Magnesium and nitrogen react in a combination reaction to produce magnesium nitride:

3 Mg + N2 → Mg3N2

In a particular experiment, a 8.33-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.

Answer:

At equilibrium the rate of the forward reaction is equal to the rate of the backward reaction.

When the product of a reaction at equilibrium is increased the equilibrium will shift left or to the reactant side. As a result the excess product will get converted to reactant. This is in accordance to Le Chatelier's principle.

Le Chatelier's principle states that when a system is subjected to stress the equilibrium will shift in a direction to minimize effect of the stress.

Thus the products added to the system at equilibrium will make the equilibrium shift to the reactant side, the rate of the reverse or backward reaction will increase.

Explanation:

Hope This Helps Amigo!

Answer:

Here's what I get  

Explanation:

A Lewis structure shows the valence electrons surrounding the atoms.

Your structure has two problems:

Here's one way to draw a Lewis structure.

1. Draw a trial structure

Make F and O terminal atoms and give each one an octet (Fig. 1).

2. Count the valence electrons in the trial structure

5 BP + 15 LP  = 10 + 30 = 40 electrons

3. Check the number of valence electrons available  

1 S =   1 × 6 =  6 electrons

1 O =   1 × 6 =   6

4 F  = 4 × 7 = 28                    

     TOTAL = 40 electrons

The trial structure has the correct number of electrons.

4. Determine the formal charge on each atom.

To get the formal charges, we cut the covalent bonds in half.

Each atom gets the electrons on its side of the cut.

Formal charge = valence electrons in isolated atom - electrons on bonded atom

FC = VE - BE  

(a) On S

VE = 6

BE = 5 bonding electrons = 5

FC = 6 - 5 = +1

(b) On O:

VE = 6

BE = 3 LP(six electrons) + 1 bonding electron  = 7

FC = 6 - 7 = -1

(c) On F:

VE = 6

BE = 3 lone pairs(6 electrons) + 1 bonding electron = 6 + 1 =7

FC = 7 - 7 = 0

5. Minimize the formal charges

We must rearrange the valence electrons so that S gets one more and O gets one fewer.

Move a lone pair from the O to make an S=O double bond (Fig. 2).

6. Recalculate the formal charges

(a) On S

VE = 6

BE =  (3 bonding electrons) = 6

FC = 6 - 6 = 0

(b) On O:

VE = 6

BE = 2 LP(four electrons) + 2 bonding electrons = 6  

FC = 6 - 6 = 0

Fig. 2 shows the Lewis structure in which all atoms have a formal charge of  zero.

The formal charge of the atoms can be concluded zero with the bond formation between the sulfur and oxygen atom.

The lewis structure can be defined as the dot structure of the valence bond with the bonded atoms. The formal charge can be calculated with the difference in the valence electrons and the bonding electrons.

The formal charge of an atom can be zero when the valence electrons and the bonding electrons are equal. In the structure of [tex]\rm OSF_4[/tex], the formal charge has been assigned zero with the bond formation resulting in the valence electrons and bonding electrons being equal.

The lewis structure with the central S atom has been attached.

For more information about lewis structure, refer to the link:

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Answer:

$11.81

Explanation:

27 lb cost $16

27/16=$1.69 per pound

$1.69*7=$11.81 for 7 lbs

Answer:

THE VOLUME OF NaOH NEEDED TO BE ADDED TO CITRIC ACID TO REACH THE EQUIVALENT POINT IS 4.725 L

Explanation:

The titration is between citric acid (H3C6H507) and NaOH

mass of citric acid = 0.306 g

Volume of citric acid = 250 mL = 250 /1000 = 0.25 L

Concentration of NaOH = 0.1000 M

Volume = unknown

First calculate the molar mass of citric acid

( 1 * 3 + 12* 6 + 1*5 + 16*7) = (4 + 72 + 5 + 112) = 193 g/mol

Since,

Concentration in moles/dm3 = concentration in g/dm3 / RMM

So the molarity of citric acid is:

Molarity = 0.306g / 0.25dm3  / Rmm

Molarity = 1.224g/dm3 / 193 g/mol

Molarity = 0.0063 M

Equation for the reaction is:

C3H5O(COOH)3 + 3NaOH → Na3C3H5O(COO)3 + 3H2O

Using the formula:

CaVa / CbVb = na/ nb

Ca = 0.0063 M

Cb = 0.1000 M

Va = 0.25 L

Vb = unknown

na = 1

nb = 3

Vb = Ca Va nb/ Cb na

Vb = 0.0063 * 0.25 * 3 / 0.1000 * 1

Vb = 0.4725 / 0.1000

Vb = 4.725 L

The volume of NaOH needed to reach the equivalent point is therefore 4.725 L

Answer:

Na₂SO₄

Explanation:

Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.

Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:

Na₂SO₄

Answer:

4.41

Explanation:

Step 1: Write the balanced equation

CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

Step 2: Calculate the respective concentrations

[tex][CO]_i = \frac{0.500mol}{5.00L} = 0.100M[/tex]

[tex][H_2]_i = \frac{1.500mol}{5.00L} = 0.300M[/tex]

[tex][H_2O]_{eq} = \frac{0.198mol}{5.00L} = 0.0396M[/tex]

Step 3: Make an ICE chart

        CO(g) + 3 H₂(g) = CH₄(g) + H₂O(g)

I       0.100      0.300        0            0

C         -x           -3x          +x          +x

E    0.100-x    0.300-3x     x            x

Step 4: Find the value of x

Since the concentration at equilibrium of water is 0.0396 M, x = 0.0396

Step 5: Find the concentrations at equilibrium

[CO] = 0.100-x = 0.100-0.0396 = 0.060 M

[H₂] = 0.300-3x = 0.300-3(0.0396) = 0.181 M

[CH₄] = x = 0.0396 M

[H₂O] = x = 0.0396 M

Step 6: Calculate the equilibrium constant (Kc)

[tex]Kc = \frac{[CH_4] \times [H_2O] }{[CO] \times [H_2]^{3} } = \frac{0.0396 \times 0.0396 }{0.060 \times 0.181^{3} } = 4.41[/tex]

Answer:

[tex]M=0.213M[/tex]

Explanation:

Hello,

In this case, for each nitrate-based salt, we compute the nitrate moles as shown below:

[tex]n_{NO_3^-}=20.00gCa(NO_3)_2*\frac{1molCa(NO_3)_2}{164.088 gCa(NO_3)_2} *\frac{2molNO_3^-}{1molCa(NO_3)_2} =0.244molNO_3^-[/tex]

[tex]n_{NO_3^-}=10.00gNaNO_3*\frac{1molNaNO_3}{84.9947 gNaNO_3} *\frac{1molNO_3^-}{1molNaNO_3} =0.118molNO_3^-[/tex]

[tex]n_{NO_3^-}=50.00gAl(NO_3)_3*\frac{1molAl(NO_3)_3}{212.996gAl(NO_3)_3} *\frac{3molNO_3^-}{1molAl(NO_3)_3} =0.704molNO_3^-[/tex]

We notice calcium nitrate has two moles of nitrate ion, sodium nitrate has one and aluminium nitrate has three. Hence we add the moles to obtain the total moles nitrate ion:

[tex]n_{NO_3^-}^{Tot}=0.244+0.118+0.704=1.066molNO_3^-[/tex]

Finally, we compute the molarity:

[tex]M=\frac{1.066molNO_3^-}{5.00L} \\\\M=0.213M[/tex]

Regards.

Answer:

1.7 moles of ethanol would be needed.

Explanation:

* Calculate the moles of ethanol needed to produce 1.70mol of water. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.

First off, we have to state the equation for the reaction.

So we know that;

ethanol + oxygen → acetic acid + water

This leads us to;

C2H5OH + O2 → CH3COOH + H2O

1                    1         1                    1

To obtain the moles of ethanol needed to produce 1.70mol of water, we look at the stoichiometry of the reaction above.

1 mol of ethanol produces 1 mole of water

x mol of thanol would produce 1.7 mol of water

Thus we have;

1 = 1

x = 1.7

x = 1.7 moles of ethanol would be needed.

y=3x + 4

Explanation:

Where 3 is the slope and 4 is the y-intercept

Answer:

The correct answer to the question is Option E (Strongly retained analytes will give broad peaks).

Explanation:

The other options are true because:

A. Initial temp = 50 °C

   Final temp =  270 °C

Differences in temp = 270 - 50 = 220°C

Rate =  10 °C/minute.

So, at 10 °C/minute,

total of 220°C /10 °C = number of minutes required to reach the final temp.

220/10 = 22 minutes

B. A column has a minimum and maximum use temperature. Solutes that are already retained would remain stationary while temperatures are low. This would only change if there is an increase in temperature. Heat transfers more energy to the liquid which would make the solute interact with the column phase.

C. Weakly retained solutes may contain larger molecules, will separate by absorbing into the solvent early in separation making the mobile phase separates out into its components on the stationary phase.

D. Retained solute's vapor pressure is higher at higher temperatures making it possible for particle to escape more from the solute when the temperature is high than when it is low.

Answer:

66.0 atm

Explanation:

We can calculate the osmotic pressure (π) using the following expression.

[tex]\pi = i \times M \times R \times T[/tex]

where,

Step 1: Calculate i

Sodium sulfate completely dissociates according to the following equation.

Na₂SO₄ ⇒ 2 Na⁺ + SO₄²⁻

Since it produces 3 ions, i = 3.

Step 2: Calculate M

We can calculate the molarity of Na₂SO₄ using the following expression.

[tex]M = \frac{mass\ of\ solute }{molar\ mass\ of\ solute\ \times liters\ of\ solution} = \frac{65.0g}{142.04g/mol \times 0.500L} =0.915M[/tex]

Step 3: Calculate T

We will use the following expression.

K = °C + 273.15

K = 20°C + 273.15 = 293 K

Step 4: Calculate π

[tex]\pi = 3 \times 0.915M \times \frac{0.08206atm.L}{mol.K} \times 293K =66.0 atm[/tex]

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:

An orbital is a region in space where there is a high probability of finding an electron.

Explanation:

The orbital is a concept that developed in quantum mechanics. Recall that Neils Bohr postulated that the electron occupied stationary states which he called energy levels. Electrons emit radiation when the move from a higher to a lower energy level. Similarly, energy is absorbed by an electron to move from a lower to a higher orbit.

This idea was upturned by the Heisenberg uncertainty principle. This principle state that the momentum and position of a particle can not be simultaneously measured with precision.

Instead of defining a 'fixed position' for the electron, we define a region in space where there is a possibility of finding an electron with a certain amount of energy. This orbital is identified by a set of quantum numbers.

Answer:

three - dimensional space that shows the probability where an electron is most likely to be found

Answer:

−153.1 J / (K mol)

Explanation:

Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)

Hg(liq) + Cl2(g) → HgCl2(s)

Given that;

The standard molar entropies of the species at that temperature are:

Sºm (Hg,liq) = 76.02 J / (K mol) ;

Sºm (Cl2,g) = 223.07 J / (K mol) ;

Sºm (HgCl2,s) = 146.0 J / (K mol)

The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]

= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]

= -153.09 J / (K mol)

= or -153.1 J / (K mol)

Hence the answer is  −153.1 J / (K mol)

Answer:

1.97kJ of energy are released.

Explanation:

The dissolution of LiOH in water is:

LiOH(s) → Li⁺(aq) + OH⁻(aq) ΔH = -23.6kJ

That means, when 1 mole of LiOH is dissolved, there are released (Because of the - in the enthalpy) 23.6kJ

2.0 g of LiOH (Molar mass: 23.95g/mol) are:

2.0g LiOH × (1 mol / 23.95g) =0.0835 moles of LiOH.

As 1 mole of LiOH release 23.6kJ, 0.0835moles release:

0.0835moles × (-23.6kJ / 1mole) = 1.97kJ of energy are released

Answer:

U = -45.557kj

Explanation:

Before we can calculate the totally internal energy change in kilojoules firstly we need to calculate W

U=q + w .

We know that

w = PΔ V

where P is the pressure of

and V is the volume

then we can calculate the work

w = 35 atm * ( 8.20L - 2.21L)

W=35atm* 5.99L

W=209.65atmJ

But 1 atm = 101.325J

then ,

w = 209.65* 101.325 J = 21242.79 J

let us convert it to Kj

But we know that 1kJ = 10^3 J .

Then w = 21.243 kJ .

Then we can now calculate the internal energy as

U = 21.243- 66.8 kJ = -45.557kj

But we know that heat was released. Theeefore, the total internal energy change was -45.557kj

Answer:solid,liquid,gas,plasma

Explanation:

This question seems incomplete. I believe the full question is as followed:

Identify the state(s) of matter that each property describes.

1.) takes the shape of its container:

O gas

O liquid

O solid

2.) fills all available space:

O gas

O liquid

O solid

3.) maintains its shape:

O gas

O liquid

O solid

4.) can be poured:

O gas

O liquid

O solid

5.) is compressible:

O gas

O liquid

O solid

6.) has a fixed volume:

O gas

O liquid

O solid

The answers to the 1st are gas and liquid.

The answer to the 2nd is gas.

The answer to the 3rd is solid.

The answer to the 4th is liquid.

The answer to the 5th is gas.

The answers to the 6th are liquid and solid.

Answer:

[tex]0.745~M[/tex]

Explanation:

In this case, we have a dilution problem. So, we have to use the dilution equation:

[tex]C_1*V_1=C_2*V_2[/tex]

Now, we have to identify the variables:

[tex]C_1~=~14.9~M[/tex]

[tex]V_1~=~25~mL[/tex]

[tex]C_2~=~?[/tex]

[tex]V_2~=~0.5~L[/tex]

Now, we have different units for the volume, so we have to do the conversion:

[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]

Now we can plug the values into the equation:

[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]

I hope it helps!

Answer:

The excess reactant is N2H4 and the leftover mass is 10.17g.

Explanation:

Step 1:

The balanced equation for the reaction.

N2O4 + 2N2H4 —> 3N2 + 4H2O

Step 2

Determination of the masses of N2O4 and N2H4 that reacted from the balanced equation:

Molar mass of N2O4 = 92.02 g/mol

Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02g

Molar mass of N2H4 = 32.05 g/mol

Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g

From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.

Step 3:

Determination of the excess reactant. This is illustrated below:

From the balanced equation above, 92.02g of N2O4 reacted with 64.1g of N2H4.

Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.

From the calculations made above, we can see that only 34.83g of N2H4 reacted out of 45g that was given. Therefore, N2H4 is the excess reactant.

Step 4:

Determination of the mass of excess reactant that is leftover.

The excess reactant is N2H4 and the leftover mass can be obtained as follow:

Mass of N2H4 given = 45g

Mass of N2H4 that reacted = 34.83g

Leftover mass of N2H4 =..?

Leftover mass of N2H4 = (Mass of N2H4 given) – (Mass of N2H4 that reacted)

Leftover mass of N2H4 = 45 – 34.83

Leftover mass of N2H4 = = 10.17g.

Answer:

No, the sample needs to be insoluble or sparingly soluble at room temperature so that the maximum amount of purified crystals form at room temperature and in the ice bath.

Explanation:

For a solvent to be adequate it must completely dissolve the substance to be purified when it is hot, that is, at boiling temperature only. It should be practically insoluble when the solvent is cold or at room temperature. This must occur in this way since impurities must be removed by hot filtering or dissolved in the mother liquor.

According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).

The ideal gas law relates the pressure, volume, number of moles and temperature of an ideal gas.

Let's consider the equation of the ideal gas law.

P . V = n . R .T

V = n . R . T / P

As we can see, there is a direct relationship between the volume and the temperature. Thus, if the temperature doubles, the volume will double as well.

According to the ideal gas law, when the temperature of a gas doubles, its volume doubles as well (Option B).

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Answer:

The correct answer is 179.94 g/mol.

Explanation:

Based on the given question, the boiling point of diethyl ether us 34.500 degree C at 1 atm pressure. The boiling point of the solution given is 35.100 degree C. The Kb of diethyl ether given is 2.02 degree C/m. The weight of the compound given is 14.94 grams, the weight of the solvent (diethyl ether) is 279.5 grams.

The molecular weight of the compound can be determined by using the formula,

deltaTb = Kb * molality

Tb-To = Kb * molality

Tb-To = Kb*wt/mol.wt*1000/w (solvent)

35.100 - 34.500 = 2.02 * 14.94 / mol. wt * 1000 g / 279.5 g

0.6 = 2.02 * 53.45/ mol.wt

mol. wt = 2.02*53.45/0.6

mol. wt = 179.94 g/mol

Hence, the molecular weight of the compound is 179.94 gram per mol.

Answer: D. The Government could take away research funds if ethical standards are not met

The government enforce ethical standards of scientific experiments

B. The government could take away research funds if ethical

standards are not met.

Ethical standards are a set of principles established by the founders of the organization to communicate its underlying moral values. This code provides a framework that can be used as a reference for decision making processes.

Politicians and bureaucrats control scientific research and research outcomes by selectively funding projects that look for potential disasters, ideally global disasters.

This analysis focuses on whether and how the statements in these eight codes specify core moral norms (Autonomy, Beneficence, Non-Maleficence, and Justice), core behavioral norms (Veracity, Privacy, Confidentiality, and Fidelity), and other norms that are empirically derived from the code statements.

To learn more about ethical standards, refer

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Answer:

Explanation:

it is based on le chatliers principles

the left side of reaction you have 4 moles , where as at the right hand side you have 2 moles,,,,

so when you increase the pressure the reaction will shift towards the lower moles producing reaction that is reaction move towards forward in you case.

Answer:

B. Mixing a solute and a solvent

Explanation:

Hello,

In this case, solutions are defined as liquid homogeneous mixtures formed when two substances having affinity are mixed. It is important to notice that the two substances are known as solute, which is added to other substance that is the solvent. Therefore, answer is B. Mixing a solute and a solvent.

Notice that when two insoluble substances are mixed no solution is formed. Furthermore, if two solutes together or a solute and a precipitate are mixed, no liquid homogeneous solution is formed, as commonly solutes are solid, nevertheless, when liquid, one should have to act as the solvent.

Best regards.

Answer:

B. Mixing a solute and a solvent

Explanation:

ap3x

Answer:

[tex]pH=10.5[/tex]

Explanation:

Hello,

In this case, the dissociation of the given weak acid is:

[tex]HA\rightleftharpoons H^++A^-[/tex]

Therefore, the law of mass action for it turns out:

[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]

That in terms of the change [tex]x[/tex] due to the reaction extent is:

[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]

Thus, by solving with the quadratic equation or solver, we obtain:

[tex]x=31.6x10^{-12}M[/tex]

Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:

[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]

Regards.

Answer:

-1648.4 kJ/mol

Explanation:

Based on Hess's law:

ΔHr = ∑n×ΔH°f(products) - ∑n×ΔH°f(reactants)

In the reaction:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

ΔHr = 2 ΔH°f {Fe₂O₃} - (4ΔH°f {Fe(s)} + 3ΔH°f{O₂(g)}

As:

ΔH°f {Fe₂O₃} = -824.2kJ/mol

ΔH°f {Fe(s)} = 0.0kJ/mol

ΔH°f{O₂(g)} = 0.0kJ/mol.

Thus,

ΔHr = 2 ₓ -824.2kJ/mol =

Answer:

-1648.4 kJ

Explanation:

The product has the only nonzero heat of formation, so it is the only value needed to calculate the enthalpy of this reaction. Normally, you would want to express the enthalpy of a reaction with respect to one mole of a chemical species, whether it is a reactant or product. However, since the balanced chemical equation contains only coefficients greater than 1, you should consider how the enthalpy relates to one mole of each substance according to the coefficients. In other words,  − 1648.4  kJ  of heat is released when 4  mol  of  Fe  reacts with 3  mol  of  O2  to produce 2  mol  of  Fe2O3 .