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sveta [45]
4 years ago
6

Phenol red is an indicator that changes color over a range from ph=6.4 to ph=8.0. what is ka of phenol red?

Chemistry
1 answer:
patriot [66]4 years ago
7 0
First, we will get the average pH of the two given values:
average pH = (6.4+8) / (2) = 7.2
At this average pH, the concentration of the acid from the phenol red is equal to the concentration of the base.
pH = 7.2
[H+] = 10^(-7.2) = 6.3 * 10^-8

Phenol red has the general formula HA, this gives us:
HA <.......> H+ + A- 
At pH = 7.2, [H+] = [A-]
<span>Ka = [H+][A-]/ [HA]
</span>Ka = [H+] = <span>6.3 x 10^-8</span>

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Suppose you have a 1:1:1 by weight mixture of three solid compounds, salicylic acid 4-nitroaniline naphthalene. You dissolve 1 g
choli [55]

Answer:

The correct answer is - 4-nitroaniline.

Explanation:

It is given that all three solid compounds salicylic acid + 4-nitroaniline + naphthalene are equal in the ratio in the mixture and then 1 gram of this mixture is dissolved in the diethyl ether and run a drop of the solution on TLC plate. This plate shows three spots.

The salicylic acid and naphthalene would stay dissolved in the diethyl ether solution due to the 4-nitroaniline could be extracted by adding aqueous acid and involve in the aqueous layer and thus spot of 4-nitroaniline would be with largest Rf value.

7 0
3 years ago
Explain why hydrogen gas is given off when a reactive metal undergoes a single-displacement reaction with water.
Kay [80]
Hydrogen gas is given off when a reactive metal undergoes a single-displacement reaction with water is because the <span>very active metals react with water. For example, calcium reacts with water to form calcium hydroxide and hydrogen gas etc. </span>
3 0
3 years ago
Which cells produce photosynthesis
AnnZ [28]
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Photosynthetic cell

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6 0
3 years ago
The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
8 0
3 years ago
1. For each of the molecules below, determine the electron geometry, molecule geometry, and bond
Alexxx [7]

Answer:

CCl4- tetrahedral bond angle 109°

PF3 - trigonal pyramidal bond angles less than 109°

OF2- Bent with bond angle much less than 109°

I3 - linear with bond angles = 180°

A molecule with two double bonds and no lone pairs - linear molecule with bond angle =180°

Explanation:

Valence shell electron-pair repulsion theory (VSEPR theory) helps us to predict the molecular shape, including bond angles around a central atom, of a molecule by examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement which tends to minimize repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom are either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far apart from each other as possible.

Lone pairs and multiple bonds are known to cause more repulsion than single bonds and bond pairs. Hence the presence of lone pairs or multiple bonds tend to distort the molecular geometry geometry away from that predicted on the basis of VSEPR theory. For instance CCl4 is tetrahedral with no lone pair and four regions of electron density around the central atom. This is the expected geometry. However OF2 also has four regions of electron density but has a bent structure. The molecule has four regions of electron density but two of them are lone pairs causing more repulsion. Hence the observed bond angle is less than 109°.

8 0
3 years ago
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