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iragen [17]
3 years ago
10

Particles in a liquid move around just as freely as particles in a solid.

Chemistry
2 answers:
andre [41]3 years ago
5 0
It is false :D I am 95% sure
Feliz [49]3 years ago
5 0

its false as only gas has that ability to do so


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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

5 0
3 years ago
22.4l of ammonia is reaxts with 1.406 mole of oxygen to produce NO and h2o .1.what volume of no is produced at ntp​
IceJOKER [234]

Answer:

The volume of NO is 22.4L at STP

Explanation:

Based on the reaction:

2NH3 + 5/2O2 → 2NO + 3H2O

<em>2 moles of NH3 react with 5/2 moles of O2 to produce 2 moles of NO.</em>

<em />

To solve this question, we need to find the moles of each reactant in order to find the limiting reactant as follows:

<em>Moles NH3 -Molar mass: -17.01g/mol-</em>

Using PV = nRT

PV/RT = n

<em>Where P is pressure = 1atm at STP</em>

<em>V is volume = 22.4L</em>

<em>R is gas constant = 0.082atmL/molK</em>

<em>T is absolute temperature = 273.15K</em>

1atm*22.4L/0.082atmL/molK*273.15K = n

n = 1.00 moles of NH3

For a complete reaction of 1.00 moles of NH3 are needed:

1.00 moles NH3 * (5/2moles O2 / 2moles NH3) = 1.25 moles of O2

As there are 1.406 moles of O2, <em>the limiting reactant is NH3</em>

<em />

The moles of NO produced are the same than moles of NH3 because 2 moles of NH3 produce 2 moles of NO. The moles of NO are 1.00 moles

And as 1.00moles of gas are 22.4L at STP:

<h3>The volume of NO is 22.4L at STP</h3>

4 0
3 years ago
What makes the results of a scientific experiment accurate?
Dmitriy789 [7]

Answer:

A

Explanation:

That's the best answer out of all of them.

6 0
3 years ago
Read 2 more answers
how many milliliters of 1.50 m hno3 contain enough nitric acid to dissolve an old copper penny with a mass of 3.94 g?
Lerok [7]
Molar mass HNO₃ = 63.0 g/mol

number of moles = 3.94 / 63.0 => 0.0625 moles

Volume = moles / molarity

V = 0.0625 / 1.50

V = 0.04166 L x 1000 = 41.66 mL

hope this helps!

5 0
3 years ago
A crime scene investigator is called in to evaluate a crime scene. The victim is lying on the floor with no apparent trauma. Wha
kodGreya [7K]
C the empty pills, because they can be linked back to see if death was a side effect.
5 0
3 years ago
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