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arlik [135]
3 years ago
15

Mrs. Sail's class wanted to observe a local pond. The class collected some pond water to analyze. When they placed a drop of the

water under a microscope, they saw many microorganisms in the water. The pond water looked clear when they collected it in the jar, so what could explain the presence of these organisms? A) They must have grown in the jar overnight. B) The microorganisms are too small to be seen with the naked eye. C) Somehow, the jar was contaminated on the way from the pond to the class. D) The microorganisms only inhabit a very small section of the lake, and the class got lucky.
Chemistry
2 answers:
olasank [31]3 years ago
8 0

B) the microorganisms are to small too see with the naked eye

LiRa [457]3 years ago
3 0

B. its more of a reliable answer because it's true. You cant see it with the naked eye if you had to put in under a microscope.



I might be wrong.

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What active non metal has 6 valence e ?
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Answer:

brainliest please

Explanation:

Examples include neon (Ne), argon (Ar), and krypton (Kr). Oxygen, like all the other elements in group 16, has six valence electrons.

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3 years ago
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The ph of a fruit juice is 2.2 . find the hydronium ion​ concentration, left bracket upper h 3 upper o superscript plus right br
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Answer:

[H₃O⁺] = 6.31 x 10⁻³ M.

Explanation:

  • To find the hydronium ion concentration [H₃O⁺], we can use the relation:

<em>pH = - log[H₃O⁺].</em>

<em></em>

∴ 2.2 = - log[H₃O⁺].

∴ log[H₃O⁺] = - 2.2

<em>∴  [H₃O⁺] = 6.31 x 10⁻³ M.</em>

3 0
4 years ago
How many degenerate orbitals are needed to contain six electrons with two of them unpaired?
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5 0
3 years ago
The side chain of tyrosine has a pKa of about 10. What percent of tyrosine side chains would be deprotonated at pH 8.5?
bezimeni [28]

Answer:

Explanation:

Let the tyrosine molecule be represented by TH . It will ionise in water as follows

TH ⇄ T⁻ + H⁺

Let C be the concentration of undissociated TH and α be the degree of dissociation

TH       ⇄       T⁻ + H⁺

c                    0       0   ( before )

c( 1-α )            αc      αc ( after ionisation)

Ka = α²c² / c( 1-α )

= α²c  ( neglect α in the denominator as it is very small )

pKa = 10

Ka  = 10⁻¹⁰

pH = 8.5

H⁺ = 10⁻⁸°⁵

αc = 10⁻⁸°⁵

α²c =Ka = 10⁻¹⁰

α x10⁻⁸°⁵ = 10⁻¹⁰

α = 10⁻¹⁰⁺⁸°⁵

= 10⁻¹°⁵ = 1 / 31.62

Percentage of dissociation = 100 / 31.62

= 3.16 %

percent of  tyrosine side chains   deprotonated

3 0
3 years ago
During which part of the cell cycle is the duplicated genetic material within the nucleus of a parent cell separated to create t
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Hope this helps! Please correct me if I'm wrong :)

3 0
3 years ago
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