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3 years ago

Is a group of deer in a field called a community or a population?

Chemistry

2 answers:

Elina [12.6K]3 years ago

8 0

Answer:

From the 2, I believe it is called a community.

timofeeve [1]3 years ago

6 0

Answer:

Explanation:

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The blackbody curve for a star named beta is shown below. The most intense radiation for this star occurs in what spectral band?

Viefleur [7K]

The answer is Infrared.

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2 years ago

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Which one allows you to obtain the most precise volume? Circle one.the first one on the right because it's too ee and also helps

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Answer:50 ml ☺️

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2 years ago

Which pairing accurately matches a biomolecules polymer with its monomer subunit

BartSMP [9]

The pairing that accurately matches a bio molecule polymer with its monomer subunit would be B. Protein and amino acids.

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3 years ago

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Which statement best explains why gneiss is composed of layers but there are no layers in granite?

Mandarinka [93]

Gneiss is a sedimentary rock and granite is a metamorphic rock.

<h3>What are rocks?</h3>

Rocks are geological hard materials that are made up of various types which include:

Sedimentary rocks: These are rocks that made up of various layers formed from sediments. Example is the gnesis.

Metamorphic rocks: These are rocks that are form from pre existing rocks that undergoes some transformation. Example is granite

Therefore, Gneiss is a sedimentary rock and granite is a metamorphic rock.

Learn more about rocks here:

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2 years ago

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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3 years ago