The aluminum in a package containing 75 ft2 of kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3 . What is the approximate thickness of the aluminum foil in millimeters?(1 ounce = 28.4g)

Answers

Answer 1

Answer:

18130 mm

Explanation:

Now we have a lot of unit conversions to do in order to correctly answer this question. We shall do these conversions gradually.

First we convert the weight in ounce to grams.

If 1 ounce = 28.4g

12 ounces = 12×28.4 = 340.8 g

Next we convert the area of aluminum from ft2 to m2

1ft2= 0.0929 m2

75 ft2= 75 × 0.0929= 6.9675m2

Now density of aluminum= 2.70 gcm-3

Density= mass/volume

But volume= area× thickness

Density= mass/ area × thickness

Density × area × thickness= mass

Thickness= mass/ density × area

Thickness= 340.8g / 2.70gcm-3 × 6.9675m2

Thickness= 340.8/18.8

Thickness= 18.13 m

Since 1000 milimeters make 1 metre

Thickness= 18130 mm


Related Questions

Click on the Delta H changes sign whan a process is reversed button within the activity and analyze the relationship between the two reactions that are displayed. The reaction that was on the screen when you started and its derivative demonstrate that the reaction enthalpy, ΔH, changes sign when a process is reversed. Consider the reaction H2O(l)→H2O(g), ΔH =44.0kJ What will ΔH be for the reaction if it is reversed?

Answers

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.

Describe why some acids are strong while other acids are weak

Answers

Answer:

I hope this help you. Mark me as brainliest and rate please

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.

When you turn on the air conditioner during a hot summer day the cooler air will sink to the floor, while warmer air rises to the
ceiling
Which type of heat transfer is this an example of?
(A) conduction
(B) convection
(C) radiation
(D)
kinetic

Answers

It’s B convection it makes air rise and sink

Give the IUPAC name for the following compound

Answers

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.

Isomer such as acetic acid and methyl formate have

Answers

Answer:

C

Explanation:

This is the definition of an isomer.

I think it’s C hope it helps

Glycine, C2H5O2N, is important for biological energy. The combustion reaction of glycine is given by the equation 4C2H5O2N(s) + 9O2(g) → 8CO2(g) + 10H2O(l) + 2N2(g) ΔH°rxn = –3857 kJ/mol Given that ΔH°f[CO2(g)] = –393.5 kJ/mol and ΔH°f[H2O(l)] = –285.8 kJ/mol, calculate the enthalpy of formation of glycine.

Answers

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

ΔH°f C₂H₅O₂N(s)  = -537.2kJ

For some hypothetical metal the equilibrium number of vacancies at 750°C is 2.8 × 1024 m−3. If the density and atomic weight of this metal are 5.60 g/cm3 and 65.6 g/mol, respectively, calculate the fraction of vacancies for this metal at 750°C.

Answers

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.  

The solubility of cadmium oxalate, , in 0.150 M ammonia is mol/L. What is the oxalate ion concentration in the saturated solution? If the solubility product constant for cadmium oxalate is , what must be the cadmium ion concentration in the solution? Now, calculate the formation constant for the complex ion

Answers

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]

[Cd²⁺] = 2.459x10⁻⁶M

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

[Cd(NH₃)₄²⁺] = 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

0.1256M = [NH₃]

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

Kf = 9.96x10⁶

Trans-4-hexen-3-ol can be synthesized starting from acetaldehyde. One of the key reagents is ethyl grignard.
1. Synthesize ethyl grignard from acetaldehyde in the steps below using the reagents provided.
2. Synthesize (trans)-4-hexen-3-ol from acetaldehyde.

Answers

find the given attachment

iron oxide + oxygen equals to ?

Answers

Answer:

It's ferric oxide Fe2O3

Explanation:

I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me plz...

Iron+ oxygen= Fe+ 3O2 hopefully this will help!

1. ______The force that keeps the nucleons bound inside the nucleus of an atom
A. Strong electrostatic force
B. Strong nuclear force
C. Strong centripetal force
D. Gravitational attraction

2._____The amount of energy needed to split the nucleus into individual protons and neutrons
A. Nuclide transfer energy
B. Nuclear binding energy
C. Mass energy equivalence
D. Nuclear energy
3._______ The difference between the mass of the nucleons and the mass of an Atom
A. Mass of nucleus
B. Mass defect
C. Atomic mass
D. Isotopic mass

Answers

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.

What is strong nuclear force ?

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.

To learn more about the strong nuclear force, follow the link;

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#SPJ2

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produced 39.61 g CO2 and 9.01 g H2O. The molar mass of equilin is 268.34 g/mol. Find its molecular formula.

Answers

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:

Step 1: Converting the given masses into moles.

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]

Step 2: Calculating the mole ratio of the given elements.

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]

Step 3: Taking the mole ratio as their subscripts.

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].

A base has a molarity of 1.5 M with respect to the hydroxyl ion (OH-) concentration. If 7.35 cm³ of this base is taken and diluted to 147 cm³, then what is the concentration of the hydroxyl ion. How many moles of hydroxyl ion are there in the 7.35 cm³? In the 147 cm³?

Answers

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

0.077M is the concentration of the hydroxyl ion

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 correct to 2 significant digits = 0.011 mol OH- ions.

8) What is the molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22011) in 35.5 mL of solution?
A) 3.52 M
B) 1.85 x 10-2M
C) 0.104 M
D) 0.0657 M
E) 1.85 M

Answers

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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what is the name of the liquid in the clinical thermometer​

Answers

Answer:I suppose it is mercury...

Explanation:

I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thnk me...

The answer is mercury

Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 4.3 g of hexane is mixed with 7.14 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to significant digits.

Answers

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

The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Determine how much heat (in kJ) is produced by the decomposition of 1.71 mol of H2O2 under standard conditions.

Answers

Answer:

[tex]Q=-361.56kJ[/tex]

Explanation:

Hello,

In this case, the decomposition of hydrogen peroxide is given by:

[tex]2H_2O_2\rightarrow 2H_2O+O_2[/tex]

Which occurs in gaseous phase, therefore the enthalpy of reaction is:

[tex]\Delta _rH=2\Delta _fH_{H_2O}-2\Delta _fH_{H_2O_2}[/tex]

Oxygen is not included as it is a pure element. The enthalpies of formation for both hydrogen peroxide and water are -136.11 and -241.83 kJ/mol respectively, so we compute the enthalpy of reaction:

[tex]\Delta _rH=2(-241.83kJ/mol)-2(-136.11kJ/mol)=-211.44kJ/mol[/tex]

Then, the total heat that is released for 1.71 mol of hydrogen peroxide is:

[tex]Q=n*\Delta _rH=1.71mol*-211.44kJ/mol\\\\Q=-361.56kJ[/tex]

Whose sign means a released heat.

Regards.

tertbutylamine and ammonia. Which is more basic

Answers

Answer:

ammonia

Explanation:

I WILL GIVE BRAINLIEST

Answers

Molarity= no. of molecules of solute /1 liter
one moles of sodium hydroxide =49 gm of sodium hydroxide
So we can say that if we want to prepare 1 molar NaOH solution then we need 40 gm NaOH dissolve in one liter of water so it can become one 1 molar NaOH solution.

What is a good title for this chart?

Answers

Answer:

pH of the acid

Explanation:

If you have 101 g of hydrogen gas (H2) and excess amount of nitrogen gas (N2), how many grams of ammonia gas (NH3) can you make?

Answers

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.

If a gas occupies 12.60 liters at a pressure of 1.50 atm, what will its pressure at a volume of 2.50 liters?

Answers

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

Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?

Answers

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

Consider 10.0 g of helium gas (He) in a rigid steel container. If you add 10.0 g of neon gas (Ne) to this container, which of the following best describes what happens? (Assume the temperature is constant.)
a) The pressure in the container doubles.
b) The pressure in the container more than doubles.
c) The volume of the container doubles.
d) The volume of the container more than doubles.
e) The pressure in the container increases but does not double.

Answers

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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A mixture of compounds containing diethylamine, phenol, ammonia, and acetic acid is separated using liquid-liquid extraction as follows: Step 1: Concentrated HCl is added followed by draining the aqueous layer. Step 2: Dilute NaOH is added to the organic layer followed by draining the aqueous layer. Step 3: Concentrated NaOH is added to the organic layer followed by draining the aqueous layer. Which compound would you expect to be extracted into the aqueous layer after the addition of dilute HCl, step 1? Group of answer choices

Answers

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]

       

Acetonitrile, CH3CN, is a polar organic solvent that dissolves many solutes, including many salts. The density of a 1.80 M acetonitrile solution of LiBr is 0.826 g/mL. Calculate the concentration of the solution in units of (a) molality; (b) mole fraction of LiBr; (c) mass percentage of CH3CN.

Answers

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.

What type of bond will be formed for atoms that have a +1 or -1 charge?

Answers

covalent bonding. example lithium bond with fluorine since lithium has a valence charge of +1 and fluorine has a valence charge of +7. they will bond together to give u a stable full electron

g The solution you created in this simulation was a 0.300M NH4Cl solution. The lab also stated that, in g/L, this concentration was 16.0 g/L. Show the calculations that prove that to be true.

Answers

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.

What is the molarity?

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.

Learn more about molarity, here:

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The three‑dimensional structure of a generic molecule is given. Identify the axial and equatorial atoms in the three‑dimensional structure. What is the shape of this molecule?

Answers

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

Given the information you now know, what is the effect of hyperventilation on blood pH?pH? During hyperventilation, the rapid in the blood CO2CO2 concentration shifts the equilibrium to the which the concentration of H+,H+, thereby the blood pH.

Answers

Answer:

When hypercapnia processes occur, where the concentration of carbon dioxide gas increases in the blood, the protonization of the blood increases, this means that the H + ions increase in concentration, thus generating metabolic acidosis.

This metabolic acidosis is regulated by various systems, but the respiratory system collaborates by generating hyperventilation, to increase blood oxygen pressures, decrease CO2 emissions, and indirectly decrease acidity.

Explanation:

This method of regulating the body is crucial, since the proteins in our body will not be altered if they do not happen.

The enzymes, the red globules, and many more fundamental things for life ARE PROTEINS, that in front of acidic media these modify their structure by denaturing themselves and ceasing to fulfill their functions. This is the reason why it seeks to neutralize the blood pH when it comes to an increase in CO2.

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