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3 years ago

Nucleotides are the __ of nucleic acids

Chemistry

1 answer:

Arte-miy333 [17]3 years ago

8 0

Nucleotides are the building blocks of nucleic acids, as they are essential for the making of the acid

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Help now please this is not to hard of a q

marishachu [46]

Answer: Thomson

Explanation: Thomson proposed the plum pudding model of the atom.

3 0

2 years ago

When the ph of an aqueous solution is changed from 1 to 2, the concentration of hydronium ions in the solution is.

kiruha [24]

Answer:

decreased by a factor of 10

Explanation:

pH is defined in such a way that;

pH= −log10(H)

Where H represents the concentration of Hydronium or Hydrogen ions

Given that pH is changed from 1 to 2,

By rearranging the above formula , we get 10−pH = H

if pH=1,H=10−1=0.1M
if pH=2,H=10−2=0.01M

Therefore, 0.1/0.01 = 10 and 0.1 > 0.01

Hence, the concentration of hydronium ions in the solution is decreased by a factor of 10

7 0

2 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

2 years ago

Which of the following elements will lose electrons to form an

photoshop1234 [79]

B and c...will lose electron(s) in forming an Ion.

P is an Anion
b..Fe. and c...Pb form Cations (+) by losing electrons.
d. Se is an Anion.

4 0

3 years ago

Halons contain halogens, which are highly reactive with oxygen. true or false

Natalija [7]

Answer:

true

Explanation:

6 0

3 years ago

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