Nitrogen is a group 15 element. What does being in this group imply about the structure of the nitrogen atom?
O A. Nitrogen has 15 valence electrons.
OB.
Nitrogen has 15 neutrons.
OC. Nitrogen has 5 valence electrons.
D.
Nitrogen has 5 neutrons.

Answer:

1). strong nuclear force 2). nuclear binding energy 3), mass defect

Explanation:

Right on Edge

1. Strong nuclear force the force that keeps the nucleons bound inside the nucleus of an atom.

2. Nuclear binding energy the amount of energy needed to split the nucleus into individual protons and neutrons.

3. Mass defect the difference between the mass of the nucleons and the mass of an Atom.

The term strong nuclear force is defined as the force that binds protons and neutrons together. It also binds them all together in a nucleus and is responsible for the energy released in nuclear reactions.

The examples of strong nuclear force are the force that hold protons and neutrons in nuclei of atoms. The elements' greater than the hydrogen atom. The fusion of hydrogen into helium in the sun's core.

Thus, 1. option B, 2. option B and 3. option B is correct.

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Answer:

ammonia

Explanation:

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is  ammonia

Explanation:

The mixture contains two base compound which are

           ammonia,

and     diethylamine

Now the addition of HCl which is  a strong acid in step 1  will cause the protonation of  the  two base compound , which makes the soluble hence resulting in them being extracted to the aqueous layer as represented in below

       [tex]NH_3 + HCl\to NH_4 ^{+} + Cl^-[/tex]

and

     [tex](CH 3CH 2) 2NH + HCl \to (CH 3CH 2) 2NH_2^{+} + Cl[/tex]

Answer: The molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=39.61g[/tex]

Mass of [tex]H_2O=9.01g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, [tex]\frac{12}{44}\times 39.61=10.80g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, [tex]\frac{2}{18}\times 9.01=1.00g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = [tex]\frac{0.9}{0.10}=9[/tex]

For Hydrogen = [tex]\frac{1}{0.10}=10[/tex]

For Oxygen = [tex]\frac{0.10}{0.10}=1[/tex]

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is [tex]C_9H_{10}O[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

[tex]n=\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex]

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

[tex]n=\frac{268.34g/mol}{134g/mol}=2[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2[/tex]

Thus, the molecular formula for the given organic compound is [tex]C_{18}H_{20}O_2[/tex].

Answer:

We can produce 6.20 grams of CO2

Explanation:

Step 1: Data given

Mass of hexane = 4.3 grams

Molar mass of hexane = 86.18 g/mol

Mass of oxygen = 7.14 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The balanced equation

2C6H14 + 19O2 → 12CO2 + 14H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles hexane = 4.3 grams / 86.18 g/mol

Moles hexane = 0.0499 moles

Moles oxygen = 7.14 grams / 32.0 g/mol

Moles oxygen = 0.2231 moles

Step 4: Calculate the limiting reactant

For 2 moles hexane we need 19 moles O2 to produce 12 moles CO2 and 14 moles H2O

Oxygen is the limiting reactant. It will completely be consumed ( 0.2231 moles). Hexane is in excess. There will react 2/19 * 0.2231 = 0.02348 moles

There will be porduced 12/19 * 0.2231 = 0.1409 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.1409 moles * 44.01 g/mol

Mass CO2 = 6.20 grams

We can produce 6.20 grams of CO2

Answer:

3–bromo–5–chloro–4–methylhexane.

Explanation:

To name the compound given in the question, the following must be observed:

1. Locate the longest continuous carbon chain. This gives the parent name of the compound. In this case, the longest chain is carbon 6 i.e Hexane.

2. Identify the substituents attached. In this case the substituents attached are:

a. Chloro i.e Cl.

b. Bromo ie Br.

c. Methyl i.e CH3.

3. Give the substituents the lowest possible count alphabetically. Bromo comes before Chloro alphabetically, so we shall consider bromo first. Their positions are given below:

Bromo i.e Br at carbon 3

Chloro i.e Cl is at carbon 5

Methyl i.e CH3 is at carbon 4

4. Combine the above to get the name of the compound.

Therefore, the name of the compound is:

3–bromo–5–chloro–4–methylhexane.

Answer:

ΔH°f C₂H₅O₂N(s)  = -537.2kJ

Explanation:

Based on the reaction:

4 C₂H₅O₂N(s) + 9O₂(g) → 8CO₂(g) + 10H₂O(l) + 2N₂(g)

ΔHrxn = ΔH°f products - ΔH°f reactants.

As:

ΔH°fO₂(g) = 0

ΔH°fCO₂(g) = -393.5kJ/mol

ΔH°fH₂O(l) = -285.8kJ/mol

ΔH°fN₂(g) = 0

The ΔHrxn is:

ΔHrxn = (8×-393.5kJ/mol + 10×-285.8kJ/mol) - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol

-6006kJ/mol - (4×ΔH°fC₂H₅O₂N(s)) = -3857kJ/mol

-4×ΔH°fC₂H₅O₂N(s) = 2149kJ/mol

ΔH°fC₂H₅O₂N(s) = 2149kJ/mol / -4

find the given attachment

Answer:

16.0473 g/L

Explanation:

0.300 M=

0.300 mol/L x 53.491 grams/mol = 16.0473 grams/L

The concentration of the 0.300M NH₄Cl solution in g/L will be equal to 16.04 g/L.

The concentration of the solution can be determined if we have the molecular formula of the compound and its molecular weight. We can easily determine the majority of a solution from the moles of solute and the volume of the solution.

The molarity of a solution can be evaluated from the number of moles of a solute per liter of a solution.

The Molarity can be determined from the formula mentioned below:

Molarity (M) = Moles of solute (n)/Solution's volume ( in L)

Given, the molarity of NH₄Cl solution = 0.300 M

We can also write it as 0.300 mol/L

It means 0.300 moles in one liter.

The molar mass of NH₄Cl  = 53.5 g/mol

Then the mass of 0.300 mol of NH₄Cl  = 0.300 ×53.5 = 16.04 g

Therefore, the concentration of NH₄Cl solution is 16.04g/L is equivalent to 0.300 M.

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Answer : The net ionic equation will be,

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

[tex]Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(aq)+BaSO_4(s)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]Ba^{2+}(aq)+2OH^-(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2H^+(aq)+2OH^{-}(aq)[/tex]

In this equation, [tex]H^+\text{ and }OH^-[/tex] are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]

Answer:

7.56 atm

Explanation:

Boyle's law states that the pressure and volume of a gas are proportional to each other

The formular for Boyle's law is

P1V1=P2V2

According to the question above, the values given are

P1=1.50 atm

P2= ?

V1=12.60 litres

V2= 2.50 litres

Let us make P2 the subject of formular

P2= P1V1/V2

P2= 1.50×12.60/2.50

P2= 18.9/2.50

P2= 7.56 atm

Hence when the volume of a gas is 2.50 litres then it's pressure is 7.56 atm

Answer:

0.077M is the concentration of the hydroxyl ion

Explanation:

If 7.35 cm3 of this base is take and diluted to 147 cm3, then what is the concentration of the hydroxyl ion?

Use the dilution equation:

M1V1 = M2V2

M1 * 147cm³ = 1.5 M * 7.35 cm³

M1 = 1.5 M * 7.35 cm³ / 147 cm³

M1 = 0.077 M

How many moles of hydroxyl ion are there in the 7.35 cm3?

1000 cm³ contains 1.5 mol OH- ions

7.35 cm³ contains : 7.35 cm³ / 1000 cm³ *1.5 mol

= 0.011025 mol

Answer: (e) The pressure in the container increases but does not double.

Explanation:

To solve this, we need to first remember our gas law, Boyle's law states that the pressure and volume of a gas have an inverse relationship. That is, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, increasing the volume in this case does not double the pressure owning to out gas law, but an increase in pressure would be noticed if temperature is constant

The pressure in the container increases but does not double.

At constant temperature and volume, the pressure of a given mass of gas is directly proportional to the number of moles of gas present.

Number of moles of He = 10 g/4 g/mol = 2.5 moles

Number of moles of Ne = 10 g/20 g/mol  = 0.5 moles

We can see that the number of moles only increases by 1/5 of its initial value therefore, the pressure in the container increases but does not double.

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Answer:

pH of the acid

Explanation:

Answer:

ΔH = - 44.0kJ

Explanation:

H2O(l)→H2O(g), ΔH =44.0kJ

In the reaction above, liquid water changes to gaseous water. This occurs through a process known as boiling. This process requires heat, hence the ΔH  is positive.

If he reaction is reversed, we have;

H2O(g)→H2O(l)

In this reaction, gaseous water changes to liquid water. This process is known as condensation. The water vapor loses heat in this reaction. Hence ΔH would be negative but still  have the same value.

Answer:

Explanation:

the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.

It is important that you don't confuse the words strong and weak with the terms concentrated and dilute.

As you will see below, the strength of an acid is related to the proportion of it which has reacted with water to produce ions. The concentration tells you about how much of the original acid is dissolved in the solution.

It is perfectly possible to have a concentrated solution of a weak acid, or a dilute solution of a strong acid.

Answer:

[Cd²⁺] = 2.459x10⁻⁶M

Kf = 9.96x10⁶

Explanation:

Solubility of CdC₂O₄ is 6.1x10⁻³M and ksp is 1.5x10⁻⁸

The ksp of CdC₂O₄ is:

CdC₂O₄(s) ⇄ Cd²⁺(aq) + C₂O₄²⁻(aq)

ksp = [Cd²⁺] [C₂O₄²⁻] = 1.5x10⁻⁸

As solubility is 6.1x10⁻³M, concentration of C₂O₄²⁻ ions is 6.1x10⁻³M. Replacing:

[Cd²⁺] = 1.5x10⁻⁸ / [6.1x10⁻³M]

All Cd²⁺ in solution is 6.1x10⁻³M and exist as Cd²⁺ and as Cd(NH₃)₄²⁺. That means concentration of Cd(NH₃)₄²⁺ is:

[Cd(NH₃)₄²⁺] + [Cd²⁺] = 6.1x10⁻³M

[Cd(NH₃)₄²⁺] = 6.1x10⁻³M - 2.459x10⁻⁶M = 6.098x10⁻³M

In the same way, the whole concentration of NH₃ in solution is 0.150M, as you have 4ₓ6.098x10⁻³M = 0.024M of NH₃ producing the complex, the concentration of the free NH₃ is:

[0.150M] = [NH₃] + 0.024M

The equilibrium of the complex formation is:

Cd²⁺ + 4 NH₃ → Cd(NH₃)₄²⁺

The kf, formation constant, is defined as:

Kf = [Cd(NH₃)₄²⁺] / [Cd²⁺] [NH₃]⁴

Replacing:

Kf = [6.098x10⁻³M] / [2.459x10⁻⁶M] [0.1256M]⁴

Answer:

The correct answer is 5.447 × 10⁻⁵ vacancies per atom.

Explanation:

Based on the given question, the at 750 degree C the number of vacancies or Nv is 2.8 × 10²⁴ m⁻³. The density of the metal is 5.60 g/cm³ or 5.60 × 10⁶ g/m³. The atomic weight of the metal given is 65.6 gram per mole. In order to determine the fraction of vacancies, the formula to be used is,  

Fv = Nv/N------ (i)  

Here Nv is the number of vacancies and N is the number of atomic sites per unit volume. To find N, the formula to be used is,  

N = NA×P/A, here NA is the Avogadro's number, which is equivalent to 6.022 × 10²³ atoms per mol, P is the density and A is the atomic weight. Now putting the values we get,  

N = 6.022 × 10²³ atoms/mol × 5.60 × 10⁶ g/m³ / 65.6 g/mol

N = 5.14073 × 10²⁸ atoms/m³

Now putting the values of Nv and N in the equation (i) we get,  

Fv = 2.8 × 10²⁴ m⁻³ / 5.14073 × 10²⁸ atoms/m^3

Fv = 5.44669 × 10⁻⁵ vacancies per atom or 5.447 × 10⁻⁵ vacancies/atom.  

Answer:

Explanation:

CHECK THE ATTACHMENT FOR THE COMPLETE QUESTION AND THE DETAILED EXPLANATION

NOTE:

Equatorial atoms are referred to atoms that are attached to carbons in the cyclohexane ring which is found at the equator of the ring.

Axial atoms are atoms that exist in a bond which is parallel to the axis of the ring in cyclohexane

Answer:

6.65dm³

Explanation:

Equation of reaction,

4C3H5(NO3)3(s) → 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)

From the equation of reaction, 4 moles of Nitroglycerin gave 29 moles of various gases.

Molar mass of nitroglycerin C₃H₅(NO₃)₃ = 908g

Since all the product of the reaction are in gaseous phase, let's assume that law of conservation of matter is held hence there's no loss in mass.

908g of C₃H₅(NO₃)₃ = 908g of products

2.70×10²g of C₃H₅(NO₃)₃ = 2.70×10²g of products

Number of moles = mass / molar mass

Molar mass of C₃H₅(NO₃)₃ = 908g/mol

Number of moles = 2.70×10² / 908

Number of moles = 0.297 moles

But 1 mole = 22.4dm³

0.297mole = x dm³

x = (0.297 × 22.4) / 1

x = 6.65dm³

The volume of gas that'll be produced when 2.70×10²g of C₃H₅(NO₃)₃ would be 6.65dm³

Answer:

It's ferric oxide Fe2O3

Explanation:

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Answer:

E) 1.85 M

Explanation:

M(C12H22O11) = 342.3 g/mol

22.5 g * 1mol/342.3 g = 0.0657 mol

35.5 mL = 0.0355 L

Molarity = mol solute/L solution = 0.0657 mol/0.0355L =1.85 mol/L = 1.85 M

The molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M

From the question,

We are to determine the molarity (that is, concentration) of the given sucrose solution

First, we will determine the number of moles present in the given mass of sucrose

Mass of sucrose = 22.5 g

Using the formula

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]

Molar mass of sucrose = 342.2965 g/mol

∴ Number of moles of sucrose present = [tex]\frac{22.5}{342.2965}[/tex]

Number of moles of sucrose present = 0.0657325 moles

Now, for the molarity (concentration) of the sucrose solution

From the formula

Number of moles = Concentration × Volume

Then,

[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]

From the question,

Volume = 35.5 mL = 0.0355 L

∴ [tex]Concentration = \frac{0.0657325}{0.0355}[/tex]

Concentration = 1.85 M

Hence, the molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M

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Answer:

Explanation:

The given chemical reaction is:

[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]

From above equation  [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.

Given that :

the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]

the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]

the volume of distilled water [tex]V_W = 15 \ mL[/tex]

The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]

Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]

Let take an integral look with the reaction between KI and AgNO₃; we have

[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]

At the end point; the moles of KI will definitely be equal to the moles of AgNO₃

So;

[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]

[tex]V_{AgNO_3} = 15 \ ml[/tex]

Thus; the volume of 0.1 M AgNO₃  needed to reach the end point is 15 mL

Answer:

(a) [tex]m=2.69m[/tex]

(b) [tex]x_{LiBr}=0.099[/tex]

(c) [tex]\% LiBr=18.9\%[/tex]

Explanation:

Hello,

In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:

(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:

[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]

Next, we compute the mass of the solution:

[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]

Then, the mass of the solvent (acetonitrile) in kg:

[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]

Finally, the molality:

[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]

(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):

[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]

Then, the mole fraction of lithium bromide:

[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]

(c) Finally, the mass percentage with the previously computed masses:

[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]

Regards.

Answer:

572. 3 g of NH3

Explanation:

Equation of the reaction: 3H2 + N2 ----> 2NH3

From the equation of reaction, 3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3.

Since N2 is in excess in the given reaction, H2 is the limiting reactant.

Molar mass of H2 = 2 g/mol

Molar mass of NH3 = 17 g/mol

Therefore 3 * 2 g of H2 reacts to produce 2 * 17 g of NH3

6 g of H2 produces 34 g of NH3

101 g of H2 will produce (34 * 101)/6 g of NH3 = 572.3 g of NH3

Therefore, 572.3 g of NH3 are produced

Answer:

572.33g of NH3.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Next, we shall determine the mass of the H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g.

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Finally, we can determine the mass of ammonia (NH3) produced by reacting 101g of H2 as follow:

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 101g of H2 will react to produce = ( 101 x 34) / 6 = 572.33g of NH3.

Therefore, 572.33g of NH3 is produced from the reaction.

Answer:I suppose it is mercury...

Explanation:

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