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frosja888 [35]
2 years ago
8

Buffer consists of undissociated acid (ha) and the ion made by dissociating the acid (a-). How does this system buffer a solutio

n against decreases in ph?]
Chemistry
1 answer:
docker41 [41]2 years ago
3 0

In buffer solution there is an equilibrium between the acid  HA and its conjugate base A⁻: HA(aq) ⇌ H⁺(aq) + A⁻(aq).

When acid (H⁺ ions) is added to the buffer solution, the equilibrium is shifted to the left, because conjugate base (A⁻) reacts with hydrogen cations from added acid, according to Le Chatelier's principle: H⁺(aq) + A⁻(aq) ⇄ HA(aq). So, the conjugate base (A⁻) consumes some hydrogen cations and pH is not decreasing (less H⁺ ions, higher pH of solution).

A buffer can be defined as a substance that prevents the pH of a solution from changing by either releasing or absorbing H⁺ in a solution.

Buffer is a solution that can resist pH change upon the addition of an acidic or basic components and it is able to neutralize small amounts of added acid or base, pH of the solution is relatively stable


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1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore
Alina [70]

Answer :

(1) The theoretical yield of product, MgO is, 1.257 grams.

(2) The percent yield of MgO is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2Mg(s)+O_2(g)\rightarrow 2MgO(s)

From the reaction, we conclude that

As, 2 mole of Mg react to give 2 mole of MgO

So, 0.03142 moles of Mg react to give 0.03142 moles of MgO

Now we have to calculate the mass of MgO

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

\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g

Theoretical yield of MgO = 1.257 g

Experimental yield of MgO = 0.8922 g

Now we have to calculate the percent yield of MgO

\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100

\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%

Therefore, the percent yield of MgO is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

4 0
2 years ago
A(n) _______________ can be formed by linking together several monosaccharides via glycosidic bonds.
mario62 [17]

Answer:

A polysaccharide (n) can be formed by linking several monosaccharides through glycosidic linkages.

Explanation:

Polysaccharides are carbohydrates or complex carbohydrates, where monosaccharides join with glucosidic bonds to form a more complex structure that would be the polysaccharide.

An example of a polysaccharide is starch, or glycogen.

Starch is found in many foods such as potatoes or rice, and glycogen is a form of energy reserve of our organism housed in muscles and liver to fulfill locomotion, physical activity, and other activities that consist of glycolysis.

Polysaccharides are degraded in our body by different stages, and several enzymes unlike monosoccharides or disaccharides, since they have more unions and a more complex structure to disarm in our body and thus assimilate it.

Polysaccharides are also part of animal structures, such as insect shells or nutritional sources, among others.

7 0
2 years ago
In the reaction of nitrogen gas with oxygen gas to produce nitrogen oxide, what is the effect of adding more oxygen gas to the i
Alborosie
<h3>Answer:</h3>

The Equilibrium would shift to produce more NO

<h3>Explanation:</h3>

The reaction is;

N₂(g) + O₂(g) ⇆ 2NO(g)

  • When a reaction is at equilibrium then the forward reaction rate will be equivalent to the reverse reaction rate. Additionally, the concentration of the reactants and products are the same.
  • From Le Chatelier's principle, additional reactants favor the formation of more products while additional products favor the formation of more reactants.
  • For example, when more oxygen is added then more Nitrogen (II) oxide will be formed.
  • Oxygen is a reactant and when increased it favors forward reaction which leads to the formation of more NO which is the product.

3 0
2 years ago
Three moles of helium gas (molar mass MM = 4.00 g/molg/mol) are in a rigid container that keeps the volume of the gas constant.
Anastaziya [24]

Answer:

The rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Explanation:

Mass of 3 moles of Helium = 3 moles × 4.00 g/mol = 12.00 g = 0.012 kg

The initial average kinetic energy of the helium atoms = (1/2)(m)(u²)

where u = initial rms speed of the gas = 850 m/s

Initial average kinetic energy of the gas = (1/2)(0.012)(850²) = 4335 J

Then, 3600 J is added to the gas,

New kinetic energy of the gas = 4335 + 3600 = 7935 J

New kinetic energy of Helium atoms = (1/2)(m)(v²)

where v = final rms speed of the gas = ?

7935 = (1/2)(0.012)(v²)

v² = (7935×2)/0.012

v² = 1,322,500

v = 1150 m/s

Hence, the rms speed of the gas atoms after 3600 J of heat energy is added to the gas = 1150 m/s.

Hope this Helps!!!

8 0
1 year ago
A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

             = 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

                                      = 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0
1 year ago
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