Solve for x:LaTeX: \frac{x^2}{0.160-x}\:=\:0.058

Answer:

Required compressor work W=8560.44  KJ/Kmol

Explanation:

Given that

Initial pressure = 1 bar

[tex]P_1=1\ bar[/tex]

Final pressure = 40 bar

[tex]P_2=40\ bar[/tex]

Process is isothermal and T=280 K

We know that ,work done in isothermal process given as

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

given taht gas is ideal so

P V =m R T

[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]

R for ethylene

R=0.296 KJ/kg.K

Now by putting the values

[tex]W=mRT\ln \dfrac{P_1}{P_2}[/tex]

[tex]W=1\times 0.296\times 280 \ln \dfrac{1}{40}[/tex]

W= -305.73 KJ/kg

Negative sign indicates that work done on the system.

Required compressor work W=305.73 KJ/kg

Molar mass of ethylene M= 28 Kg/Kmol

So W= 305.73 x 28  KJ/Kmol

W=8560.44  KJ/Kmol

Required compressor work W=8560.44  KJ/Kmol

Answer:

9.834 moles Cf.

Explanation:

The number of moles of a substance is an easy way to represents its amount. Avogadro has determined that the total amount in 1 mol is equal to 6.02x10²³(Avgadros' number), so 1 mol has 6.02x10²³ atoms, molecules, ions, or what we are measuring. So:

1 mol of Cf -------------------- 6.02x10²³ atoms

x -------------------- 5.92x10²⁴

By a simple direct three rule:

6.02x10²³x = 5.92x10²⁴

x = 5.92x10²⁴/6.02x10²³

x = 9.834 moles Cf

Answer:

c. None of the above

Explanation:

The equation is:

H₃O⁺ + CH₃COO⁻ ⇒ CH₃COOH + H₂O

An acid is a species that donates a proton. Thus, the acid is the species on the reactants' side that has one proton more than it's conjugate base species on the products's side.

The available options are:

a. CH₃COOH

b. H₂O

c. None of the above.

Both CH₃COOH and H₂O are on the right-hand side of the equation, so the answer must be c. The answer to the problem is H₃O⁺, since it has lost a proton and become H₂O.

Answer:

b metal denting

Explanation:

Answer:

Choice b) Metal denting.

Explanation:

What's the difference between chemical and physical changes? In a chemical change, new substances are created. However, the types of substances stay the same in a physical change.

When silver tarnishes, it reacts with a substance that contains sulfur (e.g., [tex]\rm SO_3[/tex]) to produce silver sulfide, which is a substance different from the other two.

Before the change:

After the change:

A new substance is created in this change. As a result, this change is chemical.

When metal is dented, atoms in the metal slide past each other. The shape of the metal might change, however the substance will still be the same the metal.

Before the change:

After the change:

No substance is created. As a result, this change is physical.

When iron rusts in the air, it reacts with oxygen to produce oxides of iron.

Before the change:

After the change:

Oxides of iron are created. As a result, this change is chemical.

When gasoline burns in the air, it reacts with oxygen to produce water and carbon dioxide (or carbon monoxide, or both.)

Before the change:

After the change:

Water and carbon dioxide/monoxide are created. As a result, this change is chemical.

Answer:

[tex]578.2\times 10^{-6}[/tex]

Explanation:

Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.

For example, [tex]1000^2[/tex] is to be written as [tex]10^6[/tex] in engineering notation.

The given number:

0.00057824 can be written as [tex]578.24\times 10^{-6}[/tex]

Answer upto 4 significant digits = [tex]578.2\times 10^{-6}[/tex]

Answer:

B

Explanation:

Limiting reagent is the one which is present in small molar amount. It got exhausted at the end of the reaction and the formation of the products is governed by it.

The given reaction:

2A + 3B → D+E

2 moles of A react with 3 moles of B

1 mole of A react with 3 / 2 moles of B

Given moles of A = 3 moles

Moles of B = 4 moles

So,

3 moles of A will react with 1.5 * 3 moles of B

Moles of B = 4.5

Since, only we have 4 moles of B. So, B is the limiting reagent.

Answer:

Explanation:

The air 9% mole% methane have an average molecular weight of:

9%×16,04g/mol + 91%×29g/mol = 27,8g/mol

And a flow of 700000g/h÷27,8g/mol = 25180 mol/h

In the reactor where methane solution and air are mixed:

In = Out

Air balance:

91% air×25180 mol/h + 100% air×X = 95%air×(X+25180)

Where X is the flow rate of air in mol/h = 20144 mol air/h

The air in the product gas is

95%×(20144 + 25180) mol/h = 43058 mol air× 21%O₂ = 9042 mol O₂ ×32g/mol = 289 kg O₂

43058 mol air×29g/mol 1249 kg air

Percent of oxygen is: [tex]\frac{289kg}{1249 kg}[/tex] =0,231 kg O₂/ kg air

I hope it helps!

Answer:

A) Mass flow rate of air = [tex]22.982[/tex] kmol/hr

B) percentage by mass of oxygen in the product gas = [tex]$22.52 \%$[/tex]

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have [tex]0.09[/tex] mole of methane(CH4) and the remaining will be the air which is [tex]$(100 \%-9 \%)=91 \%$=0.91[/tex]

We are given the average molecular weight of air = [tex]29[/tex] g/mol Thus; Average molar mass of air and methane mixture is;

M-air- [tex]$=(0.09 \times 16)+(0.91 \times 29)$[/tex]

[tex]=$27.83 \mathrm{~g} / \mathrm{mol}$[/tex]

Mass fraction of oxygen in the product gas= [tex]=43.016 \mathrm{kmol} / \mathrm{h} \times(0.21 \mathrm{molO} 2 / 1 \mathrm{~mol}$ air $) \times(32 \mathrm{~kg}$ oxygen/1kmol oxygen $) \times(1 / 1283.596 \mathrm{~kg} / \mathrm{h})=0.2252$[/tex]

Learn more about bonds refer:

Answer:

Re=309926.13

Explanation:

density=92.8lbm/ft3*(0.45kg/1lbm)*(1ft3/0.028m3)=1491.43kg/m3

viscosity=4.1cP*((1*10-3kg/m*s)/1cP)=0.0041kg/m*s

velocity=237ft/min*(1min/60s)*(0.3048m/1ft)=1.2m/s

diameter=28inch*(0.0254m/1inch)=0.71m

Re=(density*velocity*diameter)/viscosity=(1491.43kg/m3*1.2m/s*0.71m)/0.0041kg/m*s

Re=309926.13

Answer: The mass of carbon dioxide produced is 352 grams

Explanation:

We are given:

Moles of ethanol = 4 mol

For the given chemical equation:

[tex]C_2H_5OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]

By Stoichiometry of the reaction:

1 mole of ethanol produces 2 moles of carbon dioxide

So, 4 moles of ethanol will produce = [tex]\frac{2}{1}\times 4=8mol[/tex] of carbon dioxide

To calculate the mass of carbon dioxide, we use the equation:

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

Moles of carbon dioxide = 8 moles

Molar mass of carbon dioxide = 44 g/mol

Putting values in above equation:

[tex]8mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=(8mol\times 44g/mol)=352g[/tex]

Hence, the mass of carbon dioxide produced is 352 grams

Final answer:

The combustion of 4 moles of ethanol produces 352 g of CO₂, based on the stoichiometry of the balanced chemical equation.

Explanation:

The question involves a stoichiometry calculation based on the balanced chemical equation for the combustion of ethanol, C₂H₅OH (l) + 3O₂ (g) → 2CO₂ (g) + 3H₂O (g). Given that 4 moles of ethanol are combusted, we need to determine the amount of CO2 produced.

From the equation, it is clear that 1 mole of ethanol produces 2 moles of CO₂. Therefore, 4 moles of ethanol will produce 4 * 2 = 8 moles of CO₂.

To calculate the weight of these CO₂ moles, we use the molar mass of CO₂, which is approximately 44 g/mol. Hence, the weight of CO₂ produced is 8 moles * 44 g/mol = 352 g.

The metathesis reaction between potassium carbonate and iron (III) chloride results in the formation of a precipitate of Iron (III) Carbonate along with Potassium chloride. The chemical, full ionic, and net ionic equations for this reaction have been described in detail.

Firstly, the metathesis (or double replacement) reaction between potassium carbonate (K2CO3) and iron (III) chloride (FeCl3) can be represented as follows:

K2CO3(aq) + FeCl3(aq) -> Fe2(CO3)3(s) + 2 KCl(aq)

The evidence of the reaction is typically observed as the formation of a precipitate, in this case, Iron (III) Carbonate (Fe2(CO3)3).

The full ionic equation can be given as:

2 K+(aq) + CO32-(aq) + Fe3+(aq) + 3 Cl-(aq) -> Fe2(CO3)3(s) + 2 K+(aq) + 2 Cl-(aq)

From this, the net ionic equation becomes:

Fe3+(aq) + CO32-(aq) -> Fe2(CO3)3(s)

This reaction is characterized by a double exchange of ions between the reactants, resulting in the formation of a precipitate.

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The balanced molecular equation for the metathesis reaction between potassium carbonate (K₂CO₃) and iron(III) chloride (FeCl₃) is:

[tex]\[ \text{K}_2\text{CO}_3(aq) + 2\text{FeCl}_3(aq) \rightarrow 2\text{KCl}(aq) + \text{Fe}_2(\text{CO}_3)_3(s) \][/tex]

Evidence of reaction occurs in the mixture as the formation of a solid precipitate, which is iron(III) carbonate (Fe₂(CO₃)₃). The balanced ionic equation, taking into account that K₂CO₃ and FeCl₃ are strong electrolytes and dissociate completely in aqueous solution, is:

        [tex]\[ 2\text{K}^+ + \text{CO}_3^{2-} + 2\text{Fe}^{3+} + 6\text{Cl}^- \rightarrow 2\text{K}^+ + 2\text{Cl}^- + \text{Fe}_2(\text{CO}_3)_3(s) \][/tex]

Explanation:

1) Edge length of the metal in BCC unit cell = [tex]a=0.33226 nm[/tex]

Atomic radius of the metal  atom = r

For BCC unit cell, relationship between edge length and radius is given as:

[tex]r=\frac{\sqrt{3}}{4}\times a=0.4330a[/tex]

[tex]r=0.4330\times 0.33226 nm=0.144 nm[/tex]

[tex]1 nm=10^{-7} cm[/tex]

[tex]r=0.144 nm=0.1439\times 10^{-7} cm=1.44\times 10^{-8} cm[\tex]

The atomic radius of the metal  atom in BCC unit cell is [tex]1.44 \times 10^{-8} cm[/tex].

2) Edge length of the metal in FCC unit cell = [tex]a=4.3992 \AA[/tex]

Atomic radius of the metal  atom = r

For FCC unit cell, relationship between edge length and radius is given as:

[tex]r=\frac{1}{2\sqrt{2}}\times a=0.3535a[/tex]

[tex]r=0.3535\times 4.3992 \AA=1.56 \AA[/tex]

[tex]1 \AA=10^{-8} cm[/tex]

[tex]1.56\AA=1.56 \times 10^{-8} cm[/tex]

The atomic radius of the metal  atom in FCC unit cell is [tex]1.56 \times 10^{-8} cm[/tex].

Answer:

[tex]2.0489\times 10^{-6} cm[/tex] is the average thickness of the sheet.

Explanation:

Mass of gold ,m= 1.78 g

Volume of the gold = V

Density of the gold = D = [tex]19.32g/cm^3[/tex]

[tex]D=\frac{m}{V}=19.32g/cm^3=\frac{1.78 g}{V}[/tex]

[tex]V = 0.09213 cm^3[/tex]

Area of the hammered gold sheet,A = [tex]48.4 ft^2=44,965.052 cm^2[/tex]

Thickness of the hammered gold = h

([tex]1 ft^2=929.03 cm^2[/tex])

Volume = Area × thickness

[tex]V= A\times h[/tex]

[tex]0.09213 cm^3=44,965.052 cm^2\times h[/tex]

[tex]h=2.0489\times 10^{-6} cm[/tex]

[tex]2.0489\times 10^{-6} cm[/tex] is the average thickness of the sheet.

Explanation:

It is given that molarity is 0.15 M and volume is 250.0 ml. As 1 ml equals 0.001 l.

Hence, 250.0 ml = 0.25 l.

Also, molarity is equal to the number of moles present in liter of solution. Therefore, calculate number of moles as follows.

                   Molarity = [tex]\frac{\text{no. of moles}}{volume}[/tex]

                      0.15 M = [tex]\frac{\text{no. of moles}}{0.25 l}[/tex]

             No. of moles = 0.0375 mol

Thus, we can conclude that moles present in given solution are 0.0375.

Final answer:

To find the number of moles in a 250.0 mL solution of 0.15 M nitric acid, you convert the volume to liters and multiply by the molarity, resulting in 0.0375 moles of HNO3.

Explanation:

To calculate the number of moles in a 0.15 M nitric acid (HNO3) solution with a volume of 250.0 mL, you would first convert the volume from milliliters (mL) to liters (L) because concentration (Molarity, M) is typically expressed in moles per liter. Therefore, 250.0 mL is equivalent to 0.250 L (since 1 L = 1000 mL).

Using the formula for molarity:

Molarity (M) = moles of solute / liters of solution

We can rearrange the formula to solve for the number of moles:

moles of solute = Molarity (M) × liters of solution

Thus, the number of moles of nitric acid in the solution is:

0.15 moles/L × 0.250 L = 0.0375 moles of HNO3

Answer:

0.152 moles

Explanation:

Given that:

The mass of the oxygen gas produced = 4.87 g

Also, The molar mass of oxygen gas, [tex]O_2[/tex] = [tex]2\times 16\ g/mol[/tex] = 32 g/mol

The formula for the calculation of moles is shown below:

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

Thus,

[tex]Moles= \frac{4.87\ g}{32\ g/mol}[/tex]

[tex]Moles=0.152\ mol[/tex]

Both given values and the answer is in 3 significant digits.

Answer: The given mass in kilograms is 0.00399 kg

Explanation:

Gram and Kilograms are the units which are used to express the mass of a substance. These units are inter changeable.

We are given:

Mass of a substance = 3.99 g

To convert the given mass into kilograms, we use the conversion factor:

1 kg = 1000 g

Converting the given value, we get:

[tex]\Rightarrow 3.99g\times (\frac{1kg}{1000g}=0.00399kg[/tex]

Hence, the given mass in kilograms is 0.00399 kg

To determine the molar mass of an unknown protein, first calculate the molar concentration using the osmotic pressure formula. Next, calculate the moles of solute from this molarity. Finally, determine the molar mass by dividing the mass of the protein by the moles of solute.

First, we need to calculate the molarity (M) of the solution using the osmotic pressure formula II = MRT. Here, 'II' represents the osmotic pressure in atm, which is 0.0861 atm, 'R' is the gas constant (0.08206 L atm/mol K), and 'T' is the temperature in Kelvin. Convert the given temperature 25.0 °C to Kelvin by adding 273.15, which results in 298.15 K.

Substituting the known values into the formula, we get M = II / RT. Now, compute the molar concentration (M) from the osmotic pressure.

Next, the molar mass of the protein can be determined. Divide the mass of the solute by the number of moles in that mass. In this case, the mass of the solute is 400mg, which equals 0.4 g (as 1 g = 1000 mg). By using the calculated molarity, you can get the moles of solute, from which the molar mass can be determined by dividing the mass of solute by moles of solute.

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To determine the molar mass of the protein, use the osmotic pressure equation. After solving for molarity and then moles, the molar mass is found to be approximately 22,700 g/mol. This process involves converting units and applying the ideal gas constant.

To find the molar mass of the protein, we use the formula for osmotic pressure (π):

π = MRT

π (osmotic pressure) = 0.0861 atm

M (molarity) = n/V (number of moles/volume)

R (gas constant) = 0.0821 L·atm/(K·mol)

T (temperature) = 25 + 273 = 298 K

First, convert the mass of the protein to grams:

400 mg = 0.400 g

The volume of the solution is 5.00 mL, which is 0.00500 L.

We can rearrange the formula to find the molarity (M):

M = π / (RT)

M = 0.0861 atm / (0.0821 L·atm/(K·mol) × 298 K)

M = 0.00352 mol/L

Now, the number of moles of the protein (n) can be related to its molarity and the volume of the solution:

n = M × V

n = 0.00352 mol/L × 0.00500 L = 1.76 x 10⁻⁵ mol

Finally, the molar mass (M) is given by the mass of the protein divided by the number of moles:

Molar mass = mass / n

Molar mass = 0.400 g / 1.76 x 10⁻⁵ mol ≈ 22727 g/mol

Rounded to three significant digits, the molar mass of the protein is 22,700 g/mol.

Answer :

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

[tex]\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)[/tex]

where,

[tex]\Delta U[/tex] = change in internal energy = ?

n = number of moles of gas = 5 moles

[tex]C_v[/tex] = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]300^oC=273+300=573K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta U=nC_v(T_2-T_1)[/tex]

[tex]\Delta U=(5moles)\times (2R)\times (573-298)[/tex]

[tex]\Delta U=(5moles)\times 2(8.314J/mole.K)\times (573-298)[/tex]

[tex]\Delta U=22863.5J=22.86kJ[/tex]

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

[tex]\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)[/tex]

where,

[tex]\Delta H[/tex] = change in enthalpy = ?

n = number of moles of gas = 5 moles

[tex]C_p[/tex] = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]300^oC=273+300=573K[/tex]

Now put all the given values in the above formula, we get:

[tex]\Delta H=nC_p(T_2-T_1)[/tex]

[tex]\Delta H=(5moles)\times (3R)\times (573-298)[/tex]

[tex]\Delta H=(5moles)\times 3(8.314J/mole.K)\times (573-298)[/tex]

[tex]\Delta H=34295.25J=34.29kJ[/tex]

The change in enthalpy of the gas is 34.29 kJ.

Answer:

[S₂] = 1.27×10⁻⁷ M

Explanation:

2 H₂S(g) ⇄ 2 H₂(g) + S₂(g), Kc=1,625x10⁻⁷

The equation of this reaction is:

1,625x10⁻⁷ = [tex]\frac{[H_2]^2[S_2]}{[H_{2}S]^2}[/tex]

The equilibrium concentrations are:

[H₂S] = 0,162 - 2x

[H₂] = 0,184 + 2x

[S₂] = x

Replacing:

1,625x10⁻⁷ = [tex]\frac{[0,184+2x]^2[x]}{[0,162-2x]^2}[/tex]

Solving:

4x³ + 0,736x² + 0,033856x - 4,3x10⁻⁹

x = 1.27×10⁻⁷

Thus, concentration of S₂ is:

[S₂] = 1.27×10⁻⁷ M

Explanation:

A chemical formula is defined as the symbolic representation of atoms present in a compound or molecule which also depicts the ratio in which the elements are combined to each other.

Chemical formula's for the given compounds are as follows.

Final answer:

The frequency of light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level is approximately -2.06 x 10^14 Hz.

Explanation:

The frequency of light emitted by a hydrogen atom during a transition of its electron can be calculated using the formula:

f = R(1/n1^2 - 1/n2^2)

where f is the frequency, R is a constant (R = 3.29 x 10^15 Hz), and n1 and n2 are the initial and final principal energy levels, respectively.

In this case, the electron transitions from n = 4 to n = 1. Plugging the values into the formula, we get:

f = 3.29 x 10^15 (1/4^2 - 1/1^2)

f = 3.29 x 10^15 (1/16 - 1/1)

f = 3.29 x 10^15 (0.9375 - 1)

f = 3.29 x 10^15 (-0.0625)

f = -2.06 x 10^14 Hz

The frequency of the light emitted by the hydrogen atom during this transition is approximately -2.06 x 10^14 Hz.

Answer:

Every substance, whether an element or a compound that contains Avogadro's number, is said to form a mole of it. Therefore for the fluorine atoms that form the Avogadro number we say that we have one mole of fluorine atoms and this weighs 19 g, so 19 g/mol

Explanation:

You can search the F in the Periodic Table where you have the molar mass, 19 g/mol but if you don't have a Periodic Table you can calculate like this:

1 atom of F weighs 19 Da (unified atomic mass unit) and 1 Da is 1,661 x10*-24 g, so try to apply a rule of three, twice.

1 Da ........... 1,661x10*-24 g

19 Da .......... x

x= (19 Da . 1,661x10*-24 g ) / 1 Da = 3,1559x10*-23 g

1 atom F ..............  3,1559x10*-23 g

6,022x10*23 atoms F ............. x

X= (6,022x10*23 atoms F . 3,1559x10*-23 g) / 1 atom F = 19.0048 g

Answer: The average atomic mass of element X is 96.6 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]    .....(1)

Mass of isotope 1 = 95.0 amu

Percentage abundance of isotope 1 = 10.68 %

Fractional abundance of isotope 1 = 0.1068

Mass of isotope 2 = 96.0 amu

Percentage abundance of isotope 2 = 16.90 %

Fractional abundance of isotope 2 = 0.1690

Mass of isotope 3 = 97.0 amu

Percentage abundance of isotope 3 = 72.42 %

Fractional abundance of isotope 3 = 0.7242

Putting values in equation 1, we get:

[tex]\text{Average atomic mass of X}=[(95\times 0.1068)+(96\times 0.1690)+(97\times 0.7242)][/tex]

[tex]\text{Average atomic mass of X}=96.6amu[/tex]

Hence, the average atomic mass of element X is 96.6 amu.

To calculate the atomic mass of the element with three isotopes, you multiply the percentage abundance by the mass of each isotope and then add the values.

To calculate the atomic mass of the element, we need to multiply the percentage abundance of each isotope by its respective mass and then sum up the products.

For isotope X-95: (10.68/100) x 95.0 amu = 10.1376 amu

For isotope X-96: (16.90/100) x 96.0 amu = 16.224 amu

For isotope X-97: (72.42/100) x 97.0 amu = 70.2774 amu

Adding the values: 10.1376 amu + 16.224 amu + 70.2774 amu = 96.639 amu

Therefore, the atomic mass of the element is 96.639 amu (rounded to 3 significant figures).

Answer: Mass of gas is 0.001 pounds.

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{Volume}[/tex]

Given : Mass of gas = ?

Density of gas = [tex]1.29g/L[/tex]

Volume of gas = 389 ml = 0.389 L     (1L=1000ml)

Putting in the values we get:

[tex]1.29g/L=\frac{mass}{0.389L}[/tex]

[tex]mass=0.5grams[/tex]  

[tex]mass=0.5\times 0.002lb=0.001lb[/tex]       (1g =0.002 lb)

Thus the mass of gas is 0.001 pounds.

Final answer:

To find the mass in pounds of 389 mL of a gas with a density of 1.29 g/L, you convert the density to g/mL, calculate the mass in grams, and then convert that mass to pounds, resulting in approximately 0.001106 pounds.

Explanation:

To calculate the mass of the gas in pounds, we first need to convert the density from grams per liter (g/L) to grams per milliliter (g/mL) since the volume of gas given is in milliliters (mL). With the given density of 1.29 g/L, we can use dimensional analysis to calculate the mass of 389 mL of the gas:

Convert density to g/mL: 1.29 g/L = 0.00129 g/mL.

Calculate the mass: mass (in grams) = density (in g/mL)  imes volume (in mL) = 0.00129 g/mL  imes 389 mL = 0.50181 g.

Convert the mass to pounds using the conversion factor 1 lb = 453.59 g: mass (in pounds) = mass (in grams) / conversion factor = 0.50181 g / 453.59 g/lb

Carrying out the division, the mass is approximately 0.001106 pounds.

The ΔH value for the formation of Hydrogen Bromide (HBr) from Hydrogen and Bromine is -36.3 kJ, indicative of an exothermic reaction.

In the given chemical reaction, the formation of the binary compound Hydrogen Bromide (HBr) is exothermic, meaning it releases energy. This is denoted by the negative ΔH value, which refers to the change in enthalpy or total energy of the system. Given that the reaction releases 36.3 kJ, the ΔH of the reaction is -36.3 kJ. Expressing this with three significant figures, the ΔH value becomes -36.3 kJ. This value is negative which indicates that the reaction is exothermic—energy is released in the formation of the compounds.

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Explanation:

The given data is as follows.

         Thickness (dx) = 0.87 m,       thermal conductivity (k) = 13 W/m-K

        Surface area (A) = 5 [tex]m^{2}[/tex],       [tex]T_{1} = 14^{o}C[/tex]

        [tex]T_{2} = 93^{o}C[/tex]

According to Fourier's law,

                    Q = [tex]-kA \frac{dT}{dx}[/tex]

Hence, putting the given values into the above formula as follows.

                      Q = [tex]-kA \frac{dT}{dx}[/tex]

                          = [tex]-13 W/m-k \times 5 m^{2} \times \frac{(14 - 93)}{0.87 m}[/tex]

                          = 5902.298 W

Therefore, we can conclude that the rate of heat transfer is 5902.298 W.

Answer:

(A) 19 bars

Explanation:

First off, we calculate the mass of platinum contained in one 1 L bar. To do that we convert 1 L into cm³ -1 L equals to 1000 cm³-.

21.4 g/cm³ * 1000 cm³ = 21,400 g

Each bar of platinum weighs 21,400 grams.

Now we convert the maximum payload capacity of the truck, into grams (1 lb equals to 453,592 g):

[tex]900lb*\frac{453,592g}{1lb}=408232.8g[/tex]

Then we divide the weight of one bar by the maximum payload capacity:

408232.8 / 21400 =19.09

Thus the thieves could carry 19 1 L bars

Answer:

c. when the raction quotient is greater than the Ksp

Explanation:

∴ Ksp = [ A+ ]∧x  *  [ B- ]∧y.....solubility product constant

∴ I.P = xA+ * yB-............ionic product

⇒ the relationship between Ksp and I.P, will give us the precipitation conditions.

∴ I.P > Ksp ⇒ precipitation

∴ I.P < Ksp ⇒ no precipitation (disolution)

∴ I.P = Ksp ⇒ equilibrium

⇒ the ionic product is directly related to the reaction quotient Q, where Q = [A+] * [B-] / [AxBy] = PI / [AB], then the higher this product the higher the reaction quotient will be and therefore will be greater than the value of the constant and will have precipitated, the correct answer is the c

CH3OH (l)

Methanol.

This an alcohol, in ambient temperature is liquid, and in contact with water, they dissolve.

HC2H3O2 (l)

Acetic Acid.

This is a liquid acid, common in vinegar, giving them it's taste and sour flavor.

With water they do not repel, they can create a solution.

C12H22O11 (s)

Sucralose.

This is what we know as a common sugar.

It is a solid that dissolves in water.

.

The mass of chloroform the student should weigh out will be 74.0g of chloroform.

Chloroform is a name of the gas whose chemical name is nitrous oxide. It is a gas that is used to freeze the area or sense of a body part when there is any operation or treatment.

"The mass per unit volume is known as density. A scalar quantity, density. It is represented by the letter D, and the Greek letter rho is used as the sign for density". "Mass divided by volume is how density is computed."

"Mass is a physical body's total amount of matter. Mass is defined as the sum of the moles of the material and the compound's molar mass".

Density relates to mass and volume, and 1 cm⁻³ = 1 mL:

1.48 g of chloroform  1 cm⁻³ chloroform

m =  50.0 cm ⁻³ chloroform

m = 74.0 g of chloroform.

Therefore, the chemistry student will need to weigh 74.0g of chloroform.

To learn more about chloroform, refer to the below link:

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The molar mass of chloroform is calculated using the ideal gas law and the conditions provided (mass, volume, temperature, pressure). By converting the conditions to appropriate units and applying the formula, the calculated molar mass should closely approximate the given value of 119.37 amu.

To calculate the molar mass of chloroform (CHCl3), we use the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. First, we need to convert the given temperature of 99.6 °C to Kelvin by adding 273.15, resulting in 372.75 K. Next, we convert the pressure from mm Hg to atmospheres by dividing by 760. So, the pressure is 0.9765 atm. Using the ideal gas law and rearranging for n (n = PV/RT), we can find the number of moles of chloroform. Finally, we calculate the molar mass by dividing the mass of the chloroform sample (0.494 g) by the number of moles calculated.

The density of chloroform mentioned is not directly needed for calculating molar mass in this context. The provided molecular mass of chloroform, 119.37 amu, serves as a reference and validation of our calculation.

Answer:

1.0489 M is the concentration (in M) of the NaOH solution that was added

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

[tex]n_1M_1V_1=n_2M_2V_2[/tex]

where,

[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]

[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.

We are given:

[tex]n_1=1\\M_1=0.0500 M\\V_1=0.450 L\\n_2=1\\M_2=?\\V_2=21.45 mL=0.02145 L[/tex]

Putting values in above equation, we get:

[tex]1\times 0.0500 M\times 0.450 L=1\times M_2\times 0.02145 L\\\\M_2=1.0489 M[/tex]

1.0489 M is the concentration (in M) of the NaOH solution that was added