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TiliK225 [7]
3 years ago
6

Why does algae grow in a lake

Chemistry
2 answers:
pentagon [3]3 years ago
7 0

Answer:

algae will typically grow around the shoreline of a pond or lake because this is where the shallower water is.

Explanation:

skad [1K]3 years ago
5 0

Answer:

Light Exposure and Water Movement. Along with food, algae require the right amount of light to thrive. ... Filamentous algae will typically grow around the shoreline of a pond or lake because this is where the shallower water is.L

Hope that helps :)

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What is the answer of these compound
olganol [36]
BaO, Barium Oxide. 

Na2SO4, Sodium Sulfate.

CuO, Copper (II) Oxide.

P2O5, Diphosphorus Pentoxide.

HNO3, Nitric Acid.

CO32-, Molecular Formula. 

Hope this helps. :)
8 0
3 years ago
Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in
Sholpan [36]

Answer:

Redox

Explanation:

Reduction is gain of electrons

oxidation is loss of electrons

3 0
3 years ago
Which of the following statements is true?
dybincka [34]
The correct answer is C. Atoms are incredibly small and can bearly be seen with the most powerful electron microscopes. The nucleas of an atom contains protons and neutrons with electrons in orbitals around the nucleas. I hope this helps. Let me know if anything is unclear.
3 0
3 years ago
Practice It!
Ostrovityanka [42]

Answer:

Explanation:

Space exploration technology has helped us explore the Moon and see images of far off galaxies.

Space technology may help us inhabit another planet or live in space in the future.

These are the correct statements

4 0
3 years ago
Read 2 more answers
A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
Lynna [10]

Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

5 0
3 years ago
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