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QveST [7]
3 years ago
10

Convert the 183000 aK to nK

Chemistry
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

0.000183 nK

Conversion ^

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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

<span>Molar mass X = 106.67 g/mol</span>

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:

<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.</em>

<em>1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.</em>

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

<em>∴ (MV) NaOH = (xMV) H₂SO₄.</em>

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

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

4 0
3 years ago
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