Chemistry, 14.07.2020 01:01 hannahs1313

Suppose you wanted to monitor a pH change between 8.7 and 7.7. Which indicator would be most appropriate

Answers: 3

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Answer from: jaynissiagreen16

Phenol red /

Thymol blue

Explanation:

pH indicator referred to halochromic chemical compound which is added in small amounts to a solution for the determination of pH visually.

An indicator can be described as a chemical detector for both hydronium and hydrogen ions as far as Arrhenius model is concerned, an indicator are usually weak acids or bases, and when they are which dissolved in water , there would be a dissociation process and ions is formed.

There are different types of indicator used in the laboratory depending on the pH values of the solution examples are thymol blue, bromothymol blue, thymol blue, phenolphthalein and methyl red

Thymol blue is an indicator that transit between red to yellow when the pH is between the range of 1.2–2.8 and also transit from from yellow to blue when the pH 8.0–9.6. It an indicator that usually have brownish-green or reddish-brown crystalline powder

Phenol red is pH indicator which is yellow in colour when it is below 6.8 pH and has a very bright pink colour when used with pH above 8.2

Answer from: aneecia082203

Phenolphthalein

Explanation:

Phenolphthalein pH range from 8.3-10.00

So, it’s more likely to fall under that category.

Otherwise,

Methyl orange color change pH interval range from 3.1-4.4

Bromothymol blue 6.0-7.6

Bromocresol green 3.8-5.4

Alizarin yellow 10.2-12.0

Answer from: Quest

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.