Which is true about fossil fuels? A.
Coal and natural gas came from swamp environments, while oil came from open oceans.

B.
Coal came from swampy environments, while oil and natural gas came from open ocean environments.

C.
Coal, oil, and natural gas came from swamp environments.

D.
Natural gas and oil came from swamp environments, while coal came from open oceans.

Answers

Answer 1

Answer:

Which is true about fossil fuels? The statement that is true about fossil fuels is Coal came from swampy environments, while oil and natural gas came from open ocean environments. The answer is letter B. The rest of the choices is not true for fossil fuel

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Consider four elements from Group 7A: fluorine in the second period, chlorine in the third period, bromine in the fourth period, and iodine in the fifth period. Which element has the largest first ionization energy?

What is the volume of 2 mol of chlorine gas at STP?

Answers

Answer:

Volume of 2 mol of chlorine gas at STP is 44.8 L

Explanation:

Let us assume chlorine gas behaves ideally.

According to law of ideal gas- PV=nRT

where P is pressure of gas, V is volume of gas, n is number of moles of gas, R is gas constant and T is temperature in Kelvin.

At STP, P is 1 atm and T is 273 K

Here n=2 and R=0.082 L.atm/mol.K

So $V=(nRT)/(P)=(2* 0.082* 273)/(1)L=44.8 L$

Hence volume of chlorine gas at STP is 44.8 L

The conversion factor for volume at STP is $\frac {1mol}{22.4L}$ or \frac {22.4L}{1mol}. Since we want volume, we would use \frac {22.4L}{1mol}. We conclude with the following calculations:

$2molCl_(2)*\frac {22.4LCl_(2)}{1molCl_(2)} = 44.8L Cl_(2)$

The answer is 44.8L Cl2

What is the role of Sodium thiosulfate used in modified Kjeldahl procedure?

Answers

Sodium thiosulfate is used as a reducing agent in the Kjeldahl procedure. It is used to prevent the interference of certain substances when determining nitrogen content in organic compounds.

During the digestion step of the Kjeldahl procedure, the organic sample is digested with sulfuric acid. This breaks down the organic compounds and converts the nitrogen present into ammonium sulfate. Some models may contain substances that can interfere with the next analysis. This can lead to inaccuracies in the outcomes.

Using sodium thiosulfate in the modified Kjeldahl procedure minimizes interference from certain compounds. The accuracy of the nitrogen determination is improved. This helps us to ensure that the results obtained are correct and instance of the actual nitrogen content in the sample.

To learn more about Nitrogen, visit here:

brainly.com/question/31473524

Final answer:

The role of Sodium thiosulfate in the modified Kjeldahl procedure is to act as a reducing agent and prevent interference from certain substances, such as nitrites and oxidizing agents, which can affect the accuracy of the nitrogen content determination in organic compounds.

Explanation:

In the modified Kjeldahl procedure, Sodium thiosulfate (Na2S2O3) plays a crucial role as a reducing agent. This procedure is commonly used for the determination of nitrogen content in organic compounds. The addition of Sodium thiosulfate to the reaction mixture serves to prevent interference from certain substances that can react with the reducing agent used in the procedure.

When performing the modified Kjeldahl procedure, it is important to accurately determine the nitrogen content in the sample. However, certain substances, such as nitrites and other oxidizing agents, can interfere with the reaction and affect the accuracy of the results. These interfering substances can react with the reducing agent, leading to inaccurate measurements.

By adding Sodium thiosulfate to the reaction mixture, these interfering substances are neutralized. Sodium thiosulfate acts as a reducing agent, effectively preventing the interference of nitrites and other oxidizing agents. This allows for a more accurate determination of the nitrogen content in the sample.

Learn more about role of sodium thiosulfate in modified kjeldahl procedure here:

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#SPJ14

if 1200 j of heat were absorbed by a 25 g block of iron, by how much would the temperature of the iron increase?

Answers

m = mass of the block of iron = 25 g

Q = heat absorbed by the block of iron = 1200 J

ΔT = increase in the temperature of iron due to heat absorbed = ?

c = specific heat of block of iron = 0.450 J/(g C)

Heat absorbed by the block of iron is given as

Q = m c ΔT

inserting the values in the above formula

1200 = 25 (0.450) ΔT

1200 = (11.25) ΔT

ΔT = 1200/11.25

ΔT = 106.67 C

Elements that are characterized by the filling of p orbitals are classified as _____.a. Groups 3A through 8A
b. transition metals
c. inner transition metals
d. groups 1A and 2A

Answers

Answer : The correct option is, (a) Groups 3A through 8A

Explanation :

The general electronic configurations of :

Group 1A : $ns^1$

Group 2A : $ns^2$

Group 3A : $ns^2np^1$

Group 4A : $ns^2np^2$

Group 5A : $ns^2np^3$

Group 6A : $ns^2np^4$

Group 7A : $ns^2np^5$

Group 8A : $ns^2np^6$

Transition metal : $(n-1)d^((1-10))ns^((0-2))$

Inner transition metal (Lanthanoids) : $4f^((1-14))5d^((0-1))6s^2$

Inner transition metal (Actinoids) : $5f^((0-14))6d^((0-1))7s^2$

From the general electronic configurations, we conclude that the groups 3A through groups 8A elements that are characterized by the filling of p-orbitals.

Elements that are characterized by the filling of p orbitals are classified as $\boxed{{\text{a}}{\text{. Groups 3A through 8A}}}$ .

Further Explanation:

In order to make the study of numerous elements easier, these elements are arranged in a tabular form in increasing order of their atomic numbers. Such a tabular representation of elements is called a periodic table. Horizontal rows are called periods and vertical columns are called groups. A periodic table has 18 groups and 7 periods.

a. Groups 3A through 8A

The elements from group 3A to 8A has the general outermost electronic configuration of $n{s^2}n{p^(1 - 6)}$ . So the added electrons are to be filled in p orbitals.

b. Transition metals

These metals have the general valence configuration of $\left( {n - 1} \right){d^(1 - 10)}n{s^(0 - 2)}$ . This indicates that the added electrons enter either s or d orbitals.

c. Inner transition metals

These are classified as lanthanoids and actinoids. The general outermost configuration of lanthanoids is $4{f^(1 - 14)}5{d^(0 - 1)}6{s^2}$ while that of actinoids is $5{f^(0 - 14)}6{d^(0 - 1)}7{s^2}$ . In both cases, the added electron enters either d or f orbitals.

d. Groups 1A and 2A

The elements of group 1A have the general valence electronic configuration of $n{s^1}$ . It implies the last or valence electron enters in the s orbital. The group 2A elements have a general configuration of $n{s^2}$ . Here also the last electron enters the s orbital.

So elements from groups 3A to 8A are classified by the filling of p orbitals and therefore option a is correct.

Learn more:

Which ion was formed by providing the second ionization energy? brainly.com/question/1398705
Write a chemical equation representing the first ionization energy for lithium: brainly.com/question/5880605

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Periodic classification of elements

Keywords: periodic table, configuration, ns1, ns2, d, p, f, 3A, 8A, transition metals, inner transition metals, lanthanoids, actinoids, orbitals, 1A, 2A.

What is the common name for calcium silicate?

Answers

halliburton is the common name for it

What is a true statement about the atom's nucleus?

Answers

That it’s the power house and only hold protons and neutrons.

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