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kow [346]
3 years ago
12

What would happen if fertilizer ran into the lake where these plants, fish, and eagles live?

Chemistry
1 answer:
Olenka [21]3 years ago
6 0
Eutrophication will happen. Aerobic bacteria in the water will decompose the fertilizer and release a tonne of minerals that algae found in the water will then use.
The bacteria will have used up a lot of oxygen and the algae, while respiring will also use up a lot.
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Rust forms because iron and oxygen are highly ______ elements.
Oliga [24]

IT forms because they are highly reactive elements.

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3 years ago
I NEED HELP PLEASE, THANKS! :)
marin [14]

Answer:

\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}

Explanation:

1. Calculate the moles of copper(II) hydroxide

\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}

2. Calculate the molecules of copper(II) hydroxide

\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}

6 0
3 years ago
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Rainbow [258]

Answer:

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6 0
2 years ago
Read 2 more answers
Calculate the ratio of the mass ratio of SS to OO in SOSO to the mass ratio of SS to OO in SO2SO2. Consider a sample of SOSO in
Veronika [31]

Answer:

The ratio of the mass ratio of S to O; in SO, to the mass ratio of S to O; in SO₂, is  2:1

Explanation:

According to the consideration, let us first find the ratio of S and O in both the compounds

For SO:

\frac{m_{S} }{m_{O} }= \frac{32}{16}\\\\   \frac{m_{S} }{m_{O} }= 2

Let us express it as

SO_{\frac{m_{S} }{m_{O} }} = 2

For SO₂,

Due to two oxygen atoms in the molecule, the mass of oxygen will be taken two times

\frac{m_{S} }{m_{O} }= \frac{32}{(2)(16)}\\\\   \frac{m_{S} }{m_{O} }= 1

Let us express it as

SO_2_{\frac{m_{S} }{m_{O} }}= 1

Now, for the ratio of both the above-calculated ratios,

\frac{SO_{\frac{m_{S} }{m_{O} }}}{SO_2_{\frac{m_{S} }{m_{O} }}}=\frac{2}{1}

The required ratio is 2:1

3 0
3 years ago
Under which conditions of temperature and pressure does a real gas behave most like an ideal gas
Veseljchak [2.6K]
At higher temperature, and lower pressure. 
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