5 kinds of trace fossils

Given the Henry’s law constant for O2(4.34*109Pa) at 25° C, calculate the molar concentration of oxygen in air-saturated and O2s

Flauer [41]

This is an incomplete question, here is a complete question.

The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.

Answer : The molar concentration of oxygen is, $2.67\times 10^2mol/L$

Explanation :

As we know that,

$C_{O_2}=k_H\times p_{O_2}$

where,

$C_{O_2}$ = molar solubility of $O_2$ = ?

$p_{O_2}$ = partial pressure of $O_2$ = 0.2 atm = 1.97×10⁻⁶ Pa

$k_H$ = Henry's law constant = 4.34 × 10⁹ g/L.Pa

Now put all the given values in the above formula, we get:

$C_{O_2}=(4.34\times 10^9g/L.Pa)\times (1.97\times 10^{-6}Pa)$

$C_{O_2}=8.55\times 10^3g/L$

Now we have to molar concentration of oxygen.

Molar concentration of oxygen = $\frac{8.55\times 10^3g/L}{32g/mol}=2.67\times 10^2mol/L$

Therefore, the molar concentration of oxygen is, $2.67\times 10^2mol/L$

8 0

3 years ago

A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce

spin [16.1K]

Explanation:

The given data is as follows.

Volume of lake = $15 \times 10^{6} m^{3}$ = $15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}$

Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = $5.6 \times 15 \times 10^{9} mg$

= $84 \times 10^{9}$ mg

= $84 \times 10^{3}$ kg

Flow rate of river is 50 $m^{3} sec^{-1}$

Volume of water in 1 day = $50 \times 10^{3} \times 86400 liter$

= $432 \times 10^{7}$ liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are $2.9792 \times 10^{10} mg$ or $2.9792 \times 10^{4} kg$

Flow rate of sewage = $0.7 m^{3} sec^{-1}$

Volume of sewage water in 1 day = $6048 \times 10^{4}$ liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = $1.8144 \times 10^{10} mg$ or $1.8144 \times 10^{4}kg$

Therefore, total concentration of lake after 1 day = $\frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l$

= 6.8078 mg/l

$k_{D}$ = 0.2 per day

$L_{o}$ = 6.8078

Hence, $L_{liquid}$ = $L_{o}(1 - e^{-k_{D}t}$

$L_{liquid}$ = $6.8078 (1 - e^{-0.2 \times 1})$

= 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

= 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0

3 years ago

Amylose is a form of starch which has ________. only β-1,4-bonds between glucose units only α-1,4-links bonds glucose units hemi

forsale [732]

Answer: Amylose is a form of starch which has only α-1,4-links bonds glucose units.

Explanation:

Amylose is a polysaccharide made up of α(1-4) bound glucose molecules. The carbon atoms on glucose are numbered, starting at the aldehyde (C=O) carbon, so, in amylose, the 1-carbon on one glucose molecule is linked to the 4-carbon on the next glucose molecule.

5 0

3 years ago

Assuming all other variables are constant, if the pressure of a gas is increased from 2. atm to 6 atm the volume would change fr

Leona [35]

Answer:

2.0 Liters

Explanation:

Boyles law

P1 V1 = P2 V2

V2= P1 V1 / V2

7 0

3 years ago

I need help please j

Sidana [21]

Reactants Hydrogen: 5
Products Hydrogen: 5

Reactants Carbon: 3
Products Carbon: 3

Reactants Oxygen: 4
Products Oxygen: 5

3 0

3 years ago