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3 years ago

Brainliest! a. Write a short historical account of the discovery of the Avogadro’s number.

Chemistry

1 answer:

kvv77 [185]3 years ago

4 0

Answer:

a. Amadeo Avogadro (1776-1856) was the author of Avogadro's Hypothesis in 1811, which, together with Gay-Lussac's Law of Combining Volumes, was used by Stanislao Cannizzaro to elegantly remove all doubt about the establishment of the atomic weight scale at the Karlsruhe Conference of 1860. The name "Avogadro's Number" is just an honorary name attached to the calculated value of the number of atoms, molecules, etc. in a gram mole of any chemical substance. Of course if we used some other mass unit for the mole such as "pound mole", the "number" would be different than 6.022 x 1023.

b. The first person to have calculated the number of molecules in any mass of substance was Josef Loschmidt, (1821-1895), an Austrian high school teacher, who in 1865, using the new Kinetic Molecular Theory (KMT) calculated the number of molecules in one cubic centimeter of gaseous substance under ordinary conditions of temperature of pressure, to be somewhere around 2.6 x 1019 molecules. This is usually known as "Loschmidt's Constant.

Task 2

Percent composition is the percent by mass of each element present in a compound. Water, H2O, is the first example. One mole of water is 18.0152 grams. In that compound, there are two moles of H atoms and 2 x 1.008 = 2.016 grams. That's how many grams of hydrogen are present in one mole of water. this is an example. i don't know what you are describing though. i need more info for this question

b. 6.022 to 6.023 x 10^23

c. i don't know what this one is since there is nothing to describe the unknown liquid.

d. Yes a killer, but not a specific person

Explanation:

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zubka84 [21]

26. B
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7 0

3 years ago

Explain electronegativity and how it relates to both molecular and ionic bonding.

Keith_Richards [23]

Answer:

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3 0

2 years ago

For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaOH, calculate:

Leviafan [203]

1) Chemical reaction

HCl        +       NaOH      --->      NaCl + H2O

25.0 ml
0.150 M            0.250M

2) 50% completion => 0.025 l * 0.150 M * (1/2) = 0.001875 mol HCl consumed and 0.001875 mol HCl in solution

0.001875 mol HCl => 0.001875 mol H(+)

Volume = Volume of HCl solution + Volumen of NaOH solution added

Volume of HCl solution = 0.0250 l

Volume of NaOH = n / M = 0.001875 mol / 0.250M = 0.0075 l

Total volume = 0.0250 l + 0.0075 l = 0.0325 l

[H+] = 0.001875 mol / 0.0325 l = 0.05769 M

pH = - log [H+] = - log (0.05769) = 1.23

Answer: 1.23

3) Equivalence point

0.02500 l * 0.150 M = 0.250M * V

=> V = 0.02500 * 0.150 / 0.250 = 0.015 l

4) 1.00 ml NaOH added beyond the equivalence point

1.00 ml * 1 l / 1000 ml * 0.250 M = 0.00025 mol NaOH in excess

0.00025 mol NaOH = 0.00025 mol OH-

Volume of the solution = 0.02500 l + 0.015 l + 1.00/1000 l = 0.041 l

[OH-] = 0.00025 mol / 0.041 l = 0.00610 M

pOH = - log (0.00610) = 2.21

pH + pOH = 14 => pH = 14 - pOH = 14 - 2.21 = 11.76

Answer: 11.76

6 0

3 years ago

A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m

kogti [31]

Answer:

The molar mass of the unknown gas is $\mathbf{ 51.865 \ g/mol}$

Explanation:

Let assume that the gas is O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires time t₂ = 6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

$R \ \alpha \ \dfrac{1}{\sqrt{d}}$

$R = \dfrac{k}{d}$ where K = constant

If we compare the rate o diffusion of two gases;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}$

Since the density of a gas d is proportional to its relative molecular mass M. Then;

$\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}$

Rate is the reciprocal of time ; i.e

$R = \dfrac{1}{t}$

Thus; replacing the value of R into the above previous equation;we have:

$\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}$

We can equally say:

${\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}$

${\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}$

$M_2 = 32 \times ( \dfrac{6.34}{4.98})^2$

$M_2 = 32 \times ( 1.273092369)^2$

$M_2 = 32 \times 1.62076418$

$\mathbf{M_2 = 51.865 \ g/mol}$

7 0

3 years ago

What is the effect on the equilibrium when sodium formate is added to a solution of formic acid? hcooh( aq) + h ( aq) right arro

artcher [175]

There is no effect on the equilibrium when sodium formate is added to a solution of formic acid for hcooh( aq) + h ( aq) right arrow choo –( aq).

Equilibrium is characterized as a state of equilibrium or a stable situation in which conflicting forces balance one another out and no changes are taking place. In terms of economics, equilibrium occurs when supply and demand are equal. When you are composed and steady, you are in an equilibrium state.

An object is considered to be in an equilibrium condition when all of the forces acting on it are in balance. If the upward forces are equal to the descending forces and the rightward forces are equal to the leftward forces, then the forces are said to be balanced. Several instances of equilibrium include a book that is open and at rest. a vehicle that is going steadily.

To learn more about equilibrium please visit -
brainly.com/question/14602120
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4 0

1 year ago