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aleksandrvk [35]
3 years ago
9

Which element does the electron configuration 1s22s22p2 represent? B C N K

Chemistry
2 answers:
ch4aika [34]3 years ago
7 0

Answer:

Carbon

Explanation:

the electronic configuration corresponds to the number of electrons that an atom of an element has and these are compensated with the number of protons in the nucleus of the element, therefore the number of protons in the nucleus is equal to the number of electrons in the orbitals.

The number of protons in the nucleus corresponds to the number Z, this is what allows us to differentiate between one element and another  

1s^2 2s^22p^2

We add the electrons

2+2+2=6

The number of electrons is 6 that matches the number of protons (which is the number Z)

We look for in the periodic table the element Z = 6 (atomic number)

The element that corresponds to this number is carbon

antiseptic1488 [7]3 years ago
5 0
The electron configuration is 1s2 2s2 2p2, then the number of electrons is 2 + 2 + 2 = 6. The number of electrons equals the number of protons and this is the atomic number. So, the atomic number of the element is 6.<span> In a periodic table you can find that the element with atomic number 6 is carbon. So the answer is C.</span>
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Please explain:) I would really appreciate a step by step explanation if possible.
UNO [17]

The enthalpy change of the reaction, ΔH = -311 kJ

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

<h3>What is the enthalpy change for the reduction of ethyne to form ethane?</h3>

The enthalpy change for the reaction is obtained from the summation of the enthalpies of the reactions of the intermediate steps according to Hess's law.

The equation of the reaction is given below:

  • C₂H₂ + 2 H₂ → C₂H₆

The enthalpy of the reaction, ΔH = ΔH₁ + 2ΔH₂ + (-ΔH₃)

ΔH = {(-1299) + (2 * -286) + (1560)}Kj

ΔH = -311 kJ

The equation for the methanation reaction is given below:

3 H₂O + CO → CH₄ + H₂O

The enthalpy for the methanation reaction is as follows:

ΔH = 1.5ΔH₁ + 0.5*(-ΔH₂) + ΔH₃ + -ΔH₄

ΔH = (-483.6 * 1.5) + (0.5 * 221.0) + (-802.7) + (393.5)

ΔH = -1024.1 kJ/mol

Molar mass of CO = 28 /mol

Enthalpy change involved in the reaction of 300 g of CO = 300/28 * -1024.1 kJ/mol

Enthalpy change involved in the reaction of 300 g of CO = -10972.5 kJ

In conclusion, the enthalpy changes are calculated from the enthalpy values of the  intermediate reactions.

Learn more about enthalpy changes at: brainly.com/question/26991394

#SPJ1

7 0
2 years ago
The coolant in automobiles is often a 50/50 % by volume mixture of ethylene glycol, C2H6O2, and water. At 20°C, the density of e
Misha Larkins [42]

Explanation:

Let the volume of the solution be 100 ml.

As the volume of glycol = 50 = volume of water

Hence, the number of moles of glycol = \frac{mass}{molar mass}

                                                  = \frac{density \times volume}{molar mass}

                         = \frac{1.1088 \times 50}{62 g/mol}

                         = 0.894 mol

Hence, number of moles of water = \frac{50 \times 0.998}{18}

                                             = 2.77

As glycol is dissolved in water.

So, the molality = 0.894 \times \frac{1000}{49.92}

                           = 17.9

Therefore, the expected freezing point = -1.86 \times 17.9

                                                                  = -33.31^{o}C

Thus, we can conclude that the expected freezing point is -33.31^{o}C.

6 0
3 years ago
Jim wants to react hydrogen and oxygen to get 36 grams of water. If he starts with 4 grams of Hydrogen (H), then how many grams
serious [3.7K]

Answer:

Mass of Oxygen = 32 grams

Explanation:

Given:

Mass of water = 36 grams

Mass of Hydrogen = 4 grams

Find:

Mass of Oxygen

Computation:

Using Law of Conservation of mass

Mass of water = Mass of Hydrogen + Mass of Oxygen

36 grams = 4 grams + Mass of Oxygen

Mass of Oxygen = 32 grams

3 0
3 years ago
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Lina20 [59]

Answer:

option b

Explanation:

8 0
3 years ago
List two things you know about nonrenewable energy
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It is natural and u can't by it
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3 years ago
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