A liquid mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a) 20 b) 25 c) 6.75 d) 9.25

Answer:

(B) F⁻, HCOOH

Explanation:

(A) CH₄, HCOOH

(B) F⁻, HCOOH

(C) F⁻, CH₃-O-CH₃

The hydrogen bonds are formed when the hydrogen is found between two electronegative atoms such as oxygen (O), nitrogen (N) or florine (F).

O····H-O, F····H-O, O····H-N

(A) CH₄, HCOOH

- here methane CH₄ is not capable to form hydrogen bond with water

- formic acid HCOOH can form hydrogen bonds with water

H-C(=O)-O-H····OH₂

(B) F⁻, HCOOH

-both floride (F⁻) and formic acid can form hydrogen bonds with water

F····OH₂

H-C(=O)-O-H····OH₂

(C) F⁻, CH₃-O-CH₃

-  dimethyl-ether CH₃-O-CH₃ is not capable to form hydrogen bond with water

- floride (F⁻) can form hydrogen bonds with water

F····OH₂

The correct pairs of species that can form hydrogen bonds with water are option (B) F, HCOOH and option (C) F.CH, OCH. This is because hydrogen bonds are formed between hydrogen and a highly electronegative atom such as Oxygen, Nitrogen, or Fluorine.

The correct pairs of species that can form hydrogen bonds with water are option (B) F, HCOOH and option (C) F .CH, OCH. Hydrogen bonds are primarily formed between hydrogen and a highly electronegative atom such as Oxygen, Nitrogen, or Fluorine, which are present in both HCOOH (formic acid) and OCH (a group from a larger molecule such as methanol).

Hydrogen bonds form due to the attraction between the slightly positive Hydrogen of one molecule and the slightly negative Oxygen, Nitrogen, or Fluorine of another. For example, in a water molecule, the oxygen atom carries a slight negative charge due to its higher electronegativity while hydrogen atoms carry a slight positive charge.

Option (A) CH cannot form a hydrogen bond with water as it is a nonpolar molecule and lacks an electronegative atom. Also, individual fluorine atoms as given in option (B) do not form hydrogen bonds as they lack the H-F bond necessary for hydrogen bonding.

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Answer:

The answer is B

Explanation:

The answer to this question is based on the molecular kinetic theory. The more you diminish the temperature, the slower the kinetic energy of molecules (i.e their movement) will be. Therefore, if the particles aren't allowed to move freely, they'll package themselves and the mean distance between two particles will be fixed and the shortest you'd be able to find.

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

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:

You need to add 203 mL of 1,00M KH₂PO₄ and 147 mL of 1,00M K₂HPO₄.

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for 7,07 phosphate buffer is:

7,07 = 7,21 + log₁₀ [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex]

0,7244 = [tex]\frac{[HPO4^{2-}] }{[H2PO4^{-}]}[/tex] (1)

As the buffer concentration must be 1,00 M:

1,00 = [H₂PO₄⁻] + [HPO4²] (2)

Replacing (2) in (1):

[H₂PO₄⁻] = 0,5799 M

Thus:

[HPO4²] = 0,4201 M

To obtain these concentrations you need to add:

0,5799 M × 0,350 L × [tex]\frac{1L}{1mol}[/tex] = 0,203 L ≡ 203 mL of 1,00M KH₂PO₄

And:

0,4201 M × 0,350 L × [tex]\frac{1L}{1mol}[/tex] = 0,203 L ≡ 147 mL of 1,00M K₂HPO₄

I hope it helps!

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.

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

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.

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.

Answer:

Look picture

Explanation:

The empirical formula is obtained with average of each atom by mass over Atomic weight, thus:

47,5% S ÷ 32,045 g/mol = 1,48 mol S

52,5% Cl ÷ 35,45 g/mol = 1,48 mol Cl

Thus, empirical formula is:

[tex]S_{1,48} Cl_{1,48}[/tex] ≡ SCl.

The Lewis structure for this SCl molecule is in the picture. You can see one unpaired electron in S. These unpaired electrons are very unstable doing SCl an improbable molecule.

The more pausible structure with the same S:Cl ratio is S₂Cl₂ (Look picture). In this molecule you don't have unpaired electrons doing this compound more stable (In fact, exist, and its name is Disulfur dichloride)

I hope it helps!

Answer:

See explanation.

Explanation:

Hello,

In this case, one could identify the subscripts in the empirical formula, based on the given percentages as shown below:

[tex]n_S=\frac{47.5g}{32g/mol}=1.48 \\n_{Cl}=\frac{52.5g}{35.45g/mol}=1.48\\ S=\frac{1.48}{1.48} =1;Cl=\frac{1.48}{1.48} =1[/tex]

Thus, the empirical formula is:

[tex]SCl[/tex]

Nevertheless, such empirical formula does not respect the sulfur's octet, based on the first drawing on the attached picture (that is a deficiency), that is why a more plausible structure is based on the following formula:

[tex]S_2Cl_2[/tex]

Which actually respect the octet based on the second drawing on the attached picture.

Best regards.

Answer:

(a) The time until overflow is 649 s

(b) The time until overflow is 355 s

Explanation:

The volume as a function of time can be expressed as

[tex]V(t) = V_0+(q_i-q_o)*t[/tex]

If the tank is initially half full, V(0) = V0 = 4/2 = 2 m3.

With ρ=1000 kg/m3, the volume flows are

Flow in = 6.33 kg/s * 0.001 m3/kg = 0.00633 m3/s

Flow out = 3.25 kg/s * 0.001 m3/kg = 0.00325 m3/s

The time until overflow (V(t)=4 m3) is

[tex]V(t) = 2+(0.00633 - 0.00325)*t=2+0.00308*t=4\\\\t=(4-2)/0.00308 = 649.4 s=11 min[/tex]

If the flows are

Flow in = 6.83 kg/s * 0.001 m3/kg = 0.00683 m3/s

Flow out = 3.50 kg/s * 0.001 m3/kg = 0.0035 m3/s

And the tank is initially 2/3 full (V(0)=2.67 m3)

The time until overflow is

[tex]V(t) = 2.67+(0.00683 - 0.00350)*t=2.67+0.00375*t=4\\\\t=(4-2,67)/0.00375 = 354.67 s  =6 min[/tex]

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]

Answer: The volume of solution required is 1.816 L

Explanation:

To calculate the molarity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}[/tex]

We are given:

Mass of solute (KCl) = 15.00 g

Molar mass of potassium chloride = 74.6 g/mol

Molarity of solution = 0.1107 M

Putting values in above equation, we get:

[tex]0.1107M=\frac{15.00g}{74.6g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=1.82L[/tex]

Hence, the volume of solution required is 1.816 L

To determine the volume of 0.1107 M KCl solution containing 15.00 g of KCl, we can use the formula Molarity (M) = moles of solute / volume of solution (in liters). By calculating the moles of KCl and using the molarity formula, the volume of the solution is found to be 1.814 L.

To calculate the number of liters of 0.1107 M KCl that contain 15.00 g of KCl, we need to use the formula:

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

First, we need to calculate the moles of KCl using its molecular weight:

Moles of KCl = mass of KCl / molecular weight of KCl

Moles of KCl = 15.00 g / 74.6 g/mol = 0.201 moles of KCl

Now, we can use the formula to calculate the volume of solution:

Volume of solution = moles of solute / molarity

Volume of solution = 0.201 moles / 0.1107 M = 1.814 L

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.

Answer : The percent abundance of the heaviest isotope is, 78 %

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]

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = [tex]\frac{x}{100}[/tex]

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = [tex]\frac{10}{100}[/tex]

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = [tex]\frac{(90-x)}{100}[/tex]

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

[tex]48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})][/tex]

[tex]x=78\%[/tex]

Therefore, the percent abundance of the heaviest isotope is, 78 %

Answer:

78 %

Explanation:

The atomic mass is the weighted average of the atomic masses of each isotope.

In a weighted average, we multiply each value by a number representing its relative importance.

In this problem, the percent abundance represents the relative importance of each isotope.

Data:

X-47: mass = 47.00 u; abundance = 10.0 % = 0.100  

X-48: mass = 48.00 u  

X-49: mass = 49.00 u

Calculations:

                 Let x = abundance of X-49

Then 0.900 - x = abundance of X-48

[tex]\begin{array}{cccc}\\\textbf{Isotope} & \textbf{Mass/u} & \textbf{Abundance} & \textbf{Contribution/u}\\\text{X-47} & 47.00 & 0.100 & 4.700\\\text{X-48} & 48.00 & 0.900 - x & 48.00(0.900 - x)\\\text{X-49} & 49.00 & x & 49.00x\\& \text{TOTAL} & = & \mathbf{48.68}\\\end{array}[/tex]

[tex]\begin{array}{rcr}4.700 + 48.00(0.900 - x) + 49.00x & = & 48.68\\4.700 + 43.20 - 48.00x + 49.00x & = & 48.68\\47.90 +x & = & 48.68\\x & = & \mathbf{0.78}\\\end{array}[/tex]

The heaviest isotope has an abundance of 78 %.

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:

4714.950 kilograms  is the mass of a sample containing 45.0 kmol of methyl acetate.

Explanation:

Moles of methyl acetate =[tex]n_1[/tex]=45.0 kmol= 45000 mol

Mole percentage of methyl acetate = 60.0%

Total moles in the sample = n

[tex]60.0\%=\frac{45000 mol}{n}\times 100[/tex]

[tex]n=\frac{45000 mol}{60.0}\times 100=75000 mol[/tex]

Mole percentage of methyl alcohol = 20.0%

Moles of methyl alcohol = n_2

[tex]20.0\%=\frac{n_2}{75000 mol}\times 100[/tex]

[tex]n_2=15,000 mol[/tex]

Mass of methyl alcohol =  [tex]n_2\times 32.04 g/mol[/tex]

=[tex]15000 mol\times 32.04 g/mol=480,600 g[/tex]

Mole percentage of acetic acid  = 20.0%

Moles of acetic acid = n_3

[tex]20.0\%=\frac{n_3}{75000 mol}\times 100[/tex]

[tex]n_3=15,000 mol[/tex]

Mass of acetic acid= [tex]n_3\times 60.05 g/mol[/tex]

[tex]15000 mol\times 60.05 g/mol=900,750 g[/tex]

Mass of methyl methyl acetate= [tex]n_1\times 74.08 g/mol[/tex]

[tex]45000 mol\times 74.08 g/mol =3,333,600 g[/tex]

Mass of sample: 480,600 g + 3,333,600 g + 900,750 g = 4714950 g

4714950 g = 4714.950 kg

(1 kg = 1000 g)

To find the mass of a sample with 45.0 kmol of methyl acetate, multiply the kmol amount by the molar mass of methyl acetate (74.08 g/mol), resulting in 3333.6 kg.

To calculate the mass of a sample containing 45.0 kmol (kilo moles) of methyl acetate, we first need to know the molar mass of methyl acetate. Methyl acetate (C3H6O2) has a molar mass of approximately 74.08 g/mol. Knowing this, we can calculate the mass of the methyl acetate in the sample.

The calculation is as follows:

This calculation reveals that a sample containing 45.0 kmol of methyl acetate has a mass of 3333.6 kilograms.

Answer:

Part A

[tex]Rate = -\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t} = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

Part B

[tex]-\frac{\Delta B}{\Delta t}= 0.0500 M s^{-1} [/tex]

Part C

[tex]\frac{\Delta C}{\Delta t} = 0.15 M s^{-1}[/tex]

Explanation:

For a general reaction,

[tex]aA(g) + bB(g) \rightarrow cC(g)[/tex]

Rate is given by:

Rate: [tex]Rate = -\frac{1}{a}\frac{\Delta A}{\Delta t} =-\frac{1}{b}\frac{\Delta B}{\Delta t} = \frac{1}{c}\frac{\Delta C}{\Delta t}[/tex]

So, for the given reaction:

[tex]2A(g) + B(g) \rightarrow 2C(g)[/tex]

[tex]Rate = -\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t} = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

Part B

[tex]-\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t}[/tex]

Given: [tex]-\frac{\Delta A}{\Delta t} = 0.100\ Ms^{-1}[/tex]

[tex] \frac{1}{2}\frac{0.100}{\Delta t} =-\frac{\Delta B}{\Delta t}[/tex]

[tex]-\frac{\Delta B}{\Delta t}[/tex] = 0.0500 M s^-1

Part C

[tex]-\frac{\Delta B}{\Delta t} =\frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

[tex]-\frac{\Delta B}{\Delta t} =\frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

[tex]0.0500 = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

[tex]\frac{\Delta C}{\Delta t} = 3 \times 0.0500 = 0.15 M s^{-1}[/tex]

The rate equation is Rate=−1/2Δ[A]/Δt=−Δ[B]/Δt=1/3Δ[C]/Δt. The rate of decrease of B is 0.100 M⋅s−1, and the rate of increase of C is 0.150 M⋅s−1.

The correct expression for the rate of the given reaction 2A(g)+B(g)→3C(g) is: Rate=−1/2Δ[A]/Δt=−Δ[B]/Δt=1/3Δ[C]/Δt. This is because the rate of disappearance of A and B and the rate of appearance of C are all proportional to the stoichiometric coefficients in the balanced chemical equation.

For Part B, since A is decreasing at a rate of 0.100 M⋅s−1, B will be decreasing at the same rate because the rate is proportional to their coefficients in the balanced equation. So, B is decreasing at a rate of 0.100 M⋅s−1.

For Part C, if A is decreasing at a rate of 0.100 M⋅s−1, since the rate of increase of product C is 1.5 times the rate of decrease of reactant A (according to their coefficients). Hence, C is increasing at a rate of 0.150 M⋅s−1.

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Answer:

4.78 g/cm³

Explanation:

Density is expressed as mass per unit volume:

D = m/V

D = (24.02 g) / (5.02 mL) = 4.78 g/mL

The density should be expressed in g/cm³, so mL must be converted to cm³. The conversion ratio is 1 mL = 1 cm³.

(4.78 g/mL)(1 mL/1 cm³) = 4.78 g/cm³

Answer:

The density of the solid is 1,316 g/mL

Explanation:

The weight of both Toluene and the solid insoluble is 58,68 g. And the weight of the solid is 32,50 g. Thus, weight of toluene is:

58,68 g - 32,50 g = 26,18 g of Toluene

To know how much volume that toluene occupy you must use density thus:

26,18 g of toluene × ( 1 mL / 0,864 g) = 30,30 mL of toluene

The volume of both Toluene and the solid is 55,00 mL and the volume of toluene is 30,30 mL. Thus, the volume of the solid is:

55,00 mL - 30,30 mL = 24,70 mL

Knowing both volume and weight it is possible to know the density thus:

32,50 g / 24,70 mL = 1,316 g/mL

I hope it helps!

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:

3.60 mol CO₂

Explanation:

Balanced chemical reaction:

2CO + O₂ ⇒ 2CO₂

The molar ratio between CO₂ and CO is 1:1

2CO₂/2CO = CO₂/CO

Thus, the moles of CO₂ produced from 3.60 moles of CO is 3.60 moles:

(3.60 mol CO)(CO₂/CO) = 3.60 mol CO₂

Answer :

(a) The volume of unit cell is, [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]

(b) The theoretical density of Ti is, [tex]4.81g/cm^3[/tex]

Explanation :

(a) First we have to calculate the volume of the unit cell.

Formula used :

[tex]V=6r^2c\sqrt {3}[/tex]

where,

V = volume of unit cell  = ?

r = atomic radius = [tex]0.1445nm=1.445\times 10^{-8}cm[/tex]

conversion used : [tex](1nm=10^{-7}cm)[/tex]

Ratio of lattice parameter = c : a = 1.58 : 1

So, c = 1.58 a

And,  a = 2r

c = 1.58 × 2r

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

[tex]V=6\times r^2\times (1.58\times 2r)\sqrt {3}[/tex]

[tex]V=6\times r^3\times (1.58\times 2)\sqrt {3}[/tex]

[tex]V=6\times (1.445\times 10^{-8}cm)^3\times (1.58\times 2)\sqrt {3}[/tex]

[tex]V=9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]

The volume of unit cell is, [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]

(b) Now we have to calculate the density of Ti.

Formula used for density :

[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]

[tex]\rho=\frac{Z\times M}{N_{A}\times V}[/tex]     ..........(1)

where,

[tex]\rho[/tex] = density  of Ti = ?

Z = number of atom in unit cell  = 6 atoms/unit cell (for HCP)

M = atomic mass  = 47.87 g/mol

[tex](N_{A})[/tex] = Avogadro's number  = [tex]6.022\times 10^{23}atoms/mole[/tex]

[tex]a^3=V[/tex] = volume of unit cell  = [tex]9.91\times 10^{-23}cm^3/\text{ unit cell}[/tex]

Now put all the values in above formula (1), we get:

[tex]\rho=\frac{6\times 47.87}{(6.022\times 10^{23})\times (9.91\times 10^{-23})}[/tex]

[tex]\rho=4.81g/cm^3[/tex]

The theoretical density of Ti is, [tex]4.81g/cm^3[/tex]

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.

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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:

0.4694 moles of CrCl₃

Explanation:

The balanced equation is:

Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)

The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.

The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:

MCr = 52 g/mol

MCl = 35.5 g/mol

MO = 16 g/mol

So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.

The number of moles is the mass divided by the molar mass, so:

n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.

For the stoichiometry:

1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃

0.2347 mol of Cr₂O₃----------- x

By a simple direct three rule:

x = 0.4694 moles of CrCl₃

Final answer:

The number of moles of CrCl3 that can be produced from 49.4 g of Cr2O3 is calculated by dividing the mass of Cr2O3 by its molar mass and then using the stoichiometry of the balanced chemical equation, resulting in approximately 0.65 moles of CrCl3.

Explanation:

To calculate the number of moles of CrCl3 that could be produced from 49.4 g of Cr2O3, we need to first find the molar mass of Cr2O3. Chromium has an atomic mass of approximately 52.00 g/mol and oxygen has an atomic mass of approximately 16.00 g/mol, which makes the molar mass of Cr2O3 to be about (2*52.00 g/mol) + (3*16.00 g/mol) = 152.00 g/mol.

Next, we divide the mass of Cr2O3 by its molar mass to find the number of moles:

Using the balanced chemical equation, we see that 1 mole of Cr2O3 produces 2 moles of CrCl3. Therefore, 0.325 moles of Cr2O3 will produce 0.325 x 2 moles of CrCl3, which is approximately 0.65 moles of CrCl3.

Final answer:

The number of moles of CrCl3 that can be produced from 49.4 g of Cr2O3 is calculated by dividing the mass of Cr2O3 by its molar mass and then using the stoichiometry of the balanced chemical equation, resulting in approximately 0.65 moles of CrCl3.

Explanation:

To calculate the number of moles of CrCl3 that could be produced from 49.4 g of Cr2O3, we need to first find the molar mass of Cr2O3. Chromium has an atomic mass of approximately 52.00 g/mol and oxygen has an atomic mass of approximately 16.00 g/mol, which makes the molar mass of Cr2O3 to be about (2*52.00 g/mol) + (3*16.00 g/mol) = 152.00 g/mol.

Next, we divide the mass of Cr2O3 by its molar mass to find the number of moles:

Using the balanced chemical equation, we see that 1 mole of Cr2O3 produces 2 moles of CrCl3. Therefore, 0.325 moles of Cr2O3 will produce 0.325 x 2 moles of CrCl3, which is approximately 0.65 moles of CrCl3.

Answer: 26138g/mol

Explanation:

[tex]\pi =CRT[/tex]

[tex]\pi[/tex] = osmotic pressure = 12 mmHg =[tex]\frac{1}{760}\times 12=0.016[/tex] atm     (760 mmHg= 1atm)

C= concentration in Molarity

R= solution constant  = 0.0821 Latm/Kmol

T= temperature = [tex]10^0C=(10+273)K=283K[/tex]

For the given solution: 18 g of macromolecule is dissolved to make 1 L of solution.

[tex]Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}[/tex]

[tex]C=\frac{18\times 1000}{M\times 1000ml}=\frac{18}{M}[/tex]

[tex]0.016=\frac{18}{M}\times 0.0821\times 283}[/tex]

[tex]M=26138g/mol[/tex]

The molecular weight of the macromolecule in grams per mole is 26138.

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:

a) 2.6 mole of O2 per mole of methane

b) 9.78 mole of N2 per mole of methane

c) Mole fraction of H2O = 0.33

d) 6.12 mole of H2O pero mole of methane

Explanation:

The first thing you have to do is the balanced reaction of the methane combustion:

CH4  +2  O2 -->2 H2O  +  CO2

You calculate the moles of O2 needed to react with 1 mole of methane and add the excess factor

2 x 1.3 = 2.6 moles of O2

For the N2 moles you calculate it with the air composition (21 % O2, 79% N2)

2.6 moles of O2 *   0.79/0.21 = 9.78 moles of N2

With the total stream of air = 12.38 moles you add the humidity factor

12.38 * 1.5 = 18.57 moles So 6.12 are moles of H20 entering per mole of CH4.

To calculate the mole fraction yo divide the moles of water among the moles of the stream:

6.12/18.57 --> 0.33

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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