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3 years ago

Which is more prevalent in the food we eat: carbon-12 or carbon-14?

1 answer:

5 0

I think the more prevalent in the food we eat is Carbon-12 which is present in Most of the food we ingest. Both carbon-12 and carbon-14 are isotopes of carbon, they differ in the number of neutrons in each atom. Because of the different number of neutrons, carbon-12 and carbon-14 differ with respect to radioactivity. Carbon-12 is a stable isotope but carbon-14 undergoes radioactive decay.

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What characteristic frequencies in the infrared spectrum of your sodium borohydride reduction product will you look for to deter

kkurt [141]

Answer:

A)The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

B)Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.

C)If the product is wet with water there will be no change in the infrared spectrum

Explanation:

The characteristic frequency to look out for is 1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

Characteristic frequency present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.

If the product is wet with water there will be no change in the infrared spectrum

5 0

3 years ago

A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane

son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

= 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

= 75 - 67.5

= 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

= 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

= 25 - 21.25

= 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

= 64.125 moles

$CO$ formed is = 0.05 x 67.5

= 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

= 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

= 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21}$

= 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

= 2 x 64.125

= 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

= 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3 = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

= 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.

6 0

3 years ago

Pentanitrogen heptachloride formula

Marianna [84]

Answer:

Explanation:

this is the answer hope it helps!(:

4 0

2 years ago

What are the relationships between temperature and viscosity of water?

iris [78.8K]

Explanation:

Both cohesion and molecular interchange contribute to liquid viscosity. The impact of increasing the temperature of a liquid is to reduce the cohesive forces while simultaneously increasing the rate of molecular interchange. The former effect causes a decrease in the shear stress while the latter causes it to increase.

temperature?

The viscosity of liquids decreases rapidly with an increase in temperature, and the viscosity of gases increases with an increase in temperature. Thus, upon heating, liquids flow more easily, whereas gases flow more sluggishly.

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8 0

3 years ago

The active ingredient in some oral anesthetics used in sore throat sprays. what is the molar mass of phenol?

raketka [301]

Answer: Molecular mass of the phenol is 94g/mol.

Explanation:

Phenol is an active ingredient in some oral anesthetics used in sore throat sprays.It is used as anesthetics to cure minor sour throat pain , minor mouth irritations etc.

For Example : Cramex , Chloraseptic sprays contains phenol as an active ingredient.

Phenol has a molecular formula of $C_6H_5OH$

Molecular mass of the phenol = 6(mass of carbon)+5(mass of hydrogen)+1(mass of oxygen)+1(mass of hydrogen) = 6(12)+5(1)+1(16)+(1) = 94 g/mol

6 0

3 years ago

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