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3 years ago

Q2.Which is true about potassium?

Chemistry

2 answers:

maks197457 [2]3 years ago

6 0

Answer is it is very reactive

Olenka [21]3 years ago

5 0

$\bf \LARGE\rightarrow{Very \: reactive}$

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Determine the highest occupied energy level in the

aleksandrvk [35]

I think its B!!! (: (:

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2 years ago

Ergosterol, a precursor of vitamin D, has δmax = 282 nm and molar absorptivity ε = 11,900 M-1cm-1. What is the concentration of

meriva

Answer:

3.6 × 10⁻⁵ M

Explanation:

Ergosterol has a maximum absorbance at λ = 282 nm. The absorbance of an analyte is related to its concentration through the Beer-Lambert's law.

A = ε × l × c

where,

A: absorbance

ε: molar absorptivity

l: optical path length

c; molar concentration

c = A / ε × l = 0.43 / (11,900 M⁻¹cm⁻¹) × 1.00 cm = 3.6 × 10⁻⁵ M

3 0

3 years ago

A teaspoon of powdered caffeine can be as deadly as a teaspoon of ricin. True or False

Darina [25.2K]

A teaspoon of caffeine is NOT deadly as teaspoon of ricin

4 0

3 years ago

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The number of energy levels to which an electron can jump depends on the

Kitty [74]

The number of energy levels to which an electron can jump depends on the amount of energy the electron possesses. Each energy level has a specific amount of energy an electron needs to have before it can be in there. So, if an electron doesn't have enough energy to be in that energy level then it won't jump to that higher level.

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3 years ago

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Phenolphthalein has a pKa of 9.7 and is colorless in its acid form and pink in its basic form. calculate [In-}/{HIn} for the fol

zvonat [6]

Answer:

1.58x10⁻⁵

2.51x10⁻⁸

0.0126

63.10

Explanation:

Phenolphthalein acts like a weak acid, so in aqueous solution, it has an acid form HIn, and the conjugate base In-, and the pH of it can be calculated by the Handerson-Halsebach equation:

pH = pKa + log[In-]/[HIn]

pKa = -logKa, and Ka is the equilibrium constant of the dissociation of the acid. [X] is the concentrantion of X. Thus,

i) pH = 4.9

4.9 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = - 4.8

[In-]/[HIn] = $10^{-4.8}$

[In-]/[HIn] = 1.58x10⁻⁵

ii) pH = 2.1

2.1 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -7.6

[In-]/[HIn] = $10^{-7.6}$

[In-]/[HIn] = 2.51x10⁻⁸

iii) pH = 7.8

7.8 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = -1.9

[In-]/[HIn] = $10^{-1.9}$

[In-]/[HIn] = 0.0126

iv) pH = 11.5

11.5 = 9.7 + log[In-]/[HIn]

log[In-]/[HIn] = 1.8

[In-]/[HIn] = $10^{1.8}$

[In-]/[HIn] = 63.10

6 0

3 years ago