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3 years ago

Which of the following organisms would likely be found on a decaying deer carcass (dead deer)?

Chemistry

2 answers:

saw5 [17]3 years ago

6 0

Answer:

c. Bacteria and fungus

Explanation:

This would likely be found on a decaying deer carcass

TEA [102]3 years ago

3 0

Answer:

c. bacteria and fungus

Explanation:

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Currently proven global reserves are expected to be largely exhausted within the next 50 years. Uranium-238 A Coal B Natural gas

Komok [63]

Answer:

Option (D)

Explanation:

Oil (crude and petroleum) is one of the major resources on earth in which the economy of a country is largely dependent on. These are non-renewable resources, which take thousands to million years of years in order to form.

It is highly essential. Every day, a large quantity of oil is being consumed. According to the researches done in the year 2015, it has been observed that the average global consumption of oil was 95 million barrels each day.

It has a high demand, and so it is being regularly used, and it has also been predicted that the present global oil reserves would be highly exhausted within the next 50 years.

Thus, the correct answer is option (D).

6 0

3 years ago

Please help help help help

Anni [7]

Answer:

50000 dollars or 5 e5

Explanation:

Mr. Garibay has $5.0 * 10^4$ dollars in his bank.

Scientific notation represents the data in the format of expanded number or with e character.

For instance, 5.0 * 10^4 dollars = 50000 dollars or 5 e5

3 0

3 years ago

Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g CO2 and 0.6551 g

Ede4ka [16]

Answer:

C2H4O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 1.6004/44 = 0.0364

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0364= 0.4368g

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.6551/18 = 0.0364 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.0364= 0.0728

The mass of hydrogen is thus 0.0728* 1 = 0.0728g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.8009 - 0.0728 - 0.4368 = 0.2913 mole

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.2913/16 = 0.0182 mole

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of carbon and oxygen 0.0182

H = 0.0728/0.0182 = 4

C= 0.0364/0.0182 = 2

O= 0.0182/0.0182= 1

The empirical formula is thus C2H4O

3 0

3 years ago

The best determinant of how an individual measures their quality of life is _______.

Ronch [10]

C is the correct answer

4 0

3 years ago

Read 2 more answers

A solution contains 90 milliequivalents of HC1 in 450ml. What is its normality?

Rashid [163]

Answer:

Normality N = 0.2 N

Explanation:

Normality is the number of gram of equivalent of solute divided of volume of solution, where the number of gram of equivalent of solute is weight of the solute divided by the equivalent weight.

Normality is represented by N.

Mathematically, we have :

$\mathbf{Normality \ N = \dfrac{Number \ of \ gram \of \ equivalent\ of\ solute }{volume \ of \ solution}}$

Given that:

number of gram of equivalent of solute = 90 milliequivalents 90 × 10⁻³ equivalent

volume of solution (HCl) = 450 mL 450 × 10⁻³ L

$\mathbf{Normality \ N = \dfrac{90 \times 10^{-3}}{450 \times 10^{-3}}}$

Normality N = 0.2 N

7 0

3 years ago