All categories

3 years ago

Nucleotides consist offa phosphate group, a nitrogenous base, and a

Chemistry

1 answer:

Mkey [24]3 years ago

4 0

Answer:

5-Carbon sugar

Explanation:

A nucleotide is made up of 3 things: a phosphate group, a 5-carbon sugar, and a nitrogenous base.

You might be interested in

Which one of these statements best describes what a scientist is

son4ous [18]

Which of what statements

3 0

3 years ago

When 0.42 g of a compound containing C, H, and O is burned completely, the products are 1.03 g CO2 and 0.14 g H2O. The molecular

Luda [366]

Answer: The empirical and molecular formula for the given organic compound is $C_2H_2O_4$ and $C_6H_4O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.03g$

Mass of $H_2O=0.14g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 1.03 g of carbon dioxide, $\frac{12}{44}\times 1.03=0.28g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.14 g of water, $\frac{2}{18}\times 0.14=0.016g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.42) - (0.28 + 0.016) = 0.124 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.28g}{12g/mole}=0.023moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.016g}{1g/mole}=0.016moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.124g}{16g/mole}=0.00775moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00775 moles.

For Carbon = $\frac{0.023}{0.00775}=2.96\approx 3$

For Hydrogen = $\frac{0.016}{0.00775}=2.06\approx 2$

For Oxygen = $\frac{0.00775}{0.00775}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 2 : 1

Hence, the empirical formula for the given compound is $C_3H_{2}O_1=C_3H_2O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 108.10 g/mol

Mass of empirical formula = 54 g/mol

Putting values in above equation, we get:

$n=\frac{108.10g/mol}{54g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(3\times 2)}H_{(2\times 2)}O_{(1\times 2)}=C_6H_4O_2$

Thus, the empirical and molecular formula for the given organic compound is $C_3H_2O$ and $C_6H_4O_2$

6 0

3 years ago

How to convert 5.2 km to mm

ZanzabumX [31]

5.2 km to mm.
1km = 1000m
= 5.2 * 1000m.
=5200 m. 1m = 1000mm

= 5200 * 1000mm
= 5200000 mm

4 0

3 years ago

How do sand dunes move? Use the terms leeward and windward in your explanation.

gogolik [260]

As the sand mound grows, the point of maximum sand deposition on the leeward face moves closer to the summit, causing a steepening of the leeward face relative to the windward face. The steepening and growing dune now forces wind out over the top of the dune rather than down the leeward face.

7 0

3 years ago

30.0 mL of 0.50 M NaOH to neutralize 15.0 mL of HNO3, what is the concentration (molarity) of the HNO3?

vredina [299]

First find the no. of moles of NaOH :
30/1000 = 0.3 dm3 so no. of moles = 0.3*0.5 = 0.15 moles 

as NaOH reacts with HNO3 in a ratio of one to one, there must have been 0.15 moles of HNO3 too 
moles/volume = concentration 
volume= 15/1000 = 0.15 dm3 

concentration = 1.15/0.15 = 1 mol.dm-3 

The quicker way would be to realize that you used twice as much NaOH so the HNO3 had to be twice as strong

4 0

3 years ago