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3 years ago

Match the natural resources to their uses. water forests wetlands parks

Chemistry

2 answers:

spin [16.1K]3 years ago

8 0

Answer:

water - to drink

forests - to go camping and to get lost

wetlands - to see some rare wildlife

parks - to play

Anna11 [10]3 years ago

3 0

Water: drink, washing
Forests: wood for shelter, wood for fire to cook food
Wetlands: grow certain crops
Parks: little kids entertainment, relaxing

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Consider the following balanced equation:

Serga [27]

Zn⁰ + 2H⁺ ------> Zn²⁺ + H2⁰

H⁺ ion has oxidation number +1.
Zn²⁺ ion has oxidation number +2.
Atom of Zn has electric charge 0, and each hydrogen atom in the molecule H2 have oxidation number 0. So, Zn and each hydrogen atom in H2 have oxidation numbers equals "0".

Answer is
D. Zn and each hydrogen atom in H2

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

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.

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3 years ago

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Answer: Either A or D

Geologists and paleontologists constructed the geologic table based on the relative positions of different strata and fossils, and estimated the time scales based on studying rates of various kinds of weathering, erosion, sedimentation, and lithification.

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