For the following reaction, 6.95 grams of benzene (CH) are mixed with excess oxygen gas. The reaction yields 18.1 grams of carbon dioxide. benzene (C₂H) () + oxygen (g) → carbon dioxide(g) + water (g) a. What is the ideal yield of carbon dioxide? Ideal yield = grams b. What is the percent yield for this reaction? Percent yield = %

Answers

Answer 1

The ideal yield of carbon dioxide is 23.5 g, while the percent yield for this reaction is 77.02%.

Mass of benzene, CH6.95 g

Mass of carbon dioxide, CO2 = 18.1 g

The balanced chemical equation for the combustion of benzene is;

C6H6 + 15O2 → 6CO2 + 3H2O

From the chemical equation, 6 moles of carbon dioxide are produced from 1 mole of benzene.The molar mass of benzene is;

6C = 6 × 12.01 g/mol = 72.06 g/mol

6H = 6 × 1.008 g/mol = 6.048 g/mol

Total molar mass = 78.108 g/mol

The moles of benzene, CH are;

Mass = number of moles × molar mass

78.108 g/mol = 6.95 g × (1 mol/78.108 g) = 0.0889 mol of CH

The ideal yield of carbon dioxide = the number of moles of CH × number of moles of CO2 produced per mole of CH

Ideal yield of CO2 = 0.0889 mol × 6 mol/1 mol = 0.534 molCO2

The ideal yield of CO2 = number of moles of CO2 produced × molar mass of CO2

Ideal yield of CO2 = 0.534 mol × 44.01 g/mol = 23.5 g of CO2

Percent yield = (actual yield/ideal yield) × 100%

The actual yield of CO2 = 18.1 g

Percent yield of CO2 = (18.1/23.5) × 100% = 77.02 %

Therefore, the ideal yield of carbon dioxide is 23.5 g, while the percent yield for this reaction is 77.02%.

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Related Questions

Consider the reaction.
A(aq)↽−−⇀3B(aq)Kc=4.30×10−6at 500 K
If a 4.90 M sample of A is heated to 500 K, what is the concentration of B at equilibrium?
[B]= ? M
--- At a certain temperature, the equilibrium constant for the chemical reaction shown is 6.13×10−3. At equilibrium, the concentration of AB is 2.125 M, the concentration of BC is 2.925 M, and the concentration of AC is 0.250 M. Calculate the concentration of B at equilibrium.
AB(aq)+BC(aq)↽−−⇀AC(aq)+2B(aq)
[B] = ? M

Answers

From the question that we have in the problem;

1) The concentration of B is 104 M

2) The concentration of B is 0.39 M

What is equilibrium concentration?

The equilibrium constant provides information about the relative concentrations of reactants and products at equilibrium. If the equilibrium constant is large (K > 1), the products are favored at equilibrium. Conversely, if the equilibrium constant is small (K < 1), the reactants are favored at equilibrium.

1) We have that;

[tex]4.30*10^-6 = 4.9/[B]^3[/tex]

[B] = ∛4.9/[tex]4.30*10^-6[/tex]

= 104 M

2) [tex]6.13*10^-3[/tex]= (0.250) [tex][B]^2/2.125 * 2.925[/tex]

[tex][B]^2= 6.13*10^-3 * 2.125 * 2.925/ (0.250)[/tex]

= 0.39 M

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Be sure to answer all parts. The first-order rate constant for the reaction of methyl chloride (CH 3

Cl) with water to produce methanol (CH 3

OH) and hydrochloric acid (HCl) is 3.32×10 −10
s −1
at 25 ∘
C. Calculate the rate constant at 55.9 ∘
C if the activation energy is 116 kJ/mol. ×10 s −1
(Enter your answer in scientific notation.)

Answers

The rate constant at 55.9°C is approximately 4.62×10^−8 s^−1. To calculate the rate constant at 55.9°C for the reaction of methyl chloride with water, we can use the Arrhenius equation.

Given the activation energy of 116 kJ/mol and the rate constant at 25°C (3.32×10^−10 s^−1), we can determine the new rate constant. The rate constant at 55.9°C is approximately 4.62×10^−8 s^−1. The Arrhenius equation describes the temperature dependence of reaction rates. It is given by:

k = A * exp(-Ea / (R * T))

Where:

- k is the rate constant

- A is the pre-exponential factor or frequency factor

- Ea is the activation energy

- R is the gas constant (8.314 J/(mol·K))

- T is the temperature in Kelvin

To calculate the rate constant at 55.9°C, we first convert the temperatures to Kelvin. T1 = 25°C + 273.15 = 298.15 K, and T2 = 55.9°C + 273.15 = 329.05 K.

We can rearrange the Arrhenius equation to solve for the rate constant at 55.9°C:

k2 = k1 * exp((Ea / R) * ((1/T1) - (1/T2)))

Plugging in the values, k1 = 3.32×10^−10 s^−1, Ea = 116 kJ/mol, R = 8.314 J/(mol·K), T1 = 298.15 K, and T2 = 329.05 K, we can calculate k2.

k2 = (3.32×10^−10 s^−1) * exp((116,000 J/mol / (8.314 J/(mol·K))) * ((1/298.15 K) - (1/329.05 K)))

  = 4.62×10^−8 s^−1

Therefore, the rate constant at 55.9°C is approximately 4.62×10^−8 s^−1.

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A student performed this experiment and obtained the following concentration values: 0.01490 M, 0.01517 M, and 0.01461 M. a. What is the mean concentration? M b. What is the standard deviation of these results?

Answers

The mean concentration of the experiment's results, measured at 0.01490 M, 0.01517 M, and 0.01461 M, is calculated to be 0.01489 M. The standard deviation of the measurements is approximately 0.0002915 M.

To calculate the mean concentration, we sum up all the concentration values and divide by the number of measurements. In this case, the student obtained three concentration values: 0.01490 M, 0.01517 M, and 0.01461 M.

Mean concentration (M) = (0.01490 M + 0.01517 M + 0.01461 M) / 3 = 0.04468 M / 3 = 0.01489 M

Therefore, the mean concentration is 0.01489 M.

To calculate the standard deviation, we need to determine the variability of the individual data points from the mean concentration. The formula for the sample standard deviation is as follows:

Standard deviation = √(Σ(xi - x_bar)² / (n - 1))

Where:

- xi represents each concentration value

- x_bar is the mean concentration

- n is the number of measurements

Substituting the values, we get:

Standard deviation = √[((0.01490 - 0.01489)² + (0.01517 - 0.01489)² + (0.01461 - 0.01489)²) / (3 - 1)]

= √[(0.00000001 + 0.00000008 + 0.00000008) / 2]

= √(0.00000017 / 2)

= √0.000000085

= 0.0002915

Therefore, the standard deviation of the results is approximately 0.0002915 M.

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Classify the following as Homogeneous mixture, Heterogeneous
mixture or Pure substance.
HCl (aq)

Answers

HCl (aq) is classified as a homogeneous mixture. A homogeneous mixture, also known as a solution, is a uniform blend of two or more substances that appear as a single phase.

In the case of HCl (aq), hydrochloric acid is dissolved in water to form a solution. The HCl molecules are evenly dispersed throughout the water, resulting in a uniform composition and appearance.

This means that at a microscopic level, the distribution of HCl molecules is consistent throughout the entire solution. Homogeneous mixtures are characterized by their consistent properties and lack of visible boundaries between components.

In the case of HCl (aq), it exhibits these characteristics and is considered a homogeneous mixture.

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The amount of I3−​(aq) in a solution can be determined by titration with a solution containing a known concentration of S2​O32−​( aq ) (thiosulfate ion). The determination is based on the net ionic equation 2 S2​O32−​(aq)+I3−​(aq)⟶S4​O62−​(aq)+3I−(aq) Given that it requires 35.5 mL of 0.360MNa2​ S2​O3​ (aq) to titrate a 20.0 mL sample of I3−​(aq), calculate the molarity of I3−​(aq) in the solution. [I3−​]= M

Answers

The amount of I3−​(aq) in a solution can be determined by titration with a solution containing a known concentration of S2​O32−​( aq ) (thiosulfate ion) the molarity of I3- in the solution is 0.319 M.

From the balanced net ionic equation, we can see that the ratio of S2O32- to I3- is 2:1. Therefore, for every 2 moles of S2O32- used, 1 mole of I3- is consumed.

Volume of Na2S2O3 solution used: 35.5 mL

Concentration of Na2S2O3 solution: 0.360 M

Volume of I3- solution: 20.0 mL

To find the moles of S2O32- used, we can use the equation:

moles S2O32- = concentration × volume

moles S2O32- = 0.360 M × 0.0355 L

moles S2O32- = 0.01278 mol

Since the molar ratio of S2O32- to I3- is 2:1, the moles of I3- is half the moles of S2O32- used:

moles I3- = 0.01278 mol / 2

moles I3- = 0.00639 mol

To calculate the molarity of I3-, we need to divide the moles of I3- by the volume of the I3- solution in liters:

molarity of I3- = moles I3- / volume of I3- solution

molarity of I3- = 0.00639 mol / 0.0200 L

molarity of I3- = 0.319 M

Therefore, the molarity of I3- in the solution is 0.319 M.

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6. A reaction has an equilibrium constant of 9.1×10 2
at 298 K. At 600 K, the equilibrium constant is 0.84. Find Δ r

H ∘
for the reaction.

Answers

At 298 K (K1 = 9.1×10^2), ΔrH° for the reaction is approximately 867.67 J/mol (or 0.868 kJ/mol).

To find ΔrH° for the reaction, we can use the Van 't Hoff equation, which relates the equilibrium constant to the change in enthalpy with temperature:

ln(K2/K1) = (-ΔrH°/R) * (1/T2 - 1/T1)

Given that K1 = 9.1×10^2 at T1 = 298 K and K2 = 0.84 at T2 = 600 K, we can solve for ΔrH°.

ln(0.84/9.1×10^2) = (-ΔrH°/R) * (1/600 K - 1/298 K)

Simplifying the equation and plugging in the values:

-0.172 = (-ΔrH°/R) * (0.001677 K⁻¹)

Assuming R = 8.314 J/(mol·K), we can rearrange the equation to solve for ΔrH°:

ΔrH° = -0.172 * (-8.314 J/(mol·K)) / (0.001677 K⁻¹)

ΔrH° ≈ 867.67 J/mol or 0.868 kJ/mol.

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A \( 15.0 \) L cylinder contains a gas with a pressure of \( 3.73 \) atms when held at a temperature of \( 35.00{ }^{\circ} \mathrm{C} \). How many moles of gas are held in the container? \( 0.670 \)

Answers

The number of moles of gas held in the container is approximately 0.670 moles.

To find the number of moles of gas, we can use the ideal gas law equation: PV = nRT

Where:

P = pressure of the gas (in atm)

V = volume of the gas (in liters)

n = number of moles of gas

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

T = temperature of the gas (in Kelvin)

First, we need to convert the given temperature from Celsius to Kelvin by adding 273.15:

\( T = 35.00 + 273.15 = 308.15 \) K

Plugging the given values into the ideal gas law equation:

\( 3.73 \) atm × \( 15.0 \) L = \( n \) × \( 0.0821 \) L·atm/mol·K × \( 308.15 \) K

Simplifying the equation:

\( 55.95 \) = \( n \) × \( 25.325815 \)

Solving for \( n \):

\( n = \frac{55.95}{25.325815} \approx 0.670 \) moles

Therefore, the number of moles of gas held in the container is approximately 0.670 moles.

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Give formula and name of the compound you would expect if Fe+3 and Cr₂072 were to combine.

Answers

When [tex]Fe^{3+}[/tex] and [tex]Cr_2O_7^{2-}[/tex] combine, they undergo a redox reaction, resulting in the formation of iron(II) chromate ([tex]FeCr_2O_4[/tex]). This compound consists of two [tex]Fe^{2+}[/tex] ions and one [tex]Cr^{3+}[/tex] ion, with iron's oxidation state being +2 and chromium's oxidation state being +3.

When [tex]Fe^{3+}[/tex]and [tex]Cr_2O_7^{2-}[/tex] combine, they undergo a redox reaction to form a compound. The oxidation state of Fe is +3, while the oxidation state of Cr in [tex]Cr_2O_7^{2-}[/tex] is +6. To balance the charges, two [tex]Fe^{3+}[/tex] ions are needed for every [tex]Cr_2O_7^{2-}[/tex] ion.

The balanced chemical equation for the reaction is as follows:

[tex]6Fe^{3+} + Cr_2O_7^{2-} -- > 6Fe^{2+} + 2Cr^{3+}[/tex]

In this reaction, [tex]Fe^{3+}[/tex] is reduced to [tex]Fe^{2+}[/tex] by gaining three electrons, while [tex]Cr_2O_7^{2-}[/tex] is reduced to [tex]Cr^{3+}[/tex] by losing three electrons.

The compound formed as a result of this reaction is iron(II) chromate, with the chemical formula [tex]FeCr_2O_4[/tex]. It consists of two [tex]Fe^{2+}[/tex] ions and one [tex]Cr^{3+}[/tex] ion. The ratio of Fe to Cr is 2:1, and the oxidation state of iron is +2, while the oxidation state of chromium is +3.

Iron(II) chromate is a brownish solid that is sparingly soluble in water. It is used in pigments and as a corrosion inhibitor.

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A black mineral is really shiny but you not sure if its a metallic or non-metallic luster but it leaves a white to very pale gray streak, is barely able to scratch glass, you're not sure it it has cleavage or not but there are some small flat faces, looks splintery (like wood grain) is -biotite -calcium plagioclase feldspar -augite -potassium feldspar (K-spar_ -sodium plagioclase feldspar -hornblende -quartz -muscovite

Answers

Among the given options, muscovite is the best match for the described mineral characteristics.

Based on the given observations, the mineral that fits the description is "muscovite." Here's why:

Metallic or non-metallic luster: Muscovite typically exhibits a non-metallic luster. It appears shiny, but without a metallic reflection.

Streak color: Muscovite has a white to very pale gray streak, which matches the description provided.

Hardness: Muscovite has a hardness of around 2.5 to 3 on the Mohs scale, which means it is barely able to scratch glass.

Cleavage: Muscovite has excellent basal cleavage, which means it tends to break along flat, thin sheets or layers.

Splintery appearance: Muscovite often displays a splintery or micaceous appearance due to its characteristic sheet-like structure, resembling wood grain.

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Consider the reaction 2SO2​( g)+O2​( g)⟶2SO3​( g) Using the standard thermodynamic data in the tables linked above, calculate ΔGrxn​ for this reaction at 298.15 K if the pressure of each gas is 21.40 mmHg.

Answers

ΔGrxn​ for reaction at 298.15 K if the pressure of each gas is 21.40 mmHg is -142.2 kJ/mol.

For calculating ΔGrxn​ (change in Gibbs free energy) for the given reaction at 298.15 K and a pressure of 21.40 mmHg for each gas, we can use the equation:

ΔGrxn​ = ΔG°rxn + RT * ln(Q)

Where:

ΔGrxn​ is the change in Gibbs free energy for the reaction

ΔG°rxn is the standard Gibbs free energy change for the reaction

R is the gas constant (8.314 J/(mol·K))

T is the temperature in Kelvin (298.15 K)

ln(Q) is the natural logarithm of the reaction quotient (Q)

First, let's find ΔG°rxn using the standard thermodynamic data. The standard Gibbs free energy change for the reaction can be obtained from the difference in standard Gibbs free energies of the products and reactants:

ΔG°rxn = ΣnΔG°f(products) - ΣnΔG°f(reactants)

Using the thermodynamic data from the tables, we have:

ΔG°f(SO2) = -300.4 kJ/mol

ΔG°f(O2) = 0 kJ/mol

ΔG°f(SO3) = -371.5 kJ/mol

ΔG°rxn = (2 * ΔG°f(SO3)) - (2 * ΔG°f(SO2) + ΔG°f(O2))

       = (2 * -371.5 kJ/mol) - (2 * -300.4 kJ/mol + 0 kJ/mol)

       = -743 kJ/mol + 600.8 kJ/mol

       = -142.2 kJ/mol

Next, we need to calculate the reaction quotient (Q) using the given pressures. Since we are dealing with gases, we can use the partial pressures to calculate Q:

Q = (P(SO3))^2 / (P(SO2))^2 * P(O2)

P(SO3) = 21.40 mmHg

P(SO2) = 21.40 mmHg

P(O2) = 21.40 mmHg

Q = (21.40 mmHg)^2 / (21.40 mmHg)^2 * (21.40 mmHg)

 = 1

Now, we can substitute the values into the equation to calculate ΔGrxn​:

ΔGrxn​ = ΔG°rxn + RT * ln(Q)

       = -142.2 kJ/mol + (8.314 J/(mol·K) * 298.15 K) * ln(1)

       = -142.2 kJ/mol

Therefore, ΔGrxn​ for the given reaction at 298.15 K and a pressure of 21.40 mmHg for each gas is approximately -142.2 kJ/mol.

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which of the following statements correctly and most accurately describes the function of fad in the pyruvate dehydrogenase enzyme complex? a) nadh passes electrons to fad to form fadh2. b) lipoamide passes electrons through fadh2, which almost instantly passes them to nad thus forming nadh. c) fadh2 donates electrons to lipoamide thus regenerating fad. d) lipoamide oxidizes nadh to nad by passing electrons to fad. e) nad accepts electrons directly from lipoamide, which has gained them via oxidation of fadh2.

Answers

The correct statement that accurately describes the function of FAD in the pyruvate dehydrogenase enzyme complex is:

c) FADH₂ donates electrons to lipoamide, thus regenerating FAD.

Flavin adenine dinucleotide (FAD), which functions as a coenzyme in the pyruvate dehydrogenase enzyme complex, is essential to the catalytic process. When pyruvate is decarboxylated, FAD receives electrons and is reduced to FADH₂ . The oxidized form of FAD is then produced by FADH₂  transferring the electrons to lipoamide, an element of the enzyme complex.

The subsequent transfer of electrons to NAD⁺ (nicotinamide adenine dinucleotide) to create NADH, which functions as a carrier of electrons for other energy-producing events in the cell, is made possible by this electron transfer from FADH₂  to lipoamide.

The role of FAD in the pyruvate dehydrogenase enzyme complex is thus appropriately described by option c).

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3.) Voltaic cell below with resistance of 3.13 ohm. Current flow through a solution is 30.0 mA. Solve for the voltage (in volts) applied to drive the reaction.
Hg(l)|Hg2Cl2 (s)|KCl(saturated)||KCl(0.90 M) |Cl2 (grams,0.29 atm) |Pt(s)
Anode : Hg2Cl2(s) + 2e-. --> 2Hg(l) + 2Cl- E0 = 1.29 V
Cathode : Cl2(g) + 2e- --> 2Cl- E0=2.80 V

Answers

The voltage applied to drive the reaction in the given voltaic cell is approximately 4.49 V.

In a voltaic cell, the voltage applied to drive a reaction is calculated using the Nernst equation: E = E° - (RT/nF) * ln(Q)

In this case, the anode reaction is Hg₂Cl₂(s) + 2e⁻ → 2Hg(l) + 2Cl⁻ with E° = 1.29 V, and the cathode reaction is Cl₂(g) + 2e⁻ → 2Cl⁻ with E° = 2.80 V.

The total cell potential (E) can be obtained by subtracting the anode potential from the cathode potential:

E = E(cathode) - E(anode) = 2.80 V - 1.29 V = 1.51 V

Since the cell potential (E) is the sum of the anode and cathode potentials, the voltage applied to drive the reaction is equal to the cell potential (E).

However, it's important to note that the given information does not provide the necessary data to calculate the reaction quotient (Q) for the Nernst equation. Without the concentration of species involved in the reaction, a more accurate voltage calculation cannot be performed.

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what is the first precaution that you need to take for performing a grignard reaction? group of answer choices the reaction vessel needs to be pre-weighed. glassware needs to be dry! the magnesium (mg) metal needs to be accurately weighed. the bromobenzene needs to be accurately weighed. flag question: question 2 question 21 pts how will you achieve your first precaution for performing a grignard reaction? group of answer choices by having the weigh balance ready by having the magnesium (mg) ready by placing the glassware in an oven to dry by having the bromobenzene ready

Answers

To achieve this precaution, the correct option would be:

By placing the glassware in an oven to dry.

The first precaution that you need to take for performing a Grignard reaction is that the glassware needs to be dry.

To achieve this precaution, the correct option would be:

By placing the glassware in an oven to dry.

By placing the glassware in an oven to dry, you can remove any moisture or residual water present on the surface of the glassware, ensuring that it is dry before performing the Grignard reaction. This is important because moisture can react with the Grignard reagent and interfere with the reaction or even lead to undesired side reactions. Therefore, drying the glassware is an essential step to ensure the success of the Grignard reaction.

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What is the pressure (in bars) exerted by 1.00 mol of CH4(g) that
occupies a 250-mL container at 0 ° C? Assume methane is an ideal
gas in this case. (R = 0.082058 L- atm/K-mol = 8.3145 J/K-mol; 1
atm

Answers

The pressure exerted by 1.00 mol of CH4(g) that occupies a 250-mL container at 0°C is 6.834 bar.

In this problem, we have to find the pressure (in bars) exerted by 1.00 mol of CH4(g) that occupies a 250-mL container at 0°C. Let us first find the volume of 1 mol of CH4(g) using the ideal gas law: PV = nRT whereP = pressureV = volume of gasn = number of moles R = gas constantT = temperature of the gas in kelvins

The given conditions are:

P = unknown

V = 250 mL

= 0.250 L (since 1 L = 1000 mL)n

= 1 mol

R = 0.082058 L-atm/K-mol (gas constant)

T = 0°C = 273 K (since 0°C = 273 K)

Therefore, PV = nRT becomes P(0.250)

= (1)(0.082058)(273)

Solving for P, we get:

P = 6.7412 atm Since the pressure is given in bars, we have to convert the pressure from atm to bars using the conversion factor: 1 atm = 1.01325 bar

P (in bars) = 6.7412 atm x (1.01325 bar/1 atm)

P = 6.834 bar (rounded to 3 significant figures)

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A precipitate forms when a solution of lead (iD) chloride is mixed with a solution of sodium hydroxide. Write the "formula" equation describing this chemical reaction.

Answers

The formula equation describing this chemical reaction is [tex]PbCl2 + 2NaOH \rightarrow Pb(OH)2 + 2NaCl[/tex].

When a solution of lead(II) chloride (PbCl2) is mixed with a solution of sodium hydroxide (NaOH), a chemical reaction occurs.

The formula equation for this reaction is [tex]PbCl2 + 2NaOH \rightarrow Pb(OH)2 + 2NaCl[/tex]. The reaction results in the formation of a precipitate, lead(II) hydroxide (Pb(OH)2), which appears as a solid. Sodium chloride (NaCl) remains dissolved in the solution.

This reaction demonstrates a double displacement reaction, where the positive ions of the reactants swap places to form new compounds.

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A sample of gas was collected into a 275mL flask at a
temperature of 28.0oC and 1.67 atm. What volume would
this gas occupy at standard temperature and pressure? Report your
answer in liters.

Answers

The gas would occupy approximately 0.216 liters at standard temperature and pressure (STP).

The ideal gas law is PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. If we have an ideal gas at standard temperature and pressure (STP), the values are defined as follows:

STP: 0°C (273.15 K) and 1 atm (760 mmHg)

Using this information, we can solve for the volume of the gas at STP using the combined gas law, which is:

P1V1/T1 = P2V2/T2 where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2 and T2 are the new pressure and temperature, respectively. We can assume that the amount of gas, n, is constant since the sample size is not changing. We can also convert the temperature from Celsius to Kelvin by adding 273.15 to get T1 = 28.0°C + 273.15 = 301.15 K.P1V1/T1 = P2V2/T2

We know that P1 = 1.67 atm, V1 = 275 mL (or 0.275 L), and T1 = 301.15 K. We also know that P2 = 1 atm and T2 = 273.15 K since this is STP.

Solving for V2, we get:V2 = (P1V1T2)/(P2T1)= [(1.67 atm) x (0.275 L) x (273.15 K)] / [(1 atm) x (301.15 K)]≈ 0.216 L

Therefore, the answer is 0.216 L

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An unknown compound has the following composition, by mass: 46.2% C, 5.17% H, and 48.7% F. The molar mass of the compound is experimentally determined to be 468 g/mol. Determine the empirical and molecular formulas for this compound.

Answers

The empirical formula of the compound is CF₂, and the molecular formula is C₂F₄.

To determine the empirical formula, we need to find the simplest whole-number ratio of the elements present in the compound.

Given the mass percentages of carbon (C), hydrogen (H), and fluorine (F), we can assume a 100g sample of the compound to make calculations easier.

1. Convert the mass percentages to grams:

- Carbon (C): 46.2g

- Hydrogen (H): 5.17g

- Fluorine (F): 48.7g

2. Convert the grams of each element to moles using their molar masses:

- Carbon (C): 46.2g / 12.01 g/mol = 3.849 mol

- Hydrogen (H): 5.17g / 1.008 g/mol = 5.13 mol

- Fluorine (F): 48.7g / 18.99 g/mol = 2.564 mol

3. Determine the simplest whole-number ratio of the moles by dividing each by the smallest mole value:

- Carbon (C): 3.849 mol / 2.564 mol = 1.5 ≈ 1

- Hydrogen (H): 5.13 mol / 2.564 mol = 2 ≈ 2

- Fluorine (F): 2.564 mol / 2.564 mol = 1

The empirical formula of the compound is CF₂.

To determine the molecular formula, we need to know the molar mass of the compound. Given that it is 468 g/mol, we can divide it by the empirical formula mass (CF₂) to find the molecular formula ratio:

Molecular formula ratio = 468 g/mol / (12.01 g/mol + 18.99 g/mol * 2) ≈ 468 g/mol / 50.99 g/mol ≈ 9.17

Round the molecular formula ratio to the nearest whole number:

Molecular formula ratio ≈ 9

Multiply the empirical formula by the molecular formula ratio:

Empirical formula (CF₂) * Molecular formula ratio (9) = C₂F₄

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the rate for the reaction a+2b =c is rate=31.2 mol. what is the initial rate of the reaction in the mol if the concentration of A if 0.701 mol and the b is 0.651 mol

Answers

The initial rate of the reaction is 10.76 mol/L²/s, in the mol if the concentration of A is 0.701 mol and the b is 0.651 mol.

How to determine rate of the reaction?

The rate of the reaction is dependent on the concentration of the reactants. The higher the concentration of the reactants, the faster the rate of the reaction.

In this case, the concentration of A is 0.701 mol and the concentration of B is 0.651 mol. The rate of the reaction is 31.2 mol.

The initial rate of the reaction is calculated using the following formula:

Initial rate = Rate × (Concentration of A)¹ × (Concentration of B)²

Where:

Initial rate = The initial rate of the reaction in mol/s

Rate = The rate of the reaction in mol/s

Concentration of A = The concentration of A in mol/L

Concentration of B = The concentration of B in mol/L

Plugging in the values:

Initial rate = 31.2 mol/s × (0.701 mol/L)¹ × (0.651 mol/L)²

= 10.76 mol/L²/s

Therefore, the initial rate of the reaction is 10.76 mol/L²/s.

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Match the relationships of the bolded H's shown ito each of the molecules below? Br. H A. OH H H B. H C. Ill H I D. OH

Answers

The relationships of the bolded H's shown into each of the molecules are mentioned below:

A. OH: In this molecule, the Hydrogen bond is attached to the -OH group.B. H: In this molecule, the H bond is attached to the carbon atom.C. III H: In this molecule, the H bond is attached to the carbon atom.D. OH: In this molecule, the H bond is attached to the -OH group.

To match the relationships of the bolded Hydrogen bond's shown in each of the molecules, let's examine the given options:

A. OH H H: This indicates a hydroxyl group (OH) attached to a hydrogen (H) atom.

B. H C. Ill H I: This indicates a hydrogen (H) atom bonded to a carbon (C) atom in a tertiary (III) carbon center.

D. OH: This indicates a hydroxyl group (OH) without any attached hydrogen atoms.

Now, let's match these relationships to the molecules provided:

CH3OH: This molecule has a hydroxyl group (OH) attached to a carbon (C) atom. Therefore, the bolded H corresponds to option D. OH.

CH3CH2CH2OH: This molecule has a hydroxyl group (OH) attached to a carbon (C) atom, and it also has three hydrogen (H) atoms bonded to a tertiary (III) carbon center. Therefore, the bolded H corresponds to options A. OH H H and B. H C. Ill H I.

Hence, the matching relationships of the bolded H's in the given molecules are as follows:

CH3OH: D. OH

CH3CH2CH2OH: A. OH H H and B. H C. Ill H I

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What volume of water (in mL) is required to react with 28.18 g
of calcium metal to produce calcium hydroxide and hydrogen gas?
(Density H2O = 1.0 g/cm3)

Answers

The volume of water required to react with 28.18 g of calcium metal to produce calcium hydroxide and hydrogen gas is approximately 1.406 mL.

The balanced equation for the reaction is:

Ca + 2 H₂O → Ca(OH)₂ + H₂

From the equation, we can see that the stoichiometric ratio between calcium (Ca) and water (H₂O) is 1:2. This means that for every 1 mol of calcium, we need 2 moles of water.

To calculate the volume of water, we need to convert the given mass of calcium into moles. The molar mass of calcium is 40.08 g/mol.

Moles of calcium = (28.18 g) / (40.08 g/mol) ≈ 0.703 mol Ca

Since the stoichiometric ratio is 1:2, the moles of water required will be double the moles of calcium.

Moles of water = 2 × 0.703 mol Ca = 1.406 mol H₂O

Now, to convert the moles of water into volume, we need to use the density of water. Since the density of water is 1.0 g/cm³, 1 mL of water is equal to 1 g.

Volume of water = 1.406 mol H₂O ≈ 1.406 mL

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you balanced the equation. You must show your work to receive full credit. H 2

O 2

(I)+ClO 2

(aq)→ClO 2
−1

(aq)+O 2

(g)

Answers

The balanced equation for the reaction is:

2H2O2(aq) + ClO2(aq) -> ClO2-1(aq) + O2(g)

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation.

Let's start with the hydrogen atoms (H). There are 2 hydrogen atoms on the left side and 4 hydrogen atoms on the right side due to the coefficient 2 in front of H2O2.

To balance the hydrogen atoms, we need to put a coefficient of 2 in front of H2O2 on the left side:

2H2O2(aq) + ClO2(aq) -> ClO2-1(aq) + O2(g)

Now, let's balance the oxygen atoms (O). There are 4 oxygen atoms on the left side (2 from H2O2 and 2 from ClO2) and 4 oxygen atoms on the right side (2 from ClO2-1 and 2 from O2). The oxygen atoms are already balanced.

Finally, let's balance the chlorine atom (Cl). There is 1 chlorine atom on the left side (from ClO2) and 1 chlorine atom on the right side (from ClO2-1). The chlorine atom is already balanced.

Therefore, the balanced equation is:

2H2O2(aq) + ClO2(aq) -> ClO2-1(aq) + O2(g)

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Answer the following questions BEFORE the lab session and submit to your instructor upon entry into the lab. 1. What is the difference between a molecular formula and a structural formula?

Answers

A molecular formula reveals the types and quantities of atoms in a compound but lacks information about their arrangement. In contrast, a structural formula visually represents the connectivity and spatial arrangement of atoms within a molecule, offering a more detailed understanding of its structure.

A molecular formula is a concise representation of the types and numbers of atoms present in a molecule.

It provides information about the elemental composition of a compound but does not reveal the arrangement of atoms within the molecule.

For example, the molecular formula for glucose is [tex]C_6H_{12}O_6[/tex], indicating that it contains six carbon (C) atoms, twelve hydrogen (H) atoms, and six oxygen (O) atoms. However, it doesn't specify how these atoms are connected.

On the other hand, a structural formula provides more detailed information about the connectivity of atoms within a molecule.

It represents the bonds between atoms and the spatial arrangement of these bonds.

It gives a visual depiction of how the atoms are arranged and connected, providing a more comprehensive understanding of the molecule's structure.

Using the example of glucose, its structural formula shows how the carbon, hydrogen, and oxygen atoms are bonded together in a specific arrangement.

In summary, while a molecular formula provides information about the elemental composition of a compound, a structural formula goes further by illustrating the specific arrangement and connectivity of atoms within the molecule.

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6. A 50mM Tris buffer of pH7.8 is sitting on the shelf at room temperature (22 ∘
C). What will be the pH of this Tris buffer if it is to be cooled and used in an experiment at 4 ∘
C ? 7. Using the graph that you plotted for glycine titration, what are the pKa values for glycine? Compare your values with those from the literature and other students. What are the percentage errors? 8. What is the pH at the isoelectric point of glycine?

Answers

The pH of a Tris buffer decreases when cooled, the pKa values for glycine can be determined by comparing with literature values, and the isoelectric point of glycine represents the pH with no net charge.

6. The pH of the Tris buffer will slightly decrease when cooled to 4 °C due to the temperature effect on the ionization constant of water. The exact pH change can be calculated using the Henderson-Hasselbalch equation.

7. The pKa values for glycine can be determined by analyzing the inflection points on the titration curve. Compare the calculated pKa values with the literature values and calculate the percentage errors to assess the accuracy of the experiment.

8. The isoelectric point of glycine is the pH at which it has no net charge. This occurs when the number of positive and negative charges on glycine is equal. The pH at the isoelectric point can be calculated based on the pKa values of its ionizable groups.

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Calculate the pH of a solution formed when 87.55 mL of 0.5532MCsOH is titrated with 95.01 mL of 0.702MHI.

Answers

The pH of a solution is 13.424, formed when CsOH is titrated with HI.

Given information,

For CsOH,

Volume, v = 0.08755 L

Concentration = 0.5532 M

For HI

Volume, v = 95.01 L

Concentration = 0.702 M

The number of moles of CsOH and HI,

moles = volume × concentration,

moles of CsOH = volume  × concentration

= 0.08755  × 0.5532

= 0.04841 moles

moles of HI = volume  × concentration

= 0.09501  × 0.702

= 0.06671 moles

volume of resulting solution = volume of CsOH + volume of HI

V= 0.08755  + 0.09501

V = 0.18256 L

moles of OH⁻ ions = moles of CsOH / volume of solution

= 0.04841 / 0.18256

= 0.26518 M

To determine the pOH, the negative logarithm (base 10) of the OH- concentration:

pOH = -log10(0.26518)

pOH  ≈ 0.576

The pH is,

pH = 14 - pOH

pH = 14 - 0.576

pH ≈ 13.424

Hence, the pH is 13.424.

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In a 10 gram geological sample an experiment was conducted and the sample was serially diluted 1000 times. Calculate the BPG for the sample if there are 100 CFU's present. Also state whether the sample Is contaminated or not? (2) Also, if any similar 3 another Experiments if BPG are 12000,24000, 55000. Then can you predict in which phase the microbes are in the 4th experiment.

Answers

Regarding the prediction of the microbial phase in the fourth experiment based on the BPG values from the previous three experiments (BPG: 12000, 24000, 55000), it is not possible to make a direct prediction of the microbial phase.

To calculate the BPG (Most Probable Number) for the sample, we need to know the dilution factor used at each step of the serial dilution. However, in this case, we have been given the final dilution, which is 1000 times.

The BPG can be calculated using the formula:

BPG = Number of positive wells / Total number of wells

In this scenario, if the sample was serially diluted 1000 times and there are 100 CFUs (Colony-Forming Units) present, it means that the CFUs are present in the last dilution, which is 1 in 1000.

So, the number of positive wells would be 1 (since the CFUs are present) and the total number of wells would be 1000.

BPG = 1/1000 = 0.001

Now, to determine if the sample is contaminated or not, we need to compare the BPG value to the acceptable threshold for contamination. The threshold for contamination varies depending on the specific experiment, industry, or guidelines being followed. Without this information, it is not possible to determine if the sample is contaminated or not.

The BPG value alone does not provide information about the specific phase of microbial growth. Additional information about the growth conditions, duration of the experiments, and specific microorganisms being studied would be necessary to make predictions about the microbial phase in the fourth experiment.

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2. CCC Patterns Use the figure to compare the melting points of the metals in Groups 1
and 2. Describe the general pattern in the relationship between a metal's position in
these two groups and its melting point.

Answers

In Groups 1 and 2 of the periodic table, the melting points of metals generally decrease as you move down the group. This trend is known as a general pattern in the relationship between a metal's position in these groups and its melting point.

Group 1 consists of alkali metals (Li, Na, K, etc.), and Group 2 consists of alkaline earth metals (Be, Mg, Ca, etc.). As we move down these groups, the number of electron shells increases, and the atomic radius of the metals also increases. This increase in atomic radius leads to weaker metallic bonding between the atoms.

The melting point of a metal is influenced by the strength of the metallic bonds. Metallic bonding occurs when metal atoms share their outer electrons freely, forming a "sea" of delocalized electrons. These delocalized electrons are responsible for the high electrical conductivity and malleability of metals. The stronger the metallic bonding, the higher the melting point of the metal.

As we move down Groups 1 and 2, the increased atomic radius results in a greater distance between the metal ions in the crystal lattice. This increased distance weakens the metallic bonding, making it easier to break the bonds and convert the solid metal into a liquid state. Therefore, metals lower in Groups 1 and 2 have lower melting points compared to metals higher up in the groups.

Additionally, the increased number of electron shells also leads to greater shielding of the outer electrons from the positive charge of the nucleus. This reduced attraction between the outer electrons and the nucleus further contributes to the weaker metallic bonding and lower melting points as we move down the groups.

In summary, the general pattern in the relationship between a metal's position in Groups 1 and 2 and its melting point is that the melting points decrease as we move down the groups due to the increasing atomic radius, weaker metallic bonding, and reduced attraction between the outer electrons and the nucleus.

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Which statement defines the heat capacity of a sample?
the temperature of a given sample
the temperature that a given sample can withstand
the quantity of heat that is required to raise the sample’s temperature by 1°C (or Kelvin)
the quantity of heat that is required to raise 1 g of the sample by 1°C (or Kelvin) at a given pressure

Answers

Answer:

Explanation:4

Complete and balance each of the following equations for
acid-base reactions.
1. H2SO4(aq)+KOH(aq)→ Express your answer as a chemical
equation. Identify all of the phases in your answer.
2. HClO4(aq

Answers

The balanced equation for the acid-base reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH) can be written as H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l).

In this reaction, sulfuric acid (H2SO4) reacts with potassium hydroxide (KOH) to form potassium sulfate (K2SO4) and water (H2O). The coefficients in the balanced equation indicate the stoichiometric ratios between the reactants and products.

Note: (aq) represents an aqueous solution, and (l) represents a liquid phase.

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All organic compounds contain carbon and hydrogen. They can also include nitrogen, phosphorus, sulfur, oxygen and halogens (fluorine, chlorine, bromine and iodine). If there are charged entities, there may also be associated cations (sodium, calcium, and potassium). Provide the electron configuration and the number of valence shell electrons for each of the following elements (note that some of them are charged!). a. Oxygen electron configuration: H val. shell electrons: b. Fluorine electron configuration: # val. shell electrons: c. Cl −
electron configuration: # val. shell electrons: d. Magnesium electron configuration: # val. shell electrons: e. Mg g

+ electron configuration: # val. shell electrons:

Answers

In the electron configurations provided, the superscript numbers represent the number of electrons present in each energy level (shell), while the valence shell electrons refer to the electrons present in the outermost energy level (valence shell).umber of electrons present in each energy level (shell), while the valence shell electrons refer to the electrons present in the outermost energy level (valence shell).

a. Oxygen:

Electron configuration: 1s^2 2s^2 2p^4

Number of valence shell electrons: 6

b. Fluorine:

Electron configuration: 1s^2 2s^2 2p^5

Number of valence shell electrons: 7

c. Cl^-

Electron configuration: 1s^2 2s^2 2p^6 3s^2 3p^6

Number of valence shell electrons: 8

d. Magnesium:

Electron configuration: 1s^2 2s^2 2p^6 3s^2

Number of valence shell electrons: 2

e. Mg^2+

Electron configuration: 1s^2 2s^2 2p^6

Number of valence shell electrons: 0

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A car tire has a volume of 32.2 L with a pressure of 34.5 psi when the temperature is 27°C. If the temperature increases to 43° and the volume decreases to 31.04, What is the new pressure?

Answers

The new pressure is 36.09 psi.

Answer:

Using the combined gas law:

(P1 x V1) / T1 = (P2 x V2) / T2

where:

P1 = 34.5 psi (initial pressure)

V1 = 32.2 L (initial volume)

T1 = 27°C + 273.15 = 300.15 K (initial temperature in Kelvin)

V2 = 31.04 L (final volume)

T2 = 43°C + 273.15 = 316.15 K (final temperature in Kelvin)

Solving for P2:

(P1 x V1 x T2) / (V2 x T1) = P2

(34.5 psi x 32.2 L x 316.15 K) / (31.04 L x 300.15 K) = P2

P2 = 37.2 psi

Therefore, the new pressure in the tire is 37.2 psi when the temperature increases to 43°C and the volume decreases to 31.04 L.

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