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3 years ago

What step of the scientific method might be the most difficult when studying creatures that lived millions of years ago?

Chemistry

1 answer:

DiKsa [7]3 years ago

3 0

Answer:

In my opinion testing with an experiment will be the most difficult when studying ancient creatures. You have to base off your research on fossils and might not be provided with accurate data depending on what your trying to prove or learn.

Explanation:

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Prior to hosting an international soccer match, the local soccer club needs to replace the artifical turf on their field with gr

GuDViN [60]

Answer:

It will cost $68,620.5 to the club to add the grass turf to their field.

Explanation:

length of the field = l = 0.102 km = 0.102 × 1000 m

( 1km = 1000 m)

Width of the field w = 0.069 km = 0.069 × 1000 m

Area of the field , A= l × w

$A=0.102\times 1000 m\times 0.069\times 1000 m=7,038 m^2$

Cost of grass turfing = $9.75 \$/m^2$

Cost of grass turfing on field of $7,038 m^2$ :

$=7,038 m^2\times 9.75 \$/m^2=\$68,620.5$

It will cost $68,620.5 to the club to add the grass turf to their field.

4 0

3 years ago

What are conclusions based on?

kati45 [8]

They're based on the results of your testing.

7 0

3 years ago

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What two gases make up 99% of the Earth's atmosphere? A. Nitrogen and water vapor B. Carbon dioxide and oxygen C. Carbon dioxide

denpristay [2]

Of the gases listed, nitrogen, oxygen, water vapor, carbon dioxide, methane, nitrous oxide, and ozone are extremely important to the health of the Earth's biosphere. The table indicates that nitrogen and oxygen are the main components of the atmosphere by volume.

so the answer is D. Nitrogen and oxygen

hope this helps!

6 0

3 years ago

Which is the chemical formula for zinc phosphate

Orlov [11]

The Chemical fomulla for Zinc Phosphate

Zn3(PO4)2

3 0

3 years ago

If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.

vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0

3 years ago

Read 2 more answers