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2 years ago

If 14c-labeled uracil is added to the growth medium of cells, what macromolecules will be labeled?

Chemistry

2 answers:

deff fn [24]2 years ago

8 0

If 14c-labeled uracil is added to the growth medium of cells, what macromolecules will be labeled? RNA

<h3>Further explanation </h3>

14C-labeled compounds

Specifications:

Specific Activity: 50-60 mCi/mmol 1.85-2.22 GBq/mmol
Synonym: Uracil [6-14C]
Formula: Not available
Molecular Weight: 112.1
Solvent: Sterile water
Concentration: 0.1 mCi/ml
CAS Number: Not available
Shipped in dry ice: No
Exclusive: Yes
Reference: Not available

If 14c-labeled uracil is added to the growth medium of cells, the macromolecules will be labeled Ribonucleic acid (RNA).

Ribonucleic acid (RNA) is a polymeric molecule essential in various biological roles in coding, decoding, regulation and expression of genes. RNA (ribonucleic acid) is made up of four chemical bases: adenine (A), cytosine (C), guanine (G) and uracil (U) connected by a ribose backbone. RNA is a single-stranded molecule and it is not stable under alkaline conditions. RNA directly codes for amino acids and as acts as a messenger between DNA and ribosomes to make proteins.

When there are 14c-lable uracil that are added to the growth medium of cells, the macromolecules that will be labled are RNA. Uracil is a nucleobase that make up the DNA or the RNA. In RNA, uracil binds with other nucleobase (adenine) through hydrogen bonds.

<h3>Learn more</h3>

Learn more about RNA brainly.com/question/9091941

<h3>Answer details</h3>

Grade: 9

Subject: chemistry

Chapter: RNA

Keywords: RNA

ziro4ka [17]2 years ago

6 0

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If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le

ElenaW [278]

The volume of sphere can be calculated using the following formula:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of the sphere which is 3.5 cm. Putting the value,

$V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}$

This is equal to the volume of lead, density of lead is $11.34 g/cm^{3}$ thus, mass of lead can be calculated as follows:

m=d×V

Putting the values,

$m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g$

Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.

Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:

m=0.686×0.866=0.5940 g

Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

$m=\frac{92.5}{100}\times 0.5940=0.54945 g$

1 g of lead obtain from $\frac{1}{0.54945}=1.82$ grams of ore.

Thus, 2041.2 g of lead obtain from:

$2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg$

Therefore, mass of ore required to make lead sphere is 3.715 kg.

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2 years ago

24 how many moles are in 2.04 × 1024 molecules of h2o?

Sergeeva-Olga [200]

The answer is 3.39 mol.

<span>Avogadro's number is the number of molecules in 1 mol of substance.
</span><span>6.02 × 10²³ molecules per 1 mol.
</span>2.04 × 10²⁴<span> molecules per x.

</span>6.02 × 10²³ molecules : 1 mol = 2.04 × 10²⁴ molecules : x
x = 2.04 × 10²⁴ molecules * 1 mol : 6.02 × 10²³ molecules
x = 2.04/ 6.02 × 10²⁴⁻²³ mol
x = 0.339 × 10 mol
<span>x = 3.39 mol
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2 years ago

Convert 338 L at 63.0 atm to its new volume at standard pressure.

taurus [48]

The new volume at standard pressure of 1 atm is 21294 liters.

Explanation:

Data given:

Initial volume of the gas V1 = 338 liters

initial pressure on the gas P1 = 63 atm

standard pressure as P2 = 1 atm

Final volume at standard pressure V2 =?

The data given shows that Boyle's law equation is to used:

P1V1 = P2V2

rearranging the equation to calculate V2,

V2 = $\frac{P1V1}{P2}$

Putting the values in the equation:

V2 = $\frac{338X63}{1}$

= 21294 L

as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.

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2 years ago

Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass

Papessa [141]

Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:

209.96 g * 22.88/100 = 48.04 g C

209.96 g * 5.76/100 = 12.09 g H

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Now we convert those masses into moles:

48.04 g C ÷ 12 g/mol = 4.00 mol C

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Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.

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2 years ago

The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution i

bija089 [108]

Answer:

110ml

Explanation:

<em>Using the dilution equation, C1V1 = C2V2</em>

<em>Where C1 is the initial concentration of solution</em>

<em>C2 is final concentration of solution</em>

<em>V1 is intital volume of solution</em>

<em>V2 is final volume of solution.</em>

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$V2 =\frac{C1V1}{C2}$

$V2 =\frac{10*6}{0.5}$

$V2 =120ml$

volume of water added = final volume -initial volume

= 120-10

=110ml

3 0

2 years ago