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Elena L [17]
3 years ago
6

You have to prepare a pH 5.00 buffer, and you have the following 0.10 M solutions available: HCOOH (pka=3.74), HCOONa, CH3COOH (

pka=4.74), CH3COONa, HCN (pka=9.31), and NaCN. Which solutions would you use?
Chemistry
1 answer:
weeeeeb [17]3 years ago
5 0

Answer:

CH3COOH - CH3COONa since its pKa is closest to the required pH.

Explanation:

Hello there!

In this case, in agreement with the theory of buffers as solutions able to withstand severe pH changes due to the addition of acidic or basic substances, it is possible to set up a generation equilibrium expression for the acids herein given:

Ka=\frac{[A^-][H_3O^+]}{[HA]}

Which leads to the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[A^-]}{[HA]} )

Thus, since all the buffers have [A-]=[HA]=0.10M, the log part becomes 0 and therefore the best buffer will have the closest pKa to the required pH, which is CH3COOH - CH3COONa since its pKa is 4.74.

Best regards!

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Photosynthesis was another biological phenomenon that occupied the attention of the chemists of the late 18th century. The demon
balu736 [363]

Answer:

In the 1770s, the English clergyman Joseph Priestley (who is credited with the discovery of O2) established the production of oxygen by vegetables recognizing that the process was, apparently, the inverse of animal respiration, which consumed such chemical element.

Explanation:

In 1772, Joseph Priestley in his Recherches sur diversces especes d'air differentiated the air of animal respiration from that emitted by vegetables in the presence of light. Of the latter, which he called "dephlogistic air", he highlighted his purifying property of the environment indicating that: plants far from affecting the air in the same way as animal respiration, produce the opposite effects, and tend to preserve the sweet and healthy atmosphere , when it becomes harmful as a result of the life and breathing of the animals or their death and their rot.

In 1780, Jean Ingeshousz in his Experiences sur les vegetaux completed and reaffirmed the observations of Joseph Priestley. At the same time, he could deny Charles Bonnet's hypothesis, by demonstrating that the air expelled from the leaves comes from inside, and that the stimulating factor of the gaseous emission was not the heat produced by the sun, but the intensity of the light .

It was, finally, Jean Senebier that between 1782 and 1784, found that the "fixed air" dissolved in the water favors the vegetation. From these observations, he hypothesized that "fixed air" (carbon dioxide) is absorbed by the plants, which take it from the atmosphere with the humidity it has and in which it is mixed. Once this gas has been captured, both from the atmosphere and from the ground, it is decomposed in the presence of light by the leaves, releasing the "vital air" (oxygen) and leaving the carbon in the plant.

Thus, at the end of the century the participation of the atmosphere in plant dynamics was already seated, although the how and why of this participation were still unknown and no theory had been formulated to explain the nutritional process as a whole.

3 0
3 years ago
The modern view of an electron orbital in an atom is best described as
Mrrafil [7]

invisable

you probably dont want to put that...

5 0
3 years ago
In which pair do both compounds exhibit predominantly ionic bonding? A) KCl and CO2 B) SO2 and BaF2 C) F2 and N2O D) N2O3 and Rb
gtnhenbr [62]

Answer:

E) NaF and SrO

Explanation:

The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.

<em>In which pair do both compounds exhibit predominantly ionic bonding? </em>

A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.

B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.

C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.

D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.

E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.

5 0
3 years ago
How many moles of oxygen are needed for the complete combustion of 29.2 grams of acetylene?
podryga [215]

Moles of Oxygen= 2.8075 moles

<h3>Further explanation</h3>

Given

29.2 grams of acetylene

Required

moles of Oxygen

Solution

Reaction(Combustion of Acetylene) :

2 C₂H₂ (g) + 5 O₂ (g) ⇒ 4CO₂ (g) + 2H₂O (g)

Mol of Acetylene :

= mass : MW Acetylene

= 29.2 g : 26 g/mol

= 1.123

From equation, mol ratio of Acetylene(C₂H₂) : O₂ = 2 : 5, so mol O₂ :

= 5/2 x mol C₂H₂

= 5/2 x 1.123

= 2.8075 moles

7 0
3 years ago
Describe what is happening in this chemical equation: Al2(SiO3)3 + NaOH → Na(SiO3)3 + Al2OH Please Help ASAP
WITCHER [35]

Answer:

Double replacement reaction

Explanation:

This is what is known as a double replacement reaction. The two parts of each molecule separate and recombine to form two new compounds. For instance, Al_2(SiO_3)_3 separates into Al_2 and (SiO_3)_3, while NaOH separates into Na and OH, and then they recombine with the other compound.

Hope this helps!

7 0
3 years ago
Read 2 more answers
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