Select the atomic symbol that would represent an isotope with 13 protons and 14 neutrons. A. 14 16 N B. 13 6 C C. 27 14 Si D. 27 13 Al

Answers

Answer 1

Answer:

Answer:

D. 27 13 Al

Explanation:

13 protons means atomic number 13. The periodic table says element 13 is aluminum (Al). 27 is the atomic mass, made up of 13 protons (1 each) and 14 neutrons (also 1 each) for a total of 27 atomic mass.

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Which element has three unpaired electrons in its p orbital? carbon (atomic number 6) oxygen (atomic number 8) nitrogen (atomic number 7) beryllium (atomic number 4)

Answers

Answer: N (atomic number 7)

Explanation:

1) Due the electron distribution of each, taking into account Hund's rule to deal with pairing of electrons.

2) Hund's rule: each subshell orbital is occupied with single electrons of parallel spin before double occupation occurs.

That is, before any one orbital is doubly occupied, every orbital in a subshell is singly occupied with one electron, and all the electrons in singly occupied orbitals have the same spin, which means that they are unpaired.

3) Distributions:

a) C (atomic number 6) ⇒ 6 electrons:

⇒ 1s² 2s² 2p² ⇒ 1s² 2s² 2p ↑ ↑ : as per Hund's rule the two electrons on 2p are unpaired.

b) O (atomic number 8) ⇒ 8 electrons

⇒ 1s² 2s² 2p⁴ ⇒ 1s² 2s² 2p⇅ ↑ ↑: as per Hund's rule the two 2p electrons are unpaired

c) N (atomic number 7) ⇒ 7 electrons

1s² 2s² 2p³ ⇒ 1s² 2s² 2p↑ ↑ ↑: as per Hund's rule, the three 2p electrons are unpaired, so this is the answer to the question.

c) Be (atomic number 4) ⇒ 4 electrons

⇒ 1s² 2s² ⇒ none p electron.

Final answer:

Nitrogen has three unpaired electrons in its p orbital.

Explanation:

Among the given elements, nitrogen (atomic number 7) has three unpaired electrons in its p orbital.

In the electron configuration of nitrogen, it has a total of five electrons: 2 in the 1st energy level and 3 in the 2nd energy level. The p sublevel of the 2nd energy level contains three orbitals, each capable of holding 2 electrons. Therefore, the three unpaired electrons in nitrogen's p orbital are in three separate p orbitals in the 2nd energy level.

Carbon has two unpaired electrons in its p orbital, while oxygen has two unpaired electrons in its p orbital and two in its s orbital. Beryllium does not have any unpaired electrons in its p orbital.

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A gas contains a mixture of NH3(g) and N2H4(g), both of which react with O2(g) to form NO2(g) and H2O(g). The gaseous mixture (with an initial mass of 61.00 g) is reacted with 10.00 moles O2, and after the reaction is complete, 4.062 moles O2 remains. Calculate themass percent of N2H4(g) in the original gaseous mixture.

Answers

Answer:

Mass percent of N2H4 in original gaseous mixture = 31.13 %

Explanation:

Given:

Initial mass of gaseous mixture = 61.00 g

Initial mole of oxygen = 10.0 mol

Moles of oxygen remaining after the reaction = 4.062 mol

Moles of oxygen used = 10.0 - 4.062 = 5.938 mol

$4NH_3 + 7O_2\rightarrow 4NO_2 + 6H_2O$

$N_2H_4 + 3O_2\rightarrow 2NO_2 + 2H_2O$

Total oxygen used in both the reactions = 10.0 parts

out of 10 parts, 3 part react with N2H4.

$No.\;of\;moles\;of \;oxygen \;used = 5.398*(3)/(10) =1.78\; moles$

Now, consider the reaction of N2H4

$N_2H_4 + 3O_2\rightarrow 2NO_2 + 2H_2O$

3 moles of O2 react with 1 mole of N2H4

1.78 moles of oxygen will react with 1.78/3 = 0.5933 mol of N2H4

$Mass = Moles* Molecular mass$

Molecular mass of N2H4 = 32 g/mol

$Mass\;of\;N_2H_4= 0.5933* 32 = 18.99 g$

$Mass\;percent = (Mass\;of\;N_2H_4)/(Total\;mass)* 100$

Total mass = 61.0 g

$Mass\;percent = (18.99)/(61.0)* 100=31.13 \%$

Final answer:

The mass percent of N2H4 in the gaseous mixture can be determined through stoichiometric calculations and determining the limiting reactant. The initial and remaining amounts of O2 are used to calculate the reacted amount of O2, which then allows for the calculation of the amount of N2H4. This information is used in the mass percent formula.

Explanation:

The balanced reaction states that for one mole of NH3, one mole of O2 is required, while for one mole of N2H4, 3 moles of O2 are required. Thus, the initial moles of O2 were 10 moles and after reaction 4.062 moles O2 remained. Thus, the reacted amount of O2 is 10 - 4.062 = 5.938 moles. From calculating the limiting reactant and applying stoichiometry, the amount of N2H4 can be determined. We know the molar mass of N2H4 is 32 g/mole. By calculating the molar ratio, we can then calculate the mass percent of N2H4 in the mixture using the formula: (mass of N2H4 / total mass) * 100%.

Learn more about Stoichiometric Calculations here:

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A rock weighing 75.0g is placed in a graduated cylinder displacing the volume from 31.4 ml to 36.5ml. What is the density of the rock in grams /cubic centimeter.

Answers

Answer:

14.71 g/cm³

Explanation:

The density of a substance can be found by using the formula

$density = (mass)/(volume) \n$

volume = final volume of water - initial volume of water

volume = 36.5 - 31.4 = 5.1 mL

But

1 mL = 1 cm³

5.1 mL = 5.1 cm³

From the question we have

$density = (75)/(5.1) \n = 14.705882..$

We have the final answer as

14.71 g/cm³

Hope this helps you

When ________ are added to an atomic nucleus, energy is produced and this is called a fusion reaction.a. protons
b. neutrons
c. protons or neutrons
d. protons, neutrons or electrons

Answers

When neutrons are added to an atomic nucleus, energy is produced and this is called a fusion reaction. The answer is letter B. To be exact, two nuclei combine to produce a more stable nucleus. The positive charge on both nuclei result in a large electrostatic energy barrier to fusion.

The correct answer is c because the nuc are combing more protns and neutrons. The answer suposed to be protons and neutrons NOT PROTONS OR NEUTRONS?!!

What is the volume of a tank of nitrogen if it contains 17 moles of nitrogen at 34 C under 12,000 Pa?