1. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr.

Explanation:

3 0

3 years ago

Morphine, C 17H 19NO 3, is often used to control severe post-operative pain. What is the pH of the solution made by dissolving 2

uysha [10]

Answer:

pH = 9.58

Explanation:

First of all, we need to determine the molarity of the solution.

We determine the molar mass of morphine:

12g/m . 17 + 1 g/m . 19 + 14 g/m + 16 g/m . 3 = 285.34 g/m

molar mass g/m, is the same as mg/mm

25 mg . 1 mmol / 285.34 mg = 0.0876 mmoles / 100 mL = 8.76×10⁻⁴ M

In diltuted solution, we must consider water.

Mass balance for morphine = [Morphine] + [Protonated Morphine]

8.76×10⁻⁴ M = [Morphine] + [Protonated Morphine]

As Kb is too small, I can skipped, the [Protonated Morphine]

8.76×10⁻⁴ M = [Morphine]

In the charge balance I will have:

[OH⁻] = [H⁺ morphine] + [H⁺]

Let's go to the Kb expression

Morphine + H₂O ⇄ MorphineH⁺ + OH⁻ Kb

Kb = [MorphineH⁺] [OH⁻] / [Morphine]

Kb = [MorphineH⁺] [OH⁻] / 8.76×10⁻⁴ M

So now, we need to clear [MorphineH⁺] to replace it in the charge balance

Kb . 8.76×10⁻⁴ M / [OH⁻] = [MorphineH⁺]

Now, the only unknown value is the [OH⁻]

[OH⁻] = Kb . 8.76×10⁻⁴ M / [OH⁻] + Kw/[OH⁻]

Remember that Kw = [H⁺] . [OH⁻]

[H⁺] = Kw/[OH⁻]

[OH⁻]² = 1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴

[OH⁻] = √(1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴)

[OH⁻] = 3.76×10⁻⁵ → - log [OH⁻] = pOH = 4.42

pH = 14 - pOH → 14 - 4.42 = 9.58

8 0

4 years ago