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2 years ago

Can anyone help me with this? The picture may or may not show up on your screen and if not I can probably type it out

Chemistry

1 answer:

3241004551 [841]2 years ago

6 0

Because H2O has 2 H and 1O and 1 H is already in the equation, the answer should contain 1H and 1O, meaning it's either H O- or OH- however the proper way to write it is OH
An easy was to remember is by writing H2O down as HOH and when you split it up it becomes H+ + OH-

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Can You compost salad

Fynjy0 [20]

Answer:

Yes, you can compost lettuce and other salad leaves. Be careful! If the salad is covered in lots of dressing, the oils or fats in the dressing may attract rats or other animals to your compost heap.

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

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

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

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

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

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I believe the answer is A.

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2 years ago

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