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AlladinOne [14]
3 years ago
13

Examination of a cell by transmission electron microscopy reveals a high density of ribosomes in the cytoplasm. This observation

suggests that this cell is actively producing large amounts of which of the following molecules? A.Polysaccharides
B.Proteins
C.Lipids
D.Nucleic acids
Chemistry
1 answer:
True [87]3 years ago
3 0

Answer:

The answer is the B. Proteins

Explanation:

It is known that Ribosomes are the factory to synthesize proteins. This is their function.  In cell zones where ribosomes are abundant, proteins are actively produced

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Here’s all of them: kingdom, phylum, classes, order, families, genus, and species
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What are the pH of these solutions?
Tju [1.3M]

Answer:

The answer to your question is below

Explanation:

a)    HCl  0.01 M

      pH = -log [0.01]

      pH = - (-2)

     pH = 2

b)    HCl = 0.001 M

      pH = -log[0.001]

      pH = -(-3)

      pH = 3

c)    HCl = 0.00001 M

       pH = -log[0.00001]

       pH = - (-5)

      pH = 5

d) Distilled water

      pH = 7.0

e) NaOH = 0.00001 M

       pOH = -log [0.00001]

       pOH = -(-5)

       pH = 14 - 5

       pH = 9

f)  NaOH = 0.001 M

      pOH =- log [0.001]

      pOH = 3

      pH = 14 - 3

      pH = 11

g)   NaOH = 0.1 M

       pOH = -log[0.1]

       pOH = 1

       pH = 14 - 1

       pH = 13

3 0
3 years ago
Conversion of minus 1 coulomb meter in debye
Viktor [21]

Answer:

-1 Coulomb meter = -2.997 × 10²⁹ Debye

Explanation:

Given:

Coulomb meter = -1 CM

Find:

In debye

Computation:

We know that,

1 Coulomb meter = 299,792,458,178,090,000,000,000,000,000 Debye

So,

-1 Coulomb meter = -299,792,458,178,090,000,000,000,000,000 Debye

-1 Coulomb meter = -2.997 × 10²⁹ Debye

3 0
3 years ago
Pure metals tend to be weaker and more reactive than an alloy which is a
Alex17521 [72]
Alloys are supposed to give greater strength to metals, which is why gold is mixed with others to make it harder. They have greater strength and are more resistant to erosion.
5 0
3 years ago
How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2
Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
4 0
2 years ago
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