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Alla [95]
3 years ago
10

Please answer!!!!!!!!!!!!!

Chemistry
2 answers:
Oduvanchick [21]3 years ago
4 0
I believe it’s Decreased Humus
humus is the organic component of soil formed by the decomposition of leaves and other plant material by soil microorganisms
Assoli18 [71]3 years ago
4 0

Answer:

B. Decreased humus

Explanation:

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Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH3COO, to 23 g of acetic acid, CH3COOH, and dissolving
Elza [17]

This is an incomplete question, here is a complete question.

Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH₃COO, to 23 g of acetic acid, CH₃COOH, and dissolving in water to make 400. mL of solution. Hint given in feedback. The Ka for CH₃COOH is 1.8 x 10⁻⁵ M. As usual, report pH to 2 decimal places.

Answer : The pH of the solution is, 4.97

Explanation :

First we have to calculate the moles of CH₃COOH and NaCH₃COO

\text{Moles of }CH_3COOH=\frac{\text{Given mass }CH_3COOH}{\text{Molar mass }CH_3COOH}=\frac{23g}{60g/mol}=0.383mol

and,

\text{Moles of }CH_3COONa=\frac{\text{Given mass }CH_3COONa}{\text{Molar mass }CH_3COOH}=\frac{59g}{82g/mol}=0.719mol

Now we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (1.8\times 10^{-5})

pK_a=5-\log (1.8)

pK_a=4.7

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}

Now put all the given values in this expression, we get:

pH=4.7+\log (\frac{0.719}{0.383})

(As the volume is same. So, we can write concentration in terms of moles.)

pH=4.97

Therefore, the pH of the solution is, 4.97

6 0
4 years ago
What would happen when a substance changes from one phase to another? A. The substance loses or gains heat. B. The average kinet
lutik1710 [3]
I think the correct answer from the choices listed above is option B. When a substance changes from one phase to another, the average kinetic energy of the substance <span>changes since the molecules movements are changed as well. Hope this answers the question. Have a nice day.</span>
3 0
3 years ago
Can someone plzzz help with numbers 14,15,#1 plzzz show work too cuz I don’t understand this plzzz help me
vovangra [49]

Answer:

See Explanation below.

Explanation:

4 0
3 years ago
How could you interpret the events taking place on a molecular level in the images seen here
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3 years ago
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Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is pro
MAVERICK [17]

<u>Answer:</u> The percent yield of water is 46.9 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For sulfuric acid:</u>

Given mass of sulfuric acid = 72.6 g    (Assuming)

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

\text{Moles of sulfuric acid}=\frac{72.6g}{98g/mol}=0.741mol

  • <u>For NaOH:</u>

Given mass of NaOH = 77 g      (Assuming)

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{77g}{40g/mol}=1.925mol

The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:

H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

0.741 moles of sulfuric acid will react with = \frac{2}{1}\times 0.741=1.482mol of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of water

0.741 moles of sulfuric acid will react with = \frac{2}{1}\times 0.741=1.482mol of water

  • Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol\times 18g/mol)=26.67g

  • To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 12.5 g  (Assuming)

Theoretical yield of water = 26.67 g

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{12.5g}{26.67g}\times 100\\\\\% \text{yield of water}=46.9\%

Hence, the percent yield of water is 46.9 %

3 0
3 years ago
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