the molal freezing point depression constant [tex](\(K_f\))[/tex] of liquid X is approximately [tex]\(4.13 \, \text{°C/molal}\)[/tex].
To calculate the molal freezing point depression constant (\(K_f\)) of liquid X, we can use the formula:
[tex]\[ \Delta T_f = K_f \times m \][/tex]
Where:
- [tex]\( \Delta T_f \)[/tex] is the freezing point depression (given as [tex]\(0.4^\circ \text{C} - (-0.5^\circ \text{C}) = 0.9^\circ \text{C}\)[/tex]),
- [tex]\( m \)[/tex] is the molality of the solution,
- [tex]\( K_f \)[/tex] is the molal freezing point depression constant.
First, we need to calculate the molality of the solution, which is defined as the number of moles of solute per kilogram of solvent.
Given:
- Mass of urea [tex](\( \text{NH}_2\text{CO} \))[/tex] = 5.90 g
- Mass of liquid X = 450.0 g
We need to find the moles of urea first:
[tex]\[ \text{moles of urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} \][/tex]
The molar mass of urea [tex](\( \text{NH}_2\text{CO} \))[/tex] is the sum of the molar masses of nitrogen, hydrogen, carbon, and oxygen:
[tex]\[ \text{molar mass of urea} = 14.01 + 2(1.01) + 12.01 + 16.00 = 60.03 \, \text{g/mol} \][/tex]
[tex]\[ \text{moles of urea} = \frac{5.90 \, \text{g}}{60.03 \, \text{g/mol}} = 0.0983 \, \text{mol} \][/tex]
Now, we can calculate the molality of the solution:
[tex]\[ \text{molality} = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} \][/tex]
[tex]\[ \text{molality} = \frac{0.0983 \, \text{mol}}{0.450 \, \text{kg}} = 0.218 \, \text{mol/kg} \][/tex]
Now, we can rearrange the formula for [tex]\(K_f\)[/tex] and solve for it:
[tex]\[ K_f = \frac{\Delta T_f}{m} \][/tex]
[tex]\[ K_f = \frac{0.9^\circ \text{C}}{0.218 \, \text{mol/kg}} \][/tex]
[tex]\[ K_f \approx 4.13 \, \text{°C/molal} \][/tex]
Therefore, the molal freezing point depression constant [tex](\(K_f\))[/tex] of liquid X is approximately [tex]\(4.13 \, \text{°C/molal}\)[/tex].
The molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X is 4.124 °C/(mol/kg).
To calculate the molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X, we can use the formula for freezing point depression:
[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]
where:
[tex]\( \Delta T_f \)[/tex] is the freezing point depression, which is the difference between the normal freezing point of the solvent and the freezing point of the solution.
i is the van 't Hoff factor, which is the number of moles of solute particles per mole of solute dissolved. For urea, i is typically 1 because urea does not dissociate in solution.
[tex]\( K_f \)[/tex] is the molal freezing point depression constant for the solvent.
m is the molality of the solution, which is the number of moles of solute per kilogram of solvent.
First, we calculate the freezing point depression [tex]\( \Delta T_f \)[/tex]:
[tex]\[ \Delta T_f = T_f^0 - T_f = 0.4C - (-0.5C) = 0.9C \][/tex]
Next, we need to calculate the molality m of the solution. The molar mass of urea [tex]\( NH_2CONH_2 \)[/tex] is 60.06 g/mol. The number of moles of urea is:
[tex]\[ n_{urea} = \frac{\text{mass of urea}}{\text{molar mass of urea}} = \frac{5.90 \text{ g}}{60.06 \text{ g/mol}} \] \[ n_{urea} = 0.0982 \text{ mol} \][/tex]
The molality m is:
[tex]\[ m = \frac{n_{urea}}{\text{mass of solvent in kg}} = \frac{0.0982 \text{ mol}}{0.450 \text{ kg}} \] \[ m = 0.2182 \text{ mol/kg} \][/tex]
Now we can rearrange the freezing point depression formula to solve for [tex]\( K_f \)[/tex]:
[tex]\[ K_f = \frac{\Delta T_f}{i \cdot m} \] Since \( i = 1 \) for urea, we have: \[ K_f = \frac{0.9C}{1 \cdot 0.2182 \text{ mol/kg}} \] \[ K_f = 4.124 \text{ C/(mol/kg)} \][/tex]
Therefore, the molal freezing point depression constant [tex]\( K_f \)[/tex] of liquid X is 4.124 °C/(mol/kg).
A 250 mL container of CO2 exerting a pressure of 1.00 atm is connected through a valve to a 500 mL container of O2 exerting a pressure of 2.00 atm.
When the valve is opened, the gases mix, forming a 750 mL mixture of CO2 and O2.
What is the total pressure of this mixture?
Explanation:
The given data is as follows.
[tex]P_{1}[/tex] = 1 atm, [tex]P_{2}[/tex] = 2 atm
[tex]V_{1}[/tex] = 250 ml, [tex]V_{2}[/tex] = 500 ml
Total volume = [tex]V_{1} + V_{2}[/tex] = 750 ml
Therefore, total pressure will be as follows.
[tex]P_{total} V_{total} = P_{1}V_{1} + P_{2}V_{2}[/tex]
[tex]P_{total} = \frac{P_{1}V_{1} + P_{2}V_{2}}{V_{total}}[/tex]
= [tex]\frac{1 atm \times 250 ml + 2 atm \times 500 ml}{750 ml}[/tex]
= 1.66 atm
Thus, we can conclude that total pressure of the given mixture is 1.66 atm.
The synthesis of CH3OH from CO and H2 is represented by the equation below.
CO + 2 H2 ---> CH3OH ΔH < 0
Which of the following statements is true about the bond energies in this reaction?
The energy absorbed as the bonds in the reactants is broken is greater than the energy released as the bonds in the products are formed.
The energy released as the bonds in the reactants is broken is greater than the energy absorbed as the bonds in the products are formed.
The energy released as the bonds in the reactants is broken is less than the energy absorbed as the bonds in the products are formed.
The energy absorbed as the bonds in the reactants is broken is less than the energy released as the bonds in the products are formed.
The energy absorbed as the bonds in the reactants are broken is less than the energy released as the bonds in the products are formed. This is indicative of an exothermic reaction.
Explanation:In the given reaction, CO + 2 H2 ---> CH3OH ΔH < 0, this is an exothermic reaction as indicated by the negative delta H value. The negative ΔH value signifies that the total energy released when the new bonds are formed in the product (CH3OH) is more than the total energy absorbed to break the bonds in the reactants (CO and H2).
Therefore, the correct statement is: The energy absorbed as the bonds in the reactants are broken is less than the energy released as the bonds in the products are formed.
This corresponds to the fact that in an exothermic reaction, the products are at a lower energy level than the reactants, therefore, energy is released in the reaction.
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The correct statement is that the energy absorbed as the bonds in the reactants is broken is less than the energy released as the bonds in the products are formed, indicating an exothermic reaction.
The correct statement is: The energy absorbed as the bonds in the reactants is broken is less than the energy released as the bonds in the products are formed.
In this specific reaction, energy is required to break a C-O triple bond and two H-H single bonds in the reactants. Energy is then released as three C-H single bonds, a C-O single bond, and an O-H single bond are formed in the product, CH₃OH.
Since ΔH is negative, it indicates that the total energy released in bond formation is greater than the total energy absorbed in bond breaking, reaffirming that the reaction is exothermic.
A silver nitrate, AgNO3, solution of unknown concentration was discovered in the lab. To determine the concentration of the solution, a concentration cell was set up with the unknown solution in the anode compartment and a 1.0 M AgNO3 solution in the cathode compartment. The cell had a potential (E) of +0.045 V at 25°C. What is the concentration of silver in the unknown solution?
Final answer:
The concentration of silver in the unknown solution can be determined using the Nernst equation with the given cell potential of +0.045 V. By rearranging and substituting known values into the Nernst equation, we can calculate the unknown silver ion concentration at the anode.
Explanation:
To determine the concentration of silver in the unknown AgNO3 solution using a concentration cell, we can apply the Nernst equation. The reaction in a concentration cell is Ag+ (aq) → Ag+ (aq), which proceeds with a transfer of electrons from one compartment to the other. Since there is a potential difference (E) of +0.045 V, we can calculate the unknown concentration at the anode using the Nernst equation:
E = E°_cell - (RT/nF) ln(Q)
Where:
E is the cell potentialE°_cell is the standard cell potential, which is 0 V in concentration cellsR is the gas constant (8.314 J/(mol·K))T is the temperature in Kelvin (298 K)n is the number of electrons transferred per mole of reaction (1 for silver)F is the Faraday constant (96485 C/mol)Q is the reaction quotient, which is equal to [Ag+] anode / [Ag+] cathodeRearranging the Nernst equation and solving for Q will provide us the ratio of the Ag+ concentration in the anode to that in the cathode. By substituting the measured E value, R, T, n, and F into the Nernst equation and knowing that [Ag+] cathode is 1.0 M, we can solve for the unknown [Ag+] anode concentration. Finally, by taking the anti-logarithm of the result, we can find the concentration of silver in the unknown solution.
Which statement about intermolecular forces is true?
Group of answer choices
A. Only occur in ionic bonds.
B. They have to be overcome to decompose a substance.
C. These forces hold atoms together in a molecule.
D. They are responsible for the physical properties of matter.
Intramolecular forces describe those forces that are found within the molecules and are shared by the electronic bonds. They hold the atoms and molecules together.
They are responsible for the holding of atoms together in a single molecule. Thus they also include electromagnetic forces of attraction and repulsion.Hence the option C is correct.
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A sample of 0.300 mg pure chromium was added to excess hydrochloric acid to form a 10.0 mL aqueous solution of a chromium (III) salt, which has a violet hue. Exactly 1.00 mL of the resulting solution was analyzed using a spectrophotometer in a 1.00-cm cell at 575 nm, and the percent transmittance for the solution was 62.5%. What is the extinction coefficient?
extinction coefficient (ε) = 347 L·mol⁻¹·cm⁻¹
Explanation:
The chemical reaction between chromium (Cr) and hydrochloric acid (HCl):
2 Cr + 6 HCl → 2 CrCl₃ + 3 H₂
number of moles = mass / molar weight
number of moles of Cr = 0.3 × 10⁻³ (g) / 52 (g/mole)
number of moles of Cr = 5.77 × 10⁻⁶ moles
From the chemical reaction we see that 2 moles of Cr will produce 2 moles of CrCl₃ so 5.77 × 10⁻⁶ moles of Cr will produce 5.77 × 10⁻⁶ moles of CrCl₃.
molar concentration = number of moles / volume (L)
molar concentration of CrCl₃ = 5.77 × 10⁻⁶ / 10 × 10⁻³
molar concentration of CrCl₃ = 5.77 × 10⁻⁴ moles / L
Now we need to transform percent transmittance (%T) in absorbance (A) using the following formula:
A = 2 - log (%T)
A = 2 - log (62.5)
A = 2 - 1.8
A = 0.2
We know that absorbance (A) is defined in respect with extinction coefficient (ε), cell length (l) and concentration (c):
A = εlc
ε = A / lc
ε = 0.2 / (1 × 5.77 × 10⁻⁴)
ε = 0.0347 × 10⁴
ε = 347 L·mol⁻¹·cm⁻¹
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The extinction coefficient for a chromium (III) salt solution can be determined using spectrophotometry, if the absorbance, path length, and concentration are known. The formula A = εcl, integrates these variables, but without the concentration of the chromium (III) in the analysed solution, it is impossible to accurately calculate the extinction coefficient.
Explanation:The question pertains to the measurement of the extinction coefficient for a chromium (III) salt solution using spectrophotometry. The percent transmittance provided allows us to calculate the absorption of light by the solution, which is related to the extinction coefficient. Absorbance can be calculated using the formula A = -log10(T), where T represents transmittance.
The formula for the extinction coefficient, ε, is formulated as A = εcl and includes absorbance (A), path length (l), and concentration (c) variables. However, we don't have the concentration of chromium (III) in the 1.00 mL of the solution we're examining. Therefore, we can't provide an accurate answer to this question without that piece of information.
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Which of the following chemical reactions has positive entropy change?
A. N2(g) + 3 H2(g) \longrightarrow ⟶ 2NH3(g)
B. Br2(g) \longrightarrow ⟶ Br2(l) N2O4(g)
C. \longrightarrow ⟶ 2 NO2(g)
D. NaOH(aq) + CO2(g) \longrightarrow ⟶ NaHCO3(aq)
Answer: [tex]N_2O_4(g)\longrightarrow 2NO_2(g)[/tex]
Explanation:
Entropy is the measure of randomness or disorder of a system. If a system moves from an ordered arrangement to a disordered arrangement, the entropy is said to decrease and vice versa.
[tex]\Delta S[/tex] is positive when randomness increases and [tex]\Delta S[/tex] is negative when randomness decreases.
a) [tex]N_2(g)+3H_2(g)\longrightarrow 2NH_3(g)[/tex]
4 moles of gas are converting to moles of gas, hence the entropy decreases.
b) [tex]Br_2(g)\longrightarrow Br_2(l)[/tex]
1 mole of gas is converting to 1 mole of liquid, hence the entropy decreases.
c) [tex]N_2O_4(g)\longrightarrow 2NO_2(g)[/tex]
1 mole of gas is converting to 2 moles of gas, hence the entropy increases.
d) [tex]NaOH(aq)+CO_2(g)\longrightarrow NaHCO_3(aq)[/tex]
1 mole of gas is disappearing , hence the entropy decreases.
Thus [tex]N_2O_4(g)\longrightarrow 2NO_2(g)[/tex] has positive entropy change.
Given only the following data, what can be said about the following reaction?3H2(g) + N2(g)---> 2NH3(g) ΔH=-92kJA.) The enthalpy of products is greater than the enthalpy of reactantsB.) The total bond energies of products are greater than the total bond energies of reactantsC.) The reaction is very fastD.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia
Answer:
D.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.
Explanation:
For the reaction:
3H₂(g) + N₂(g) → 2NH₃(g)
The enthalpy change is ΔH = -92kJ
This enthalpy change is defined as the enthalpy of products - the enthalpy of reactants. As the enthalpy is <0, The enthalpy of products is lower than the enthalpy of reactants.
Also, it is possible to obtain the enthalpy change from the bond energies of products - bond energies of reactants, thus, The total bond energies of products are lower than the total bond energies of reactants.
The rate of the reaction couldn't be determined using ΔH.
As the bond energy of ammonia is lower than bonds of nitrogen and hydrogen, D. Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.
I hope it helps!
Final answer:
The correct statement is that the enthalpy of products is less than the enthalpy of reactants, indicating that heat is released and bond energies are lower in the products than in the reactants. Reaction speed and relative bond stability cannot be inferred from the given data.
Explanation:
From the data provided, several conclusions about the chemical reaction forming ammonia can be drawn. It is evident that the reaction is exothermic since the enthalpy change (ΔH) is negative, meaning that heat is released when hydrogen and nitrogen react to form ammonia. Specifically, ΔH is -92kJ, which indicates that the enthalpy of the reactants is higher than the enthalpy of the products, and thus, the correct statement is:
A.) The enthalpy of products is less than the enthalpy of reactants because energy is released in the formation of ammonia from nitrogen and hydrogen.
Moreover, a negative ΔH suggests that the total bond energies of products are less than that of reactants because forming stronger bonds in the products releases energy. This also aligns with the principle that an exothermic reaction strengthens product bonds compared to reactant bonds, as seen in the enthalpy change provided.
However, the claim that the reaction is very fast (C.) cannot be substantiated by the given data. In fact, the synthesis of ammonia from nitrogen and hydrogen is known to be slow under ambient conditions and requires specific industrial processes, such as the Haber process, to enhance its rate. As for the last option (D.), while nitrogen and hydrogen do form very stable bonds, the statement that they are more stable compared to the bonds in ammonia is not specified by the given ΔH value and requires additional bond energy data for a direct comparison.
Interstellar space has an average temperature of about 10 K and an average density of hydrogen atoms of about one hydrogen atom per cubic meter. Calculate the mean free path of hydrogen atoms in interstellar space. Take d = 100 pm for a hydrogen atom.
Answer:
The mean free path is [tex]0.0000373631 m[/tex]
Explanation:
The formula for mean free path is :
λ = [tex]\frac{V}{\sqrt{2}\pi d^{2}N }[/tex]
where,
λ - is the mean free path distance
V - volume of the gas
d - the diameter of the molecule
N - number of molecules.
now ,
density [tex]D[/tex] = [tex]\frac{mass}{volume}[/tex] = [tex]\frac{M}{V}[/tex] ;
mass of the gas = (number of molecules)[tex]*[/tex](mass of one molecule) ;
as it's atomic hydrogen
[tex]M = N*m \\m=1.66*10^{-24}\\M=N*1.66*10^{-24}[/tex]
∴
[tex]D[/tex] = [tex]\frac{N*1.66*10^{-24}}{V}[/tex]
∴
[tex]\frac{V}{N*1.66*10^{-24}} = \frac{1}{ D}[/tex]
⇒ λ = [tex]\frac{1}{\sqrt{2}\pi d^{2}D }[/tex]
⇒ λ = [tex]\frac{1.66*10^{-24}}{\sqrt{2}\pi (100*10^{-12})^{2}*1 }[/tex]
⇒ λ = [tex]0.0000373631 m[/tex]
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN−. Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution?
Answer:
The concentration of Cd2+ is 0.0175 M
The concentration of Mn2+ is 0.0305 M
Explanation:
Step 1: Data given
A 50.0 mL sample contains Cd2+ and Mn2+
volume of 0.05 M EDTA = 56.3 mL
Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.
Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+
Step 2: Calculate mole ratio
The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+ in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed in the back titration with Ca2+:
Step 3: Calculate total mol of EDTA
Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA
Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA
Step 4: Calculate total moles of CD2+ and Mn2+
So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol
Step 5: Calculate remaining moles of Cd2+
The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction with cyanide:
Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.
Step 6: Calculate remaining moles of Mn2+
The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+
Step 7: Calculate initial concentrations
The initial concentrations must have been:
(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+
(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+
The concentration of Cd2+ is 0.0175 M
The concentration of Mn2+ is 0.0305 M
In the laboratory a student determines the specific heat of a metal. He heats 19.5 grams of copper to 98.27 °C and then drops it into an insulated cup containing 76.3 grams of water at 24.05 °C. When thermal equilibrium is reached, he measures the final temperature to be 25.69 °C. Assuming that all of the heat is transferred to the water, he calculates the specific heat of copper to be__________ J/g°C.
Answer:
The specific heat of copper is 0.37 J/g°C
Explanation:
Step 1: Data given
Mass of copper = 19.5 grams
Initial temperature of copper = 98.27 °C
Mass of water = 76.3 grams
Initial temperature of water = 24.05 °C
Final temperature of water and copper = 25.69 °C
Step 2: Calculate specific heat of copper
Qgained = -Qlost
Q = m*c*ΔT
Qwater = -Qcopper
m(water) * c(water) * ΔT(water) = - m(copper) * c(copper) *ΔT(copper)
⇒ with m(water) = 76.3 grams
⇒ with c(water) = 4.184 J/g°C
⇒ with ΔT(water) = T2-T1 = 25.69 - 24.05 = 1.64
⇒ with m(copper) = 19.5 grams
⇒ with c(copper) = TO BE DETERMINED
⇒ with ΔT(copper) = T2-T1 = 25.69 - 98.27 = -72.58
76.3 * 4.184 * 1.64 = - 19.5 * c(copper) * -72.58
523.552 = 1415.31 * c(copper)
c(copper) = 0.37 J/g°C
The specific heat of copper is 0.37 J/g°C
Calculate the lattice energy for CaCl2 from the following information: Energy needed to vaporize one mole of Ca(s) is 192kJ. For calcium the first ionization energy is 589.5kJ/mol and the second ionization energy is 1146kJ/mol. The electron affinity of Cl is-348kJ/mol. The bond energy of Cl2 is 242.6kJ/mol of CI-Cl bonds. The standard heat of formation of CaCl2 is -795kJ/mol. (include the sign and your numerical answer, do not include units, do not use scientific notation, and round your answer to 1 decimal point)
Calculate the total energy for individual process, including energy to vaporize Ca(s), first and second ionization energies of Ca, bond energy of Cl2 and electron affinity of Cl. The lattice energy is the result of subtracting standard heat of formation of CaCl2 from the total energy calculated, which equals to 2635.9 kJ.
Explanation:The lattice energy for CaCl2 can be calculated using the given information and by understanding that the process of forming a salt like CaCl2 involves several different energies. First, there is the energy required to vaporize Ca(s), then the energy for first and second ionization of the calcium atom, followed by the electron affinity for Cl and the bond energy of Cl2.
To start calculating, first add the energy to vaporize Ca(s), the first and second ionization energies for Ca: 192kJ + 589.5kJ + 1146kJ = 1927.5kJ. When 2 Cl react with Ca, this process includes the bond energy for Cl2 and twice the electron energy for Cl: 242.6kJ + 2 * -348kJ = -453.4kJ. To calculate the energy required to form lattice, subtract the standard heat of formation of CaCl2 from the total energy calculated: 1927.5kJ - (-453.4kJ + -795kJ) = 2635.9 kJ
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Write a net ionic equation to show how triethylamine, (C2H5)3N, behaves as a base in water. The substance 2-methylpiperidine is a weak nitrogenous base like ammonia. Complete the following equation that shows how 2-methylpiperidine reacts when dissolved in water.
You may find bellow the net ionic equations.
Explanation:
Net ionic equation of triethylamine in water:
(C₂H₅)₃N (l) + H₂O (l) → (C₂H₅)₃NH⁺ (aq) + OH⁻ (aq)
where:
(l) - liquid
(aq) - aqueous
You may find in the attached picture the ionization of 2-methylpiperidine in water.
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Triethylamine and 2-methylpiperidine act as weak bases in water, accepting protons to form hydroxide ions and their respective conjugate acids. This is an example of base ionization. It was not possible to provide an exact equation for 2-methylpiperidine without a specific structure.
Explanation:When triethylamine, also a weak nitrogen base like ammonia, is dissolved in water, it accepts a proton from a water molecule to form a hydroxide ion (OH-) and an ethylammonium ion ((C2H5)3NH+). This is represented by the equation: (C2H5)3N(aq) + H2O(l) ⇌ (C2H5)3NH+(aq) + OH-(aq).
Similarly, 2-methylpiperidine being a weak base, will also react with water to form hydroxide ions (OH-) and 2-methylpiperidinium ions. However, without a specific structure for 2-methylpiperdine, it's challenging to write the exact equation.
These are examples of base ionization, where a base reacts with water to produce hydroxide ions and a conjugate acid. Both triethylamine and 2-methylpiperidine act as Brønsted bases, accepting protons from water molecules.
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For which of the following reactions will the reactant experience the largest degree of decomposition
upon reaching equilibrium at 500 K?
A) 2NO2F(g)--> 2NO2(g) + F2(g); Kp = 6.6 10-22
B) 2SO3(g)--> 2SO2(g) + O2(g); Kp = 1.3 10-5
C) 2NOF(g)--> 2NO(g) + F2(g); Kp = 1.2 10-26
D) 2NOCl(g)--> 2NO(g) + Cl2(g); Kp = 1.7 10-2
E) 2NO2(g)--> 2NO(g) + O2(g); Kp = 5.9 10-5
Answer:
D) 2 NOCl(g) ⇄ 2 NO(g) + Cl₂(g); Kp = 1.7 × 10⁻²
Explanation:
In order to compare the degree of decomposition of these reactions, we have to compare the equilibrium constant Kp. Kp is equal to the partial pressure of the products raised to their stoichiometric coefficients divided by the partial pressure of the reactants raised to their stoichiometric coefficients. The higher the Kp, the more products and fewer reactants at equilibrium. Among these reactions, D is the one that has the highest Kp, therefore the one experiencing the largest degree of decomposition.
Here is the information: A 1.2516 gram sample of a mixture of CaCO3 and Na2SO4 was analyzed by dissolving the sample and adding C2O42- to completely precipitate the Ca2+ as CaC2O4. The CaC2O4 was dissolved in sulfuric acid and the resulting H2C204 was titrated with a standard KMnO4 solution.
This is the balanced equation: 3H2C2O4 + 2H+ + MnO4- ---> Mn2+ + 4H20 + 6CO2
Another part: The titration of the H2C2H4 obtained required 35.62 milliliters of .1092 molar MnO4- solution. Calculate the number of moles of H2C2O4 that reacted with the MnO4-. i think the correct answer is .01167 mol H2C2O4. This is not my question.
Next Part: Calculate the number of moles of CaCO3 in the original sample. Based on the answer to the part above, the answer is .003883 mol CaCO3 (i think this is correct, however this is not my question)
Calculate the percentage by weight of CaCO3 in the original sample.
FInd this based off of the .01167 mol H2C2O4 and the .003883 mol CaCO3. Basically just use the answers above to solve for this.
Answer:
93,32 % (w/w)
Explanation:
The Ca²⁺ of CaCO₃ was completely precipitate to CaC₂O₄ that results in H₂C₂O₄. The titration of this one is:
3H₂C₂O₄ + 2H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O + 6CO₂
As you required 35,62mL of 0,1092M MnO₄⁻, the moles of H₂C₂O₄ are:
0,03562L×[tex]\frac{0,1092M}{L}[/tex] =
3,890x10⁻³mol of MnO₄⁻×[tex]\frac{3molH_{2}C_{2}O_{4}}{1mol MnO_{4}^-} [/tex] = 0,01167 mol of H₂C₂O₄
The number of moles of CaCO₃ are the same number of moles of H₂C₂O₄ because every Ca²⁺ was converted in CaC₂O₄ that was converted in H₂C₂O₄. That means: 0,01167 mol of CaCO₃
0,01167 mol of CaCO₃ are:
0,01167 mol of CaCO₃×[tex]\frac{100,0869g}{1mol}[/tex] = 1,168 g of CaCO₃
As the mass of the initial mixture is 1,2516 g, the percentage by weight of CaCO₃ is:
[tex]\frac{1,168g}{1,2516g}[/tex]×100 = 93,32 % (w/w)
I hope it helps!
A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y , that is, X more readily reacts to form a cation than Y does. Classify the descriptions by whether they apply to the X or Y electrode.
(a) anode
(b) cathode
(c) electrons in the wire flow toward
(d) electrons in the wire flow away
(e) cations from salt bridge flow toward
Answer:
(a) X electrode
(b) Y electrode
(c) Y electrode
(d) X electrode
(e) Y electrode
Explanation:
A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y, that is, X more readily reacts to form a cation than Y does.
In the X electrode occurs the oxidation whereas in the Y electrode occurs the reduction.
Oxidation: X(s) → X⁺ⁿ(aq) + n e⁻
Reduction: Y⁺ˣ(aq) + x e⁻ → Y(s)
Classify the descriptions by whether they apply to the X or Y electrode.
(a) anode. Is where the oxidation takes place (X electrode).
(b) cathode. Is where the reduction takes place (Y electrode).
(c) electrons in the wire flow toward. Electrons in the wire flow toward the cathode (Y electrode).
(d) electrons in the wire flow away. Electrons in the wire flow away from the anode (X electrode).
(e) cations from salt bridge flow toward. Cations from the salt bridge flow toward the cathode (Y electrode) to maintain the electroneutrality.
In a galvanic cell, the more reactive metal X is the anode where electrons flow away from, and the less reactive metal Y is the cathode where electrons flow toward. Cations from the salt bridge migrate toward the anode, maintaining electrical neutrality.
Explanation:In a galvanic (voltaic) cell, the generic metal X is more reactive than Y, indicating that X acts as the anode where oxidation takes place, and Y acts as the cathode where reduction occurs. The characteristics of the electrodes in a voltaic cell are as follows:
(a) Anode - X electrode since it is more reactive and readily oxidized.(b) Cathode - Y electrode since it is less reactive and reduction occurs here.(c) Electrons in the wire flow toward the Y electrode (cathode).(d) Electrons in the wire flow away from the X electrode (anode).(e) Cations from the salt bridge flow toward the X electrode (anode) because it is the site of oxidation.These characteristics are essential to the proper functioning of the galvanic cell, ensuring the flow of electrons from the anode to the cathode, and allowing the cell to generate electricity through spontaneous redox reactions.
Thermal decomposition of a rail car load of limestone to lime and carbon dioxide requires 2.97 x 106 kJ of heat.
Convert this energy to calories.
Once you have the value, determine its log (base 10).
Answer: a) [tex]68.3\times 10^6cal[/tex]
b) [tex]log_{10}(68.3\times 10^6cal)=7.83[/tex]
Explanation:
Heat is defined as a spontaneous flow of energy from one object to another. It is measured in Joules, calories, kilo Joules etc.
These units of energy are inter convertible.
We are given:
a) Energy absorbed by limestone = [tex]2.97\times 10^6kJ[/tex]
Converting this unit of temperature into [tex]calories[/tex] by using conversion factor:
1 kJ = 239.006 calories
[tex]2.97\times 10^6kJ=\frac{239.006}{1}\times 2.97\times 10^6=68.3\times 10^6cal[/tex]
Thus the energy in calories is [tex]68.3\times 10^6cal[/tex]
b) The value of log base 10 of [tex]68.3\times 10^6cal[/tex] is:
[tex]log_{10}(68.3\times 10^6cal)=7.83[/tex]
The slowest step of a reaction mechanism is called the
A.) Elementary Step
B.) Inhibitor
C.) Rate Law
D.) Rate-Determining Step
Answer:
D.) Rate-Determining Step
Explanation:
Rate - Determining Step -
The slowest step of any of the chemical reaction helps to find the overall rate of the chemical reaction , and , hence is known as the rate - determining step .
There are two type of reaction possible , one is elementary reaction and a complex reaction .
Elementary reaction is a single step reaction , and hence , that very step determined the rate of the reaction and hence is known as the rate determining step .
And ,
In case of a complex reaction , the reaction is preceded by many steps , and hence , the slowest step among other steps is known as the rate determining step of the reaction .
A cylinder and piston assembly (defined as the system) is warmed by an external flame. The contents of the cylinder expand, doing work on the surroundings by pushing the piston outward against the external pressure.
If the system absorbs 563 J of heat and does 498 J of work during the expansion, what is the value of ΔE?
Explanation:
According to the first law of thermodynamics, energy can neither be created nor it can be destroyed. It can only be converted from one form to another.
Formula given by first law of thermodynamics is as follows.
U = q + w
The given values are as follows.
q = +563 J
w = -498 J (as work done by the system is negative)
Hence, putting the given values into the above formula and calculate the internal energy as follows.
U = q + w
= 563 J + (-498 J)
= 65 J
As internal energy is represented by [tex]\Delta U[/tex] or [tex]\Delta E[/tex]. Therefore, the value of [tex]\Delta E[/tex] for the given system is 65 J.
A liquid of density 1290 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.91 m/s and the pipe diameter d 1 is 10.5 cm . At Location 2, the pipe diameter d 2 is 16.7 cm . At Location 1, the pipe is Δ y = 9.01 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ P between the fluid pressure at Location 2 and the fluid pressure at Location 1.
Answer:
114 kPa
Explanation:
By Bernoulli's equation when a fluid flows steadily through a pipe:
P + ρ*g*y + v² = constant in the pipe, where P is the pressure, ρ is the density of the fluid, g is the gravity acceleration (9.8 m/s²), y is the high, and v the velocity.
By the continuity equation, the liquid flow must be constant in the pipe, and then:
A1*v1 = A2*v2
Where A is the area, v is the velocity, 1 is the point 1, and 2 the point 2 in the pipe. The are is the circle area: π*(d/2)². So:
π*(0.105/2)²*9.91 = π*(0.167/2)²*v2
0.007v2 = 0.027
v2 = 3.9 m/s
Then:
P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²
ρ*g*y1 - ρ*g*y2 + v1² - v2² = P2 - P1
ρ*g*Δy + v1² - v2² = ΔP
ΔP = 1290*9.8*9.01 + 9.91² - 3.9²
ΔP = 113,987.42 Pa
ΔP = 114 kPa
The balanced equation for the reaction of ammonia and oxygen is the following. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) The standard molar entropies for the products and reactants are listed below. Calculate the change in standard molar entropy for this reaction at 298.0 K and standard pressure, in J/mol·K. (Enter your answer to the tenths place. Include the sign of the value in your answer.)
Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
Final answer:
The question requests the calculation of the change in standard molar entropy for a specific chemical reaction, but specific entropy values are not provided, making it impossible to complete the calculation as asked.
Explanation:
The question asks to calculate the change in standard molar entropy (ΔS°) for the reaction of ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO) and water (H2O) at 298.0 K and standard pressure. The formula to use is ΔS° = ∑S°(products) - ∑S°(reactants). Unfortunately, the standard molar entropies (S°) for the reactants and products are not provided in the question. Normally, these values would be found in a standard thermodynamic table. Once you have these values, you would multiply the entropy of each substance by its coefficient in the balanced equation, sum these values for both reactants and products, and then subtract the sum for reactants from the sum for products. This will give you the change in entropy for the reaction.
If 44.5 mL of 0.0989M sodium hydroxide solution is needed to titrate the amount of acetic acid in a 5.00 mL sample of vinegar, what is the weight/weight percent (w/w %) of acetic acid in the vinegar.
(MM acetic acid = 60.04 g) Do not include units in your answer. Use the correct number of significant figures.
Answer:
5.28 %
Explanation:
To solve this problem we are going to calculate the moles of NaOH required to titrate the vinegar. Since this reaction is one mole NaOH to 1 mol acetic acid we will have the moles of acetic acid present in the 5 mL sample . By multiplying the number of moles of acetic acid by its molecular weight we obtain the grams of acid.
Now since we are being asked the w/w % we need the weight of solution which is not given in the problem. However we can assume that its value will be close to that of pure water and give it the value of 5 g since the density of water is one g/mL.
moles NaOH = 44.5 mL x 1L/1000 mL x 0.098 mol/L = 0.00440
mol acetic acid = 0.00440
mass acetic acid = 0.00440 mol x 60.04 g/mol = 0.264 g acetic acid
percent (w/w) = (0.264 g/ 5 g ) x 100 = 5.28 %
What is the mass percent concentration of solution containing 30 g kcl and 160 of water assume a density of 1.00g/ml for water?
Answer:
The mass percent = 15.79 %
Explanation:
Step 1: Data mass
Mass KCl = 30 grams
volume of water = 160 mL
density of water = 1.00 g/mL
Step 2: Calculate mass of water
mass water = volume * density of water
mass water = 160 mL / 1.00 g/mL
mass = 160 grams
Step 3: Calculate total mass of solution
30 grams + 160 grams = 190 grams
Step 4: Calculate mass % :mass KCl / Mass of solution
% mass = (30 grams / 190 gram) *100%
% mass = 15.79 %
The mass percent = 15.79 %
Which of the reactions are exothermic?
A. 2Mg(s) + O2( g ) -----> 2MgO (s) Δ H = − 1203 kJ/mol
B. NH3 (g) + HCl (g)-----> NH4Cl (s) Δ H = − 176 kJ/mol
C. AgCl (s) ------> Ag + (aq) + Cl − (aq) Δ H = 127 kJ/mol
D. 2 Fe2O3 (s) + 3C (s)-------> 4Fe (s) + 3CO2 (g) Δ H = 468 kJ/mol
E. C(graphite) + O2 (g) -------> CO2 (g) Δ H = − 393.5 kJ/mol
F. CH4 (g) + 2O2 (g) -------> CO2 (g) + 2H2O (l) Δ H = − 891 kJ/mol
Answer:
Reaction A, B, E and F are exothermic reactions.
Explanation:
Exothermic reactions are defined as the reactions in which energy of reactants is more than the energy of the products. In these reactions, energy is released by the system.
The total enthalpy of the reaction [tex](\Delta H)[/tex] comes out to be negative.
Endothermic reactions are defined as the reactions in which energy of products is more than the energy of the reactants. In these reactions, energy is absorbed by the system.
The total enthalpy of the reaction [tex](\Delta H)[/tex] comes out to be positive.
So, from the given option reaction which are exothermic are with negative value [tex](\Delta H)[/tex] that enthalpy of reaction and those are:
1) [tex]2Mg(s) + O_2( g )\rightarrow 2MgO (s)[/tex],ΔH = -1203 kJ/mol
2)[tex]NH_3 (g) + HCl (g)\rightarrow NH_4Cl (s)[/tex],ΔH =-176 kJ/mol
3) [tex]C(graphite) + O_2 (g)\rightarrow CO_2 (g) [/tex],ΔH = -393.5 kJ/mol
4) [tex]CH_4 (g) + 2O2 (g)\rightarrow CO_2 (g) + 2H_2O (l) [/tex],ΔH =-891 kJ/mol
Reactions A, B, and E are exothermic because they have negative ΔH values, indicating that energy is released during these reactions.
Explanation:Exothermic reactions are characterized by the release of heat, which is indicated by a negative change in enthalpy (ΔH). In the provided reactions, reactions A (2Mg(s) + O2( g ) -----> 2MgO(s) ΔH = -1203 kJ/mol), B (NH3(g) + HCl(g) -----> NH4Cl (s) ΔH = -176 kJ/mol), and E (C(graphite) + O2(g) -----> CO2(g) ΔH = -393.5 kJ/mol) are exothermic because they have negative ΔH values, indicating that energy is released during these reactions. Conversely, reactions C and D are endothermic (they absorb heat), as indicated by their positive ΔH values.
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A student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, t c , was found to be 216.7 s with a baseline peak width, w c , of 13.6 s. The retention time for aspartame, t a , was 267.1 s with a baseline peak width, w a , of 18.9 s. The retention time for the unretained solvent methanol was 45.2 s.
Calculate the average plate height, H , in micrometers for this separation, given that it was performed on a 24.1 cm long column.
H = μ m
Calculate the resolution, R , for this separation using the widths of the peaks.
R =
Calculate the resolution if the number of theoretical plates were to increase by a factor of 1.5 .
R 1.5 N =
To calculate the average plate height and resolution for a chromatographic separation, we need to find the number of theoretical plates and use specific formulas.
Explanation:To calculate the average plate height, H, we can use the formula: H = L/N, where L is the length of the column and N is the number of theoretical plates. Given that the column length is 24.1 cm, we first need to find the number of theoretical plates.
The number of theoretical plates can be calculated using the formula: N = (t_r / w)^2, where t_r is the retention time and w is the peak width at the base. For caffeine, N_c = (216.7 s / 13.6 s)^2 and for aspartame, N_a = (267.1 s / 18.9 s)^2.
After finding the values for N_c and N_a, we can calculate the average plate height using H = L / (N_c + N_a). Finally, to calculate the resolution, R, we can use the formula: R = 1.18 * (t_a - t_c) / (w_a + w_c).
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The answers are: 1). The average plate height (H) for the separation is 46.63 micrometers. 2). The resolution between caffeine and aspartame is 3.10. 3). The new resolution is 3.80.
1). To calculate the average plate height (H) in micrometers (µm) for this separation, we first determine the number of theoretical plates (N) using the formula:
[tex]\[ N = 16 \left( \frac{t_r}{w} \right)^2 \][/tex]For caffeine:
[tex]\[ N_c = 16 \left( \frac{216.7}{13.6} \right)^2 = 4010.66 \][/tex]For aspartame:
[tex]\[ N_a = 16 \left( \frac{267.1}{18.9} \right)^2 = 6323.53 \][/tex]The average number of theoretical plates [tex](N_{avg})[/tex] is:
[tex]\[ N_{avg} = \frac{4010.66 + 6323.53}{2} = 5167.1 \][/tex]The column length (L) is 24.1 cm (or 241,000 µm). Therefore, the plate height (H) is:
[tex]\[ H = \frac{L}{N_{avg}} = \frac{241,000 \, \mu m}{5167.1} = 46.63 \, \mu m \][/tex]2). Resolution (R):
The resolution (R) between caffeine and aspartame can be calculated using the formula:
[tex]\[ R = 2 \left( \frac{t_a - t_c}{w_a + w_c} \right) \][/tex]Substituting the given values:
[tex]\[ R = 2 \left( \frac{267.1 - 216.7}{18.9 + 13.6} \right) = 2 \left( \frac{50.4}{32.5} \right) = 3.10 \][/tex]3). Resolution with Increased Number of Plates:
If the number of theoretical plates increases by a factor of 1.5, the new average number of plates [tex](N_{new})[/tex] is:
[tex]\[ N_{new} = 5167.1 \times 1.5 = 7750.65 \][/tex]The new resolution [tex](R_{1.5N})[/tex] using the formula:
[tex]\[ R_{1.5N} = R \times \sqrt{1.5} = 3.10 \times 1.225 = 3.80 \][/tex]Complete Question: -
A student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, [tex]\( t_c \)[/tex], was found to be 216.7 s with a baseline peak width, [tex]\( w_c \)[/tex], of 13.6 s. The retention time for aspartame, [tex]\( t_a \)[/tex], was 267.1 s with a baseline peak width, [tex]\( w_a \)[/tex], of 18.9 s. The retention time for the unretained solvent methanol was 45.2 s.
1. Calculate the average plate height, H, in micrometers for this separation, given that it was performed on a 24.1 cm long column. [tex]\[ H = \, \mu m \][/tex]
2. Calculate the resolution, R, for this separation using the widths of the peaks. R = ?
3. Calculate the resolution if the number of theoretical plates were to increase by a factor of 1.5. [tex]\[ R_{1.5N} = \]?[/tex]
A 75.0 mL 75.0 mL aliquot of a 1.70 M 1.70 M solution is diluted to a total volume of 278 mL. 278 mL. A 139 mL 139 mL portion of that solution is diluted by adding 165 mL 165 mL of water. What is the final concentration? Assume the volumes are additive.
Answer:
0.210 M
Explanation:
A 75.0 mL aliquot of a 1.70 M solution is diluted to a total volume of 278 mL.
In order to find out the resulting concentration (C₂) we will use the dilution rule.
C₁ × V₁ = C₂ × V₂
1.70 M × 75.0 mL = C₂ × 278 mL
C₂ = 0.459 M
A 139 mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive.
Since the volumes are additive, the final volume V₂ is 139 mL + 165 mL = 304 mL. Next, we can use the dilution rule.
C₁ × V₁ = C₂ × V₂
0.459 M × 139 mL = C₂ × 304 mL
C₂ = 0.210 M
Answer:
The correct answer is 0.21 M
Explanation:
We have an initial solution of concentration 1.70 M and we dilute it twice. In each dilution we can calculate the final concentration (Cf) from the initial concentration (Ci) and the final and initial volumes respectively (Vf and Vi) as follows:
Cf x Vf= Ci x Vi
Cf= Ci x Vi/ Vf
In the first dilution, Ci is 1.70 M, the initial volume we take is 75 ml and the final volume is 278 ml.
Cf= 1.70 M x 75 ml / 278 ml = 0.46 M
In the second dilution, the initial concentration is the previously obtained (Ci= 0.46 M), the initial volume is Vi= 139 ml and the final volume is the addition of 139 ml and 165 ml (because we add 165 ml to 139 ml). Thus:
Cf= (0.46 M x 139 ml)/ (139 ml + 165 ml) = 0.21 M
Marianne gets dressed and puts on some of her favorite perfume. While applying the perfume, she accidentally gets some in her mouth and notices that it has abitter taste. After getting dressed, she heads to the kitchen to prepare some breakfast. She fills a small bowl with plain yogurt and takes a taste. The yogurt is too sour for her liking, so she adds honey to it for some sweetness. When she is done eating, she washes her bowl with soap and water. The soap is so slippery that she accidentally drops the bowl while cleaning it, causing it to break on the kitchen floor. Frustrated, Marianne sweeps up the broken glass and puts it in the trash, and then heads to work. Which substance displays a characteristic of an acid? A:soap B:yogurt C:perfume D:honey
Answer:
Yogurt displays characteristics of an acid.
Explanation:
The characteristics of an Acid are as follows:
It has an pH of less than 7It has a sour aftertaste(though one shouldn't try it)It increases the [tex]H^{+}[/tex] concentration of water.It reacts with Metals to liberate Hydrogen Gas.It gives out [tex]H^{+}[/tex] ions in aqueous solution.It turns blue litmus to red.Only Yogurt displays one of these properties ie. sour taste, therefore it is the only material having characteristic of acid.
Bittery taste and slippery touch are characteristics of bases.
Answer:
its c
Explanation:
Which pair of samples contains the same number of oxygen atoms in each compound?0.20 mol Ba(OH)2 and 0.20 mol H2SO40.20 mol Br2O and 0.20 mol HBrO0.10 mol Fe2O3 and 0.50 mol BaO0.10 mol Na2O and 0.10 mol Na2SO4
Answer:
0.20 mol Br2O and 0.20 mol HBrO have the same number of oxygen atoms
Explanation:
0.20 mol Ba(OH)2 and 0.20 mol H2SO4
In Ba(OH)2 there are 2 moles of O atoms in every mol of Ba(OH)2.
Number of O atoms in 0.20 moles Ba(OH)2 = 2*0.20 = 0.40 moles O atom
In H2SO4 there are 4 moles of O atoms for every mol of H2SO4.
Number of O atoms in 0.20 moles H2SO4 = 4*0.20 = 0.80 moles O atom
⇒ 0.20 mol Ba(OH)2 and 0.20 mol H2SO4 do not have the same number of oxygen atoms.
0.20 mol Br2O and 0.20 mol HBrO
In Br2O there is 1 mol of O atoms in every mol Br2O
Number of O atoms in 0.20 moles Br2O = 0.20*1 = 0.20 moles O atom
In HBrO there is 1 mol of O atom in every mol HBrO
Number of O atoms in 0.20 moles HBrO = 0.20 *1 = 0.20 moles O atom
⇒ in 0.20 moles Br2O and 0.20 moles HBrO we have the same number of oxygen atoms
0.20 moles of Oygen contains: = 0.20 * 6.022*10^23 = 1.2 *10^23 O atoms
0.10 mol Fe2O3 and 0.50 mol BaO
In Fe2O3 there are 3 moles of O atoms in every mol of Fe2O3
Number of O atoms in 0.10 moles Fe2O3 = 0.10 * 3 = 0.30 moles O atom
In BaO there is 1 mol of O atoms in every mol BaO
Number of O atoms in 0.50 mol BaO = 1*0.50 = 0.50 moles O atom
⇒ 0.10 mol Fe2O3 and 0.50 mol BaO do not have the same number of oxygen atoms.
0.10 mol Na2O and 0.10 mol Na2SO4
In Na2O there is 1 mol of O atoms in every mol of Na2O
Number of O atoms in 0.10 mol Na2O = 1*0.10 = 0.10 moles O atom
In Na2sO4 there are 4 moles of O atoms in every mol of Na2SO4
Number of O atoms in 0.10 mol Na2SO4 = 4*0.10 = 0.40 moles O atom
⇒ 0.10 mol Na2O and 0.10 mol Na2SO4 do not have the same number of oxygen atoms.
The pair of samples that contains the same number of oxygen atoms in each compound is 0.20 mol H2SO4 and 0.20 mol HBrO.
Explanation:To determine which pair of samples contains the same number of oxygen atoms in each compound, we need to calculate the number of oxygen atoms in each sample.
0.20 mol Ba(OH)2 contains 2 oxygen atoms since each Ba(OH)2 molecule has 2 hydroxide groups, and each hydroxide group has 1 oxygen atom.0.20 mol H2SO4 contains 4 oxygen atoms since each H2SO4 molecule has 4 oxygen atoms.0.20 mol Br2O contains 1 oxygen atom since each Br2O molecule has 1 oxygen atom.0.20 mol HBrO contains 2 oxygen atoms since each HBrO molecule has 2 oxygen atoms.Therefore, the pair of samples that contains the same number of oxygen atoms in each compound is 0.20 mol H2SO4 and 0.20 mol HBrO.
A chemistry student needs 10.0 mL of acetone for an experiment. By consulting the CRC Handbook of Chemistry and Physics, the student discovers that the density of acetone is 0.790 g cm .
Calculate the mass of acetone the student should weigh out. Be sure your answer has the correct number of significant digits.
Answer:
The student has to weigh 7.9 ±0.1 grams
Explanation:
Density of acetone : 0.790 g/cm³
If the student has to measure 10.0 mL of acetone, he can weigh an specific mass.
Density of acetone = Mass of acetone / Volume of acetone
cm³ = mL
0.790 g/cm³ = Mass of acetone / 10 cm³
0.790 g/cm³ . 10cm³ = 7.9 g
Final answer:
To find the mass of acetone needed for the experiment, the student multiplies the density of acetone (0.790 g/mL) by the required volume (10.0 mL) to get 7.90 g, considering significant figures.
Explanation:
The student needs to calculate the mass of acetone required for their experiment. Given that the density of acetone is 0.790 g/cm³ (which is equivalent to 0.790 g/mL since 1 cm³ is equal to 1 mL), and the volume needed is 10.0 mL, they can use the formula density = mass/volume to find the mass.
To find the mass, the student should rearrange the formula to mass = density × volume. Using the given values, the calculation is as follows:
mass = 0.790 g/mL × 10.0 mL = 7.90 g
Therefore, the student should weigh out 7.90 g of acetone for their experiment, making sure to use the correct number of significant digits as provided in the question.
In which of the following reactions does a decrease in the volume of the reaction vessel at constant
temperature favor formation of the products?
A) 2H2(g) + O2(g)--> 2H2O(g)
B) NO2(g) + CO(g)--> NO(g) + CO2(g)
C) H2(g) + I2(g)--> 2HI(g)
D) 2O3(g)--> 3O2(g)
E) MgCO3(s)--> MgO(s) + CO2(g)
Answer:
The correct option is: A) 2H₂(g) + O₂(g) → 2H₂O(g)
Explanation:
According to the Le Chatelier's principle, change in the volume of the reaction system causes equilibrium to shift in the direction that reduces the effect of the volume change.
When the volume decreases then the pressure of the reaction vessel increases, then the equilibrium shifts towards the reaction side that produces less number of moles of gas.
A) 2H₂(g) + O₂(g) → 2H₂O(g)
The number of moles of reactant is 3 and number of moles of product is 2.
Therefore, when volume decreases, the equilibrium shifts towards the product side, thereby favoring the formation of products.
B) NO₂(g) + CO(g) → NO(g) + CO₂(g)
The number of moles of reactant and product both is 2.
Therefore, when the volume decreases, the equilibrium does not shift in any direction.
C) H₂(g) + I₂(g) → 2HI(g)
The number of moles of reactant and product both is 2.
Therefore, when the volume decreases, the equilibrium does not shift in any direction.
D) 2O₃(g) → 3O₂(g)
The number of moles of reactant is 2 and number of moles of product is 3.
Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.
E) MgCO₃(s) → MgO(s) + CO₂(g)
The number of moles of reactant is 1 and number of moles of product is 2.
Therefore, when volume decreases, the equilibrium shifts towards the reactant side, thereby favoring the formation of reactants.
Consider a closed flask containing a liquid and its vapor. Which statement is incorrect?
A.) The vapor exerts a pressure called the vapor pressure
B.) Increasing the temperature of the liquid would lead to a great vapor pressure
C.) Evaporation and condensation will eventually cease after a constant pressure has been attained
D.) Increasing the volume of the container at a constant temperature would cause increased condensation until the pressure of the vapor was once eagain the same as it had been
Answer:
C). Evaporation and condensation will eventually cease after a constant pressure has been attained
Explanation:
A) The vapor exerts a pressure called the vapor pressure is correct statement B) On increasing the temperature more vapor will be formed as thus leading to a higher vapor pressure.
C) This is an incorrect statement, Evaporation and condensation never stops what ever may be the pressure. Although they may achieve an equilibrium that is the rate of evaporation may be equal to the rate of condensation.
d) This is again a correct statement as vapor pressure is an extrinsic property that it depends upon the volume or mass of liquid. Greater the volume greater will be the vapor pressure.
Final answer:
Increasing the temperature of the liquid would lead to a greater vapor pressure. Evaporation and condensation continue until a dynamic equilibrium is reached. Increasing the volume of the container causes increased condensation until the pressure of the vapor returns to its original value.
Explanation:
When the liquid in a closed container is heated, more molecules escape the liquid phase and evaporate. This increase in the number of vapor molecules leads to an increase in pressure, known as the vapor pressure (A). Increasing the temperature of the liquid would actually lead to a greater vapor pressure (B), as more molecules would have enough energy to escape from the liquid phase. Evaporation and condensation processes continue in a closed flask until a constant pressure has been attained, referred to as a dynamic equilibrium (C). Increasing the volume of the container at a constant temperature would cause increased condensation until the pressure of the vapor was once again the same as it had been (D). Therefore, the incorrect statement is option B.