Draw the following isomers of the alkanes of molecular formula C-H16

ArbitrLikvidat [17]

The process of thermal decomposition of organic materials (carbon based materials) at very high temperature and absence of oxygen is known as pyrolysis.

The temperature for such process is generally 430 °C or 800 °F. This causes simultaneous change in physical phase and chemical composition of the organic material. The process is irreversible in nature.

The resultant products mostly have more useful properties than the original one. Thus, to bring the greater value to the common organic material and waste, the idea of pyrolysis is important.

6 0

4 years ago

It is necessary to make 225 mL of 0.222 M solution of nitric acid. Looking on the shelf, you see only 16 M nitric acid. How much

nalin [4]

Explanation:

The required concentration of $HNO_3$ M1 =0.222 M.

The required volume of $HNO_3$ is V1 =225 mL.

The standard solution of $HNO_3$ is M2 =16 M.

The volume of standard solution required can be calculated as shown below:

Since the number of moles of solute does not change on dilution.

The number of moles $n=molarity * volume$

$M_1.V_1=M_2.V_2$

$V2=\frac{M_1.V_1}{M_2} \\=0.222M x 225 mL / 16 M\\=3.12 mL$

Hence, 3.12 mL of 16 m nitric acid is required to prepare 0.222 M and 225 mL of nitric acid.

6 0

3 years ago

Your body uses about 24 grams of glucose to run cellular respiration. How many molecules does your body need for cellular respir

stealth61 [152]

10000000000303033830330291

3 0

3 years ago

1. A 9.941 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 14.57 grams of CO2 an

salantis [7]

Answer:

1) empirical formula = CH2O ; the molecular formula = C2H402

2) empirical formule = C3H404 = Molecular formula

Explanation:

CxHyOz → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 14.57 grams of CO2 there is: (12g Carbon / 44g CO2) * 14.57 g CO2 = 3.9736 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom This is 2g H/ 16g O

In 5.966 g of water there are: [2 g H / 18 g H2O] * 5.966 g H2O = 0.6629 g H

⇒The original sample had 9.941 g of Sample - 3.9736 g of C - 0.6629 g of H = 5.3045 g of O

2) Calculate number of moles

C: 3.9736 g / 12.0 g/mol = 0.3311 mol

H: 0.6629 g / 1.0 g/mol = 0.6629 mol

O: 5.3045g / 16.0 g/mol = 0.3315 mol

3) Now we should divide each number of mole by the smallest number (0.3311) to find the proportion of the elements

C: 0.3311/ 0.3311= 1

H: 0.6629 / 0.3311 = 2

O: 0.3315/ 0.3311 = 1

This gives us the empirical formula of CH2O

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 12 + 2*1 + 16 = 30g

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

60 / 30 = 2 = n

If we multiply CH2O by 2 we'll get: C2H4O2

If we control this by calculating the molar mass:

2*12 + 4*1.01 + 2*16 = 60.04 g/mole

Then, the molecular formula is C2H4O2

CxHyOz → CO2 + H2O

⇒ So for CO2 we can say that for each 1 C-atom (12g) there are 2 O-atoms (32g).

This means the ratio is 12g C/ 32g O

CO2 has a molar mass of 44 g/mole

in 17.97 grams of CO2 there is: (12g Carbon / 44g CO2) * 17.97 g CO2 = 4.90 g Carbon

⇒For H2O we can do the same:

The ratio of each element in H2O is: 2 H-atoms/ 1 O-atom This is 2g H/ 16g O

In 4.905 g of water there are: [2 g H / 18 g H2O] * 4.905 g H2O = 0.545 g H

⇒The original sample had 14.16 g of Sample - 4.90 g of C - 0.545 g of H = 8.715g of O

2) Calculate number of moles

C: 4.90 g / 12.0 g/mol = 0.4083 mol

H: 0.545 g / 1.0 g/mol = 0.545 mol

O: 8.715g / 16.0 g/mol = 0.5447 mol

3) Now we should divide each number of mole by the smallest number (0.4083) to find the proportion of the elements

C: 0.4083/ 0.4083= 1

H: 0.545 / 0.4083 =1.33

O: 0.5447/ 0.4083 = 1.33

We should multiply everything by 3

This gives us the empirical formula of C3H4O4

4) Calculate the mass of the empirical formula

Molar mass of Carbon = 12g /mole

Molar mass of Hydrogen = 1g /mole

Molar mass of Oxygen = 16g /mole

mass of the empirical formule = 3*12 + 4*1 + 4*16 = 104

5) Calculate the molecular formula

mass of molecular formule / mass of empirical formula = n

We have to multiply the empirical formula by n to get the molecular formula.

104 / 104 = 1 = n

This means the empirical formula = molecular formula = C3H4O4

4 0

3 years ago

Which one of the following processes produces a decrease in the entropy of the system?a, precipitation of AgCl(s) from Ag+(aq) a

Usimov [2.4K]

Answer: precipitation of $AgCl(s)$ from $Ag^+(aq)$ and $Cl^-(aq)$ ions in solution

Explanation:

Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.

$\Delta S$ is positive when randomness increases and $\Delta S$ is negative when randomness decreases.

a) precipitation of $AgCl(s)$ from $Ag^+(aq)$ and $Cl^-(aq)$ ions in solution : As ions are getting solidified, entropy decreases.

b) dissolution of $LiOH(s)$ in water: The compounds dissociates into ions, entropy increases.

c) melting solid gold into liquid gold: The randomness increases, entropy increases.

d) evaporation of Hg(l) to form Hg(g) : The randomness increases, entropy increases.

e) mixing of two gases into one container : The randomness increases, entropy increases.

4 0

3 years ago