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NARA [144]
3 years ago
12

1. How did the change of stress (adding or removing reactants or products) cause a shift in the equilibrium system of your solut

ions? Use data to support your answer. Make sure you discuss all four stress changes: a. Adding a reactant b. Adding a product c. Removing a reactant d. Removing a product
Chemistry
2 answers:
tatyana61 [14]3 years ago
8 0

Answer:

<em><u>When additional reactant is added, the equilibrium shifts to reduce this stress: it makes more product. When additional product is added, the equilibrium shifts to reactants to reduce the stress. 2. Concentration is just one stress that can affect the equilibrium position.</u></em>

Explanation:

<h2>HOPE IT WILL HELP YOU✌✌✌✌✌</h2>
Charra [1.4K]3 years ago
7 0

Answer:

If reactant or product is removed, the equilibrium shifts to make more reactant or product, respectively, to make up for the loss.

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Calculate the pH of a solution prepared by mixing: (Show your work for these calculations) pk of acetic acid is 4.75 a. two mole
kupik [55]

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where <em>CH₃COOH </em>is the acid and <em>CH₃COO⁻ </em>is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ \frac{[CH_{3}COO^-]}{[CH_{3}COOH]}

a) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[2 mol]}

<em>pH = 4,75</em>

<em></em>

b) The pH is:

pH = 4,75 + log₁₀ \frac{[2 mol]}{[1mol]}

<em>pH = 5,05</em>

<em></em>

I hope it helps!

7 0
3 years ago
To three significant digit,what is the mass percentage of iron in the compound Fe2O3​
svlad2 [7]

69.9%

Explanation:

To find the mass percentage of iron in the compound in Fe₂O₃, we would go ahead to express the given molar mass of the iron to that of the compound.

 Mass percentage  = \frac{molar mass of Fe}{Molar mass of Fe_{2}O_{3}  }  x 100

Molar mass of Fe = 55.85g/mol

Molar mass of O = 16g/mol

Molar mass of Fe₂O₃ = 2(55.85) + 3(16) = 159.7‬g/mol

Mass percentage  = \frac{2(55.85)}{159.7}  x  100   = 69.94% = 69.9%

learn more:

Mass percentage brainly.com/question/8170905

#learnwithBrainly

4 0
3 years ago
PLZZZZZZ HELP ME I GIVE BRAINLIEST PLZ NO FILES!!!
natka813 [3]

Answer:

The light is most intense where it strikes Earth _ directly_______ to its surface.

Explanation:

Hope this helps :)

3 0
3 years ago
Read 2 more answers
PLEASE HELPPPP!!!!!
Likurg_2 [28]

Answer : The mass of nitric acid is, 214.234 grams.

Solution : Given,

Moles of nitric acid = 3.4 moles

Molar mass of nitric acid = 63.01 g/mole

Formula used :

\text{Mass of }HNO_3=\text{Moles of }HNO_3\times \text{Molar mass of }HNO_3

Now put all the given values in this formula, we get the mass of nitric acid.

\text{Mass of }HNO_3=(3.4moles)\times (63..01g/mole)=214.234g

Therefore, the mass of nitric acid is, 214.234 grams.


6 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
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