Convert the 183000 aK to nK

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

4 years ago

Element x reacts with oxygen to produce x2o3 in an experiment it is found that 1.0000 g of x produces 1.1xxx g of x2o3 what is t

8_murik_8 [283]

Since X is 1 g, therefore O must be 0.1 g. Therefore:

moles O = 0.1 g / (16 g / mol) = 0.00625 mol

We can see that for every 3 moles of O, there are 2 moles of X, therefore:

moles X = 0.00625 mol O (3 moles X / 2 moles O) = 0.009375 mol

Molar mass X = 1 g / 0.009375 mol

Molar mass X = 106.67 g/mol

4 0

3 years ago

Choose the element that is more reactive: oxygen or argon.

Mazyrski [523]

I choose oxygen because

7 0

3 years ago

When heated, calcium hydroxide and ammonium chloride react to produce ammonia gas, water vapor, and solid calcium chloride.

NNADVOKAT [17]

Answer:

13.73g

Explanation:

mass of reactants = mass of products.

Mass reactants = 5.00 g + 10.00 g = 15.00 g

Mass products = 1.27g + mass of ammonia and water vapor

Mass of ammonia and water vapor

15.00g – 1.27 g = 13.73 g

3 0

3 years ago

How many liters of 4.0 M NaOH solution will react with 0.60 liters 3.0 M H2SO4?

slava [35]

Answer:

A. 0.90 L.

Explanation:

NaOH solution will react with H₂SO₄ according to the balanced reaction:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.

For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.

∴ (MV) NaOH = (xMV) H₂SO₄.

x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.

∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH = 2(0.6 L)(3.0 M)/(4.0 M) = 0.90 L.

4 0

3 years ago