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3 years ago

How might you think it is possible to convert from moles to particles or grams to particles?

Chemistry

1 answer:

nevsk [136]3 years ago

4 0

Explanation:

To convert moles to particles or grams to particles, let us have a firm understanding of what a mole is.

A mole is the unit of measuring quantity of particles.

It is the amount of substance that contains the Avogadro's number of particles.

The particle can be atoms, molecules, formula units, electrons, protons, neutrons, etc.

So, to convert from moles to particles;

1 mole of a substance contains 6.02 x 10²³ particles

To convert from grams to particles;

First convert to moles;

number of moles = $\frac{mass}{molar mass}$

So, 1 mole of a substance contains 6.02 x 10²³ particles

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Archy [21]

Salt and water is formed as product, when Arrhenius acid and Arrhenius base reacts and the reaction is known as neutralization reaction.

<h3>What is Arrhenius acid-base reaction?</h3>

The concept of acid and base based on the theory of ionization which was first proposed in 1884 by Svante Arrhenius
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1 year ago

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3 years ago

Suppose that you place exactly 100 bacteria into a flask containing nutrients for the bacteria and that you find the following d

tresset_1 [31]

Answer:

The order of the production of bacteria is 1.

12,787 bacteria will be present after 112 min.

$0.04331 min^{-1}$ is the rate constant for the process.

Explanation:

Rate of formation of bacteria when there 100 bacteria: R

Rate of formation of bacteria when there 200 bacteria: R' =2R

Rate of formation of bacteria is directly proportional to the number of bacteria:

$\frac{R}{R'}=\frac{k[200]^x}{k[400]^x}$

$\frac{R}{2R}=(\frac{200}{400})^x$

x = 1

The order of the production of bacteria is 1.

The half life of the process = $t_{1/2}=16 minutes$

Rate constant of the process = k

For first order process k is only dependent on half life:

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{16 min}=0.04331 min^{-1}$

For the first order process the increment numbers is given ;

$N=N_o\times e^{kt}$

Where:

$N_o$ = Initial population

N = Population after time t.

We have ;

$N_o=100$

t = 112 minutes

$N=100\times e^{0.04331 min^{-1}\times 112 min}$

$N = 12,786.82\approx 12,787$

12,787 bacteria will be present after 112 min.

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5 0

3 years ago

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium

lilavasa [31]

Answer:

This question is incomplete, here's the complete question:

1. Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer.

Buffer A

Mass of sodium acetate used: 0.3730 g

Actual ph of the buffer 5.27

volume of the buffer used in buffer capacity titration 20.0 mL

Concentration of standardized NaOH 0.100M

moles of Naoh needed to change the ph by 1 unit for the buffer 0.00095mol

the buffer capacity 0.0475 M

Buffer B

Mass of sodium acetate used 1.12 g

Actual pH of the buffer 5.34

Volume of the buffer used in buffer capacity titration 20.0 mL

Concentration if standardized NaOH 0.100 M

moles of Naoh needed to change the ph by 1 unit 0.0019 mol

the buffer capacity 0.095 M

2.) A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium you calculated in #1 by a factor of 10 for each buffer. Are there any differences between your experimental results and the theoretical calculation?

3.) which buffer had a higher buffer capacity? Why?

Explanation:

Formula,

moles = grams/molar mass

molarity = moles/L of solution

1. Buffer A

molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M

molarity of HC2H3O2 = 0. 1 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.23/0.1)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

Buffer B

molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M

molarity of HC2H3O2 = 0.3 M

Initial pH

pH = pKa + log(base/acid)

= 4.74 + log(0.68/0.3)

= 5.10

pH = -log[H3O+]

[H3O+] = 7.91 x 10^-6 M

In 20 ml buffer,

moles of H3O+ = 7.91 x 10^-6 M x 0.02 L

= 1.58 x 10^-7 mol

2. let x moles of NaOH is added,

Buffer A,

pH = 5.10

[H3O+] = 7.91 x 10^-6 M

new pH = 4.10

new [H3O+] = 7.91 x 10^-5 M

moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L

= 1.42 x 10^-6 mol

3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A

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3 years ago

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choli [55]

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3 years ago

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