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1 year ago

In an experiment, a scientist compares the effect of adding acid rain to

Chemistry

2 answers:

Sveta_85 [38]1 year ago

6 0

Answer: Lake Minnetonka water is buffered instead of Upper Kintla Lake water that doesn´t have molecules to "muffle" the acid rain added.

Explanation:

When Adding Acid or a Base to a sample, and its pH doesn´t change, it will tell us that something is happening and <u>there must be some molecules in the sample that are reacting with the Acid or Base added </u>and that is why, the sample´s pH stay the same. These "molecules" muffle the Acid or Base, and they are called "buffer".

As you can see in the image below, there is an example with a typical Bicarbonate Buffer when adding HCl, the Bicarbonate molecules will react with the HCl, making it to change in other molecule, and losing its acid power.

<em>When the Acid Rain was added to the Upper Kintla Lake sample,</em> the pH changed instantly, because there were any molecule to react with the HCl added and HCl molecules remain the same, making the pH to drop from 7.5 to 5.2

lilavasa [31]1 year ago

3 0

Lake Minnetonka in Southern Minneosta is buffer solution

Explanation:

One can conclude from the observation that Lake Minnetonka is a buffer solution. A buffer solution is a special solution that resist changes in pH or pOH when a small amount of an acid or base is added.

Buffers are solutions in equilibrium with a fairly constant pH.
They are aqueous solutions containing weak acids and their salts or weak bases and their salts.
Since the pH of water sample from Lake Minnetonka did not change, it is a buffer.

Learn more:

buffer brainly.com/question/4431162

blood as a buffer brainly.com/question/12855542

#learnwithBrainly

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A solution contains 90 milliequivalents of HC1 in 450ml. What is its normality?

Rashid [163]

Answer:

Normality N = 0.2 N

Explanation:

Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.

Normality is represented by N.

Mathematically, we have :

$\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}$

Given that:

number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent

volume of solution (HCl) = 450 mL 450 × 10⁻³ L

$\mathbf{Normality \ N = \dfrac{90 \times 10^{-3}}{450 \times 10^{-3}}}$

Normality N = 0.2 N

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1 year ago

A man pours some hot coffee into a small cup from a large jug that contains enough coffee for several cups. Which statement corr

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For which of the following aqueous solutions would one expect to have the largest van’t Hoff factor (i)? a. 0.050 m NaCl b. 0.50

ira [324]

Answer:

The van't hoff factor of 0.500m K₂SO₄ will be highest.

Explanation:

Van't Hoff factor was introduced for better understanding of colligative property of a solution.

By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.

a) For NaCl the van't Hoff factor is 2

b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]

Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.

c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.

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1 year ago

8. The protein in a 1.2846-g sample of an oat cereal is determined by a Kjeldahl analysis. The sample is digested with H2SO4, th

docker41 [41]

Answer: The %w/w protein in the sample is 15.2 %

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}$

$\text{Moles of} HCl={0.09552\times 50.00}{1000}=0.0047moles$

$\text{Moles of} NaOH={0.05992\times 37.84}{1000}=0.0023moles$

$HCl+NaOH\rightarrow NaCl+H_2O$

According to stoichiometry :

1 mole of $NaOH$ require 1 mole of $HCl$

Thus 0.0023 moles of $NaOH$ will require= $\frac{1}{1}\times 0.0023=0.0023moles$ of $HCl$

moles of HCl used = (0.0047-0.0023) = 0.0024

$NH_3+HCl\rightarrow NH_4Cl$

1 mole of HCl uses = 1 mole of ammonia

Thus 0.0024 moles uses = $\frac{1}{1}\times 0.0024=0.0024moles$ of ammonia

Mass of ammonia= $moles\times {\text {Molar mass}}=0.0024\times 17g/mol=0.0408g$

17 g of ammonia contains = 14 g of Nitrogen

Thus 0.0408 g of ammonia contains = $\frac{14}{17}\times 0.0408=0.034 g$ of Nitrogen

Now 17.45 g of Nitrogen is present in = 100 g of protein

Thus 0.034 g of Nitrogen is present in = $\frac{100}{17.45}\times 0.034=0.195g$ of protein

Now % w/w of protein = $\frac{0.195}{1.2846}\times 100=15.2\%$

Thus %w/w protein in the sample is 15.2%

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1 year ago