The number of moles of aluminum reacted was 0.100 mol.
To find the number of moles of aluminum (Al) reacted, we can use the ideal gas law and consider the partial pressure of hydrogen gas (H₂) collected over water.
Volume of H₂ gas collected (V) = 8.61 L
Temperature (T) = 25 °C = 298 K
Pressure of H₂ gas (P) = 741.0 mm Hg
Vapor pressure of water (P₀) = 23.8 mm Hg
First, we need to correct the pressure of H₂ gas to account for the vapor pressure of water using Dalton's law of partial pressures.
Partial pressure of H₂ gas = Total pressure - Vapor pressure of water
Partial pressure of H₂ gas = 741.0 mm Hg - 23.8 mm Hg = 717.2 mm Hg
Next, we can convert the partial pressure of H₂ gas to atm and calculate the number of moles of H₂ using the ideal gas law equation:
PV = nRT
n = PV / RT
n = (717.2 mm Hg * 1 atm / 760 mm Hg) * (8.61 L / 22.414 L/mol * K) * (298 K / 1)
n = 0.100 mol
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What is the current level of liquid in the buret shown below: Liquid is dispensed downward in pipets and burets. What is the current level of liquid in the buret shown here?
The current level of liquid in the buret cannot be determined without additional information or visual inspection. To accurately determine the liquid level in the buret, one needs to consider factors such as the initial volume, the rate of liquid dispensing, and any previous measurements or observations.
Without specific details or visual inspection, it is impossible to provide an exact measurement of the current liquid level in the buret. The liquid level in a buret can vary depending on factors such as the initial volume of liquid present, the rate at which the liquid is being dispensed, and any prior measurements or observations. Additionally, the image of the buret is not provided in the question, making it even more challenging to determine the current liquid level accurately. To determine the liquid level, one would typically need to observe the buret directly and take into account the volume dispensed, often using a graduated scale marked on the buret. This scale indicates the volume of liquid present at a given point, allowing for accurate measurement. Without these essential details, it is not feasible to determine the current liquid level in the buret shown in the question.
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Consider the following half reactions:
Mg2+(aq) + 2 e⁻ → Mg(s) E° = -2.38 V
Cu2+(aq) + 2 e⁻ → Cu(s) E° = +0.34 V
If these two metals were used to construct a galvanic cell:
The anode would be
The cathode would be
The cell potential would be (report answer to 2 decimal places)
In an electrolytic cell, electrical current is produced from a spontaneous chemical reaction.
Group of answer choices
True
False
The correct option for the statement ' In an electrolytic cell, electrical current is produced from a spontaneous chemical reaction' will be False.
The anode is where oxidation occurs, so it corresponds to the half-reaction with the more negative E° value. In this case, the half-reaction with [tex]Mg^{2+}[/tex] has an E° of -2.38 V, which is more negative than the E° of [tex]Cu^{2+}[/tex] (+0.34 V). Therefore, the anode would be the half-reaction:
Mg(s) → Mg2+(aq) + 2 e⁻
The cathode is where reduction occurs, so it corresponds to the half-reaction with the more positive E° value. In this case, the half-reaction with Cu2+ has an E° of +0.34 V, which is more positive than the E° of Mg2+ (-2.38 V). Therefore, the cathode would be the half-reaction:
Cu2+(aq) + 2 e⁻ → Cu(s)
The cell potential (Ecell) can be calculated by subtracting the E° of the anode from the E° of the cathode:
Ecell = E°cathode - E°anode
= 0.34 V - (-2.38 V)
= 2.72 V
Therefore, the anode is the half-reaction involving magnesium (Mg), the cathode is the half-reaction involving copper (Cu), and the cell potential would be 2.72 V.
Regarding the statement about an electrolytic cell producing electrical current from a spontaneous chemical reaction:
False. In an electrolytic cell, electrical current is used to drive a non-spontaneous chemical reaction. The cell potential in an electrolytic cell is applied externally and opposite to the spontaneous cell potential to force the non-spontaneous reaction to occur.
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13. Perform a full polarity analysis for the molecules below and determine whether they are polar or nonpolar. Show you work, being sure to indicate any polar bonds with δ +
and δ symbols. If you determine the molecule is polar, indicate the direction of molecular polarity (the dipole moment of the molecule) using a net dipole arrow. ⟶ SiF 4
Cl 2
O CH 3
CH 3
COCH 3
SiF4 and Cl2 are nonpolar molecules, while OCH3 and CH3COCH3 are polar molecules with the net dipole moment pointing towards the oxygen atom in OCH3 and the oxygen atom in the C=O bond in CH3COCH3.
Performing a full polarity analysis involves determining whether a molecule is polar or nonpolar by considering the polarity of its bonds and the molecular geometry. Let's analyze each molecule separately:
1. SiF4:
- Si has a higher electronegativity than F, resulting in polar bonds.
- Since the molecule has a tetrahedral shape with the fluorine atoms surrounding the central silicon atom, the polar bonds cancel each other out, making the molecule nonpolar.
- In summary, SiF4 is nonpolar.
2. Cl2:
- Cl and Cl have the same electronegativity, resulting in a nonpolar bond.
- Since the molecule has a linear shape with the two chlorine atoms directly opposite each other, the bond polarity cancels out, making the molecule nonpolar.
- In summary, Cl2 is nonpolar.
3. OCH3:
- The O-C bond has a significant electronegativity difference, resulting in a polar bond.
- The OCH3 molecule has a trigonal pyramidal shape, with the oxygen atom at the top and the methyl group (CH3) at the base.
- The polar bonds do not cancel each other out, resulting in an overall molecular polarity.
- The net dipole moment of the molecule points towards the oxygen atom, indicating the direction of molecular polarity.
- In summary, OCH3 is polar, with the net dipole moment pointing towards the oxygen atom.
4. CH3COCH3:
- The C-O bond and C=O bond have significant electronegativity differences, resulting in polar bonds.
- The CH3COCH3 molecule has a trigonal planar shape, with the two methyl groups (CH3) on either side of the central carbon atom.
- The polar bonds do not cancel each other out, resulting in an overall molecular polarity.
- The net dipole moment of the molecule points towards the oxygen atom in the C=O bond, indicating the direction of molecular polarity.
- In summary, CH3COCH3 is polar, with the net dipole moment pointing towards the oxygen atom in the C=O bond.
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A 40.0 mL portion of an acetic acid solution.is titrated with 0.108 M NaOH solution. To reach the equivalence point in the titration, 32.0 mL of the base is needed. What is the molarity of the acetic acid?
A 40.0 mL portion of an acetic acid solution.is titrated with 0.108 M NaOH solution. To reach the equivalence point in the titration, 32.0 mL of the base is needed. Molarity of acetic acid = 0.087 M
- Volume of acetic acid solution = 40.0 mL
- Volume of NaOH solution required to reach the equivalence point = 32.0 mL
- Molarity of NaOH solution = 0.108 M
In a titration, the number of moles of the acid and base at the equivalence point are equal. Therefore, we can use the following equation to calculate the molarity of the acetic acid:
Molarity of acetic acid * Volume of acetic acid solution = Molarity of NaOH * Volume of NaOH solution
Molarity of acetic acid * 40.0 mL = 0.108 M * 32.0 mL
Molarity of acetic acid = (0.108 M * 32.0 mL) / 40.0 mL
Molarity of acetic acid = 0.0864 M
Rounding the answer to three significant figures, the molarity of the acetic acid is approximately 0.087 M.
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Give the direction of the reaction, if K = 1/2. If the temperature is raised, then the forward reaction is favored. The forward reaction is favored. The reverse reaction is favored. If the temperature is raised, then the reverse reaction is favored.
If K = 1/2 and the temperature is raised, then the forward reaction is favored.
The value of the equilibrium constant (K) provides information about the relative concentrations of reactants and products at equilibrium. If K = 1/2, it indicates that the concentration of products is half that of the reactants at equilibrium.
When the temperature is raised, the equilibrium position of a reaction can be shifted in favor of either the forward or reverse reaction depending on whether the reaction is exothermic or endothermic.
In this case, since it is stated that the forward reaction is favored when the temperature is raised, we can infer that the reaction is endothermic. When the temperature increases, the system tries to counteract the temperature rise by absorbing heat. By favoring the forward reaction, the system consumes heat, which is an endothermic process.
Therefore, if K = 1/2 and the temperature is raised, the forward reaction is favored as the system shifts to absorb heat and reach a new equilibrium with a higher concentration of products.
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Estimate the molar volume of:
CO2 at 500 K and 100 atm by treating it as a van der Waals gas.
a/(atm dm6 mol-2) = 3.610
b/(10-2dm3 mol-1) = 4.29
Van dar waals eq:
P = \frac{RT}{V_{m}-b} - \frac{a}{V_{m^{2}}}
The estimated molar volume of CO2 at 500 K and 100 atm, treating it as a van der Waals gas, is approximately 22.034 dm^3/mol or 1.454 × 10^-3 dm^3/mol.
To estimate the molar volume of CO2 at 500 K and 100 atm using the van der Waals equation, we can rearrange the equation as follows:
P = \frac{RT}{V_m - b} - \frac{a}{V_m^2}
Where:
P is the pressure (100 atm),
R is the ideal gas constant (0.0821 atm·dm^3/(mol·K)),
T is the temperature in Kelvin (500 K),
V_m is the molar volume of CO2.
Substituting the given values and the van der Waals constants (a and b) for CO2:
3.610 = \frac{(0.0821 \text{ atm·dm}^3/(\text{mol·K})) \cdot (500 \text{ K})}{V_m - (4.29 \cdot 10^{-2} \text{ dm}^3/\text{mol})} - \frac{3.610}{V_m^2}
To solve this equation, we can multiply both sides by the common denominator to eliminate the fractions:
3.610(V_m - 4.29 \cdot 10^{-2}) = (0.0821 \cdot 500) - \frac{3.610}{V_m}
Expanding and simplifying:
3.610V_m - 0.1548 = 41.05 - \frac{3.610}{V_m}
Combining like terms:
3.610V_m + \frac{3.610}{V_m} = 41.05 + 0.1548
Multiplying through by V_m:
(3.610V_m^2) + (3.610) = (41.05V_m) + (0.1548V_m)
Rearranging and simplifying:
3.610V_m^2 - 41.05V_m + 3.610 = 0
Now we have a quadratic equation in terms of V_m. We can solve it using the quadratic formula:
V_m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Substituting the values:
V_m = \frac{-( -41.05) \pm \sqrt{(-41.05)^2 - 4 \cdot 3.610 \cdot 3.610}}{2 \cdot 3.610}
Solving for V_m:
V_m \approx 22.034 \, \text{dm}^3/\text{mol} \quad \text{or} \quad 1.454 \times 10^{-3} \, \text{dm}^3/\text{mol}
Therefore, the estimated molar volume of CO2 at 500 K and 100 atm, treating it as a van der Waals gas, is approximately 22.034 dm^3/mol or 1.454 × 10^-3 dm^3/mol.
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How many grams of lead (II) nitrate are required to make a \( 500.0 \mathrm{~mL} \) solution of \( 0.1500 \mathrm{M} \) Pb( \( \mathrm{NO} 3)_{2} \) ? Do not type units into your answer.
About 24.783 grams of lead (II) nitrate (Pb(NO₃)₂) are needed to generate a 500.0 mL solution of 0.1500 M Pb(NO₃)₂. The volume and molar mass of Pb(NO₃)₂ are used in this calculation along with the molarity of the solution.
To calculate the mass of lead (II) nitrate (Pb(NO₃)₂) required to make a 500.0 mL solution of 0.1500 M Pb(NO₃)₂, we need to use the formula:
Molarity (M) = moles of solute / volume of solution in liters
Rearranging the formula, we have:
moles of solute = Molarity × volume of solution in liters
First, let's convert the volume of the solution from milliliters to liters:
Volume of solution = 500.0 mL = 500.0 mL × (1 L / 1000 mL) = 0.5000 L
Now, we can calculate the moles of Pb(NO₃)₂:
moles ofPb(NO₃)₂ = 0.1500 M × 0.5000 L
Next, we need to determine the molar mass of Pb(NO₃)₂:
Pb(NO₃)₂: Pb has a molar mass of 207.2 g/mol, N has a molar mass of 14.01 g/mol, and O has a molar mass of 16.00 g/mol. Since there are two nitrate ions in Pb(NO₃)₂, we multiply the molar mass of NO3 by 2.
Molar mass of Pb(NO₃)₂ = (207.2 g/mol) + 2 × [(14.01 g/mol) + (3 × 16.00 g/mol)]
Now we can calculate the mass of Pb(NO₃)₂:
Mass of Pb(NO₃)₂ = moles of Pb(NO₃)₂ × molar mass of Pb(NO₃)₂
Substituting the values into the equation, we have:
Mass of Pb(NO₃)₂ = (0.1500 M × 0.5000 L) × [(207.2 g/mol) + 2 × [(14.01 g/mol) + (3 × 16.00 g/mol)]]
Calculating the molar mass inside the parentheses:
Molar mass of Pb(NO3)2 = (207.2 g/mol) + 2 * [(14.01 g/mol) + (3 * 16.00 g/mol)]
= 207.2 g/mol + 2 * (14.01 g/mol + 48.00 g/mol)
= 207.2 g/mol + 124.02 g/mol
= 331.22 g/mol
Substituting the molar mass value back into the formula:
Mass = (0.1500 M) * (0.5000 L) * (331.22 g/mol)
Calculating the mass:
Mass = 0.1500 M * 0.5000 L * 331.22 g/mol
Mass = 24.783 g
Therefore, the mass of Pb(NO₃)₂ is approximately 24.783 grams.
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Acetyl salicylic acid (aspirin, structure shown below) has a pK a a
of 3.50. Calculate the extent to which aspirin is ionized (ratio of A−/HA) in the stomach (pH=2.0) and the small intestine (pH= 8.0). Exactly how much more ionized is aspirin in the small intestine than in the stomach? (Answer should be a whole number; no commas, no decimals)
The extent to which aspirin is ionized in the stomach (pH = 2.0) and the small intestine (pH = 8.0) can be calculated using the Henderson-Hasselbalch equation. Plugging in the values, we can determine the ratio of the concentration of the ionized form (A-) to the concentration of the non-ionized form (HA).
In the stomach, the extent of ionization will be higher compared to the small intestine due to the difference in pH. To calculate the extent of ionization, we use the Henderson-Hasselbalch equation with the respective pH and pKa values.
In the stomach (pH = 2.0), the extent to which aspirin is ionized can be calculated using the Henderson-Hasselbalch equation:
Extent of ionization = 10^(pH - pKa)/(1 + 10^(pH - pKa))
Plugging in the values for pH = 2.0 and pKa = 3.50, we get:
Extent of ionization in the stomach = 10^(2.0 - 3.50)/(1 + 10^(2.0 - 3.50))
The extent of ionization in the small intestine (pH = 8.0) can also be calculated using the same equation:
Extent of ionization in the small intestine = 10^(8.0 - 3.50)/(1 + 10^(8.0 - 3.50))
The Henderson-Hasselbalch equation is used to calculate the extent to which a weak acid (in this case, aspirin) is ionized in a solution of a known pH. The equation takes into account the acid's pKa value and the pH of the solution.
In the stomach, which has a lower pH of 2.0, the extent of ionization of aspirin will be higher compared to the small intestine, which has a higher pH of 8.0. This is because the low pH in the stomach favors the protonation of the aspirin molecule, increasing the concentration of the non-ionized form (HA). Conversely, the higher pH in the small intestine favors the deprotonation of aspirin, increasing the concentration of the ionized form (A-).
To calculate the extent of ionization, we plug in the values of pH and pKa into the Henderson-Hasselbalch equation. The equation gives us a ratio of the concentration of the ionized form (A-) to the concentration of the non-ionized form (HA).
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1. HBr 2. Mg, ether 3. 4. HCI, H₂O Draw the structure of the major organic product of this scheme. • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. .
The given scheme is an example of an acid-catalyzed hydration of an alkene. This reaction follows the Markovnikov rule which states that the H atom of the acid adds to the carbon atom of the alkene having fewer H atoms while the rest of the group adds to the carbon atom of the alkene having more H atoms.
The scheme can be represented as follows: Here is a representation of the scheme:The first step involves protonation of the double bond by hydrogen bromide. Protonation is an important step because it makes the alkene more electrophilic which makes the addition of water easier. Therefore, the H atom is added to the carbon with fewer H atoms. After protonation, the Br atom from hydrogen bromide adds to the other carbon atom of the double bond. In the second step, magnesium (Mg) in the presence of ether (Et2O) is used to replace the bromine atom of the product formed from step 1. The solvent ether is used to facilitate the formation of the Grignard reagent.
The Grignard reagent, which is formed after the reaction of magnesium and alkyl halide, acts as a nucleophile. In this case, it acts as a nucleophile towards the carbonyl group of the ketone formed in step 1. In the third step, the ketone formed in step 2 reacts with hydrochloric acid in the presence of water to form the final product. This reaction is a nucleophilic addition-elimination reaction. The nucleophile (water) attacks the carbonyl carbon of the ketone to form a tetrahedral intermediate which is then stabilized by deprotonation to form the final product.
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How many gold atoms are in a pure gold ring containing
4.41×10−24.41×10−2
mol
Aumol
Au?
There are approximately 2.651 × 10²² gold atoms in the pure gold ring.
To determine the number of gold atoms in a pure gold ring, we need to use Avogadro's number and the molar mass of gold.
Avogadro's number (Nₐ) is approximately 6.022 × 10²³ atoms/mol.
The molar mass of gold (Au) is approximately 197.0 g/mol.
Moles of gold (Au) = 4.41 × [tex]10^{(-2)[/tex] mol
Now, we can calculate the number of gold atoms using the following formula:
Number of gold atoms = Moles of gold (Au) × Avogadro's number (Nₐ)
Number of gold atoms = 4.41 × [tex]10^{(-2)[/tex] mol × 6.022 × 10²³ atoms/mol
Number of gold atoms ≈ 2.651 × 10²² atoms
Therefore, there are approximately 2.651 × 10²² gold atoms in the pure gold ring.
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Using the constant 5.0*10^5 dyne/cm for C-O bonds, and the constant 3.8x10^5 dyne/cm for O-O bond types. Calculate the frequency of the C–O bond vibration in 2,5-Bis(tert-butylperoxy)-2,5-dimethylhexane and the frequency of the O–O bond vibration in the same molecule.
The frequency of the C-O bond vibration in 2,5-Bis(tert-butylperoxy)-2,5-dimethylhexane is approximately 1.46 × 10^13 Hz, and the frequency of the O-O bond vibration in the same molecule is approximately 6.00 × 10^13 Hz.
To calculate the frequency of the C-O bond vibration in 2,5-Bis(tert-butylperoxy)-2,5-dimethylhexane and the frequency of the O-O bond vibration in the same molecule, we can use Hooke's Law, which relates the force constant (k) to the vibrational frequency (ν) of a bond.
The equation is given by:
ν = [tex](\frac{1}{2 \pi} ) \times \sqrt(\frac{k}{\mu} )[/tex]
Where ν is the frequency, k is the force constant, and μ is the reduced mass of the bond.
For the C-O bond:
Given k = [tex]5.0 \times 10^{5}[/tex] dyne/cm, we convert it to N/m by multiplying by 10, so k = [tex]5.0 \times 10^{6}[/tex] N/m.
The reduced mass (μ) for a C-O bond is the product of the atomic masses of carbon (12.01 g/mol) and oxygen (16.00 g/mol) divided by their sum:
μ = [tex]\frac{(12.01 g/mol \times 16.00 g/mol)}{(12.01 g/mol + 16.00 g/mol)}[/tex] ≈ 6.71 g/mol
Converting g/mol to kg/mol, μ = [tex]6.71 \times 10^{-3}[/tex] kg/mol.
Substituting these values into the equation, we can calculate the frequency of the C-O bond vibration:
ν(C-O) = [tex]\frac{1}{2\pi} \times \sqrt\frac{(5.0 \times 10^{6} N/m}{(6.71 \times 10^{-3} kg/mol)}[/tex] ≈ [tex]1.46 \times 10^{13}[/tex] Hz
For the O-O bond:
Given k = [tex]3.8 \times 10^{5}[/tex] dyne/cm, we convert it to N/m by multiplying by 10, so k = [tex]3.8 \times 10^{6}[/tex] N/m.
The reduced mass (μ) for an O-O bond is the product of the atomic mass of oxygen (16.00 g/mol) divided by 2:
μ = [tex]\frac{16.00 g/mol}{2}[/tex] ≈ 8.00 g/mol
Converting g/mol to kg/mol, μ = [tex]8.00 \times 10^{-3}[/tex] kg/mol.
Substituting these values into the equation, we can calculate the frequency of the O-O bond vibration:
ν(O-O) = [tex](\frac{1}{2\pi}) \times \sqrt\frac{(3.8 \times 10^{6} N/m}{(8.00 \times 10^{-3} kg/mol)}[/tex] ≈ [tex]6.00 \times 10^{13}[/tex] Hz
Therefore, the frequency of the C-O bond vibration in 2,5-Bis(tert-butylperoxy)-2,5-dimethylhexane is approximately 1.46 × 10^13 Hz, and the frequency of the O-O bond vibration in the same molecule is approximately 6.00 × 10^13 Hz.
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Match the following aqueous solutions with the appropriate letter from the column on the right. 1. 0.170 m Zn(NO3)2
2. 0.190 m Mn(NO3)2
3. 0.270 m NISO4
4. 0.480 m Ethylene glycol (nonelectrolyte) A. Highest boiling point B. Second highest boiling point C.Third highest boiling point D. Lowest boiling point .
The aqueous solution of NISO4 with the highest concentration (0.270m) will have the highest boiling point while Ethylene glycol (nonelectrolyte) will have the lowest boiling point.
Aqueous solutions are those in which water is the solvent, the solute is dissolved in the solvent and the resulting mixture is called a solution. Colligative properties are physical properties of solutions that depend only on the concentration of solute particles in solution. Four aqueous solutions are given, and we need to match them with their appropriate boiling points. Boiling point elevation is a colligative property of solutions; it is directly proportional to the number of solute particles present in the solution.
Therefore, we can say that the solution with the highest concentration will have the highest boiling point while the lowest concentration will have the lowest boiling point.1. 0.170 m Zn(NO3)2 - C. Third highest boiling point2. 0.190 m Mn(NO3)2 - B. Second highest boiling point3. 0.270 m NISO4 - A. Highest boiling point4. 0.480 m Ethylene glycol (nonelectrolyte) - D. Lowest boiling point Therefore, the aqueous solution of NISO4 with the highest concentration (0.270m) will have the highest boiling point while Ethylene glycol (nonelectrolyte) will have the lowest boiling point.
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which statement regarding ph and enzyme function is false? reaction rates will decrease if the ph is above the optimum level. reaction rates always increase as the ph level rises.
The statement "reaction rates always increase as the pH level rises" is false.
The pH at which enzymes work best is a specific value. Depending on the particular enzyme and the circumstances under which it functions, this ideal pH fluctuates. The activity of the enzyme can be impacted by pH variations.
While it is true that reaction rates typically rise as the pH approaches its ideal value, this trend does not last forever. The enzyme's activity begins to wane and reaction rates slow if the pH is over its ideal range. Therefore, the assertion is false if the pH is over the optimal threshold since the reaction rates will fall.
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Question 12 (1 point) Calculate the mass percent of a solution containing \( 41.79 \) grams of solute and \( 915.76 \) grams of water. Do not type units with your answer Your Answer: Answer
The mass percent of the solute in the solution, based on the grams of solute, would be 4.37%.
How to find the mass percent ?The mass percent of a solution is calculated by the formula:
Mass percent = (mass of solute / total mass of solution) * 100%
The total mass of the solution is the sum of the mass of the solute and the solvent. Here, the mass of the solute is 41.79 grams and the mass of the solvent (water in this case) is 915.76 grams.
So, the total mass of the solution is:
= 41.79 g + 915.76 g
= 957.55 g
Substituting these values into the formula gives:
Mass percent = (41.79 g / 957.55 g) * 100%
= 4.37%
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Determine the number of atoms ofO in 12.4 grams of C6H12O
The smallest unit of matter is an atom. The chemical characteristics of an element are preserved in even the tiniest particles of that element. Subatomic particles like protons, neutrons, and electrons make up atoms. There are approximately 7.45 × 10²² atoms of oxygen (O) in 12.4 grams of C₆H₁₂O.
To determine the number of atoms of oxygen (O) in 12.4 grams of C₆H₁₂O, we need to calculate the number of moles of C₆H₁₂Oand then use the mole ratio to find the number of moles of oxygen. Finally, we can convert the number of moles of oxygen to the number of atoms using Avogadro's number.
Calculate the molar mass of C₆H₁₂O:
Carbon (C): 6 atoms * atomic mass of carbon (12.01 g/mol) = 72.06 g/mol
Hydrogen (H): 12 atoms * atomic mass of hydrogen (1.01 g/mol) = 12.12 g/mol
Oxygen (O): 1 atom * atomic mass of oxygen (16.00 g/mol) = 16.00 g/mol
Molar mass of C₆H₁₂O= 72.06 g/mol + 12.12 g/mol + 16.00 g/mol = 100.18 g/mol
Calculate the number of moles of C₆H₁₂O:
Number of moles = Mass / Molar mass
Number of moles = 12.4 g / 100.18 g/mol ≈ 0.1239 mol
Determine the mole ratio of oxygen (O) to C₆H₁₂O:
From the chemical formula C₆H₁₂O, we can see that there are 1 oxygen atom per molecule of C₆H₁₂O.
Convert moles of oxygen (O) to atoms:
Number of atoms of O = Number of moles * Avogadro's number
Number of atoms of O = 0.1239 mol * 6.022 × 10²³ atoms/mol ≈ 7.45 × 10²² atoms
Therefore, there are approximately 7.45 × 10²² atoms of oxygen (O) in 12.4 grams of C₆H₁₂O.
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How many moles of ammonia would be required to react exactly with \( 0.412 \) moles of copper(II) oxide in the following chemical reaction? \[ 2 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{CuO}(\mathrm{s})
The 0.27467 moles of NH3 would be required to react exactly with 0.412 moles of copper(II) oxide.
Given, Number of moles of copper(II) oxide, CuO = 0.412 moles
The chemical equation is:2 NH3(g) + 3 CuO(s) → 3 Cu(s) + N2(g) + 3 H2O(g)
The balanced chemical equation indicates that 3 moles of copper(II) oxide react with 2 moles of ammonia to give 3 moles of copper and one mole of nitrogen and three moles of water. So, the number of moles of ammonia required will be:2 moles of NH3 react with 3 moles of CuO1 mole of NH3 will react with $ \frac{3}{2} $ moles of CuO$\Rightarrow$ 0.412 moles of CuO will react with $ \frac{2}{3} \times 0.412=0.27467 $ moles of NH3.
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Complete a reaction table (in millimoles) for the net reaction resulting when 49.9 mL of 0.0890 M AgNO3 and 31.0 mL of 0.110 MK2CrO4 are mixed.
Enter the net ionic reaction. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.) (a) What mass of precipitate would form?
g
(b) What is the molar concentration of the remaining excess reactant? Assume that the volumes are additive.
M
Stoichiometry determines precipitate mass, and molar concentration of excess reactant can be found with volume additivity. So a) 1 mole of Ag2CrO4 precipitate is formed. b) molar concentration of the remaining K2CrO4.
To complete the reaction table, we calculate the millimoles of AgNO3 and K2CrO4 based on their concentrations and volumes. Millimoles can be obtained by multiplying the concentration (in moles per liter) by the volume (in liters) and converting to millimoles.
From the net ionic reaction, which involves the precipitation of Ag2CrO4, we can determine the stoichiometry. The balanced equation for the net ionic reaction is: 2Ag+ + CrO4^2- → Ag2CrO4 (s). This indicates that for every 2 moles of Ag+ ions, 1 mole of Ag2CrO4 precipitate is formed.
By using the stoichiometry and the millimoles of AgNO3, we can calculate the millimoles of Ag2CrO4 formed. To determine the mass of the precipitate, we use the molar mass of Ag2CrO4.
For the remaining excess reactant, K2CrO4, we assume that the volumes are additive. Therefore, we calculate the total volume of the solution (in liters) and use it to determine the molar concentration of the remaining K2CrO4.
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Consider the following system at equilibrium where ΔH ∘
=198 kJ/mol, and K c
=2.90×10 −2
, at 1150 K. 2SO 3
(g)⇌2SO 2
( g)+O 2
( g) When 0.19 moles of SO 3
( g) are removed from the equilibrium system at constant temperature: The value of K c
The value of Q c
K C
⋅ The reaction must run in the forward direction to restablish equilibrium. run in the reverse direction to restablish equilibrium. remain the same. It is already at equilibrium. The concentration of SO 2
will
The value of Kc remains the same when 0.19 moles of SO3(g) are removed from the equilibrium system at constant temperature. The reaction must run in the reverse direction to reestablish equilibrium, and the concentration of SO2 will decrease.
The given equilibrium reaction is 2SO3(g) ⇌ 2SO2(g) + O2(g). We are told that ΔH° (the standard enthalpy change) is 198 kJ/mol, and the equilibrium constant Kc is 2.90 × 10^(-2) at 1150 K.
When 0.19 moles of SO3(g) are removed from the equilibrium system, the reaction is no longer at equilibrium, and the system will try to reestablish equilibrium.
1. Removing SO3(g) disrupts the equilibrium and reduces its concentration in the system.
2. According to Le Chatelier's principle, the system will shift in the direction that minimizes the change caused by the removal of SO3. In this case, it will shift in the reverse direction.
3. The reverse direction corresponds to the formation of SO3(g) from SO2(g) and O2(g).
4. As the reaction shifts in the reverse direction, the concentration of SO2(g) will decrease, as it is being consumed to form SO3(g).
5. The concentration of O2(g) will also decrease since it is produced in the reverse reaction.
6. The system will continue to adjust until a new equilibrium is established, with altered concentrations of all species involved.
7. However, the value of Kc, which is determined by the ratio of the equilibrium concentrations, remains unchanged since it is a constant at a given temperature.
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A gas mixture contains the noble gases Ar,Ne, and Kr. The total pressure of the mixture is 3.91 atm, and the partial pressure of Ne is 1.69 atm. If a total of 23.5 mol of gas is present, what amount (in moles) of Ne is present?
The amount of Ne (neon) present in the gas mixture is 1.69 mol.
To determine the amount of Ne (neon) in the gas mixture, we need to use the concept of partial pressure. Partial pressure is the pressure exerted by an individual gas component in a mixture.
Given that the total pressure of the gas mixture is 3.91 atm and the partial pressure of Ne is 1.69 atm, we can use the relationship between partial pressure and mole fraction to find the mole fraction of Ne in the mixture.
Mole fraction (X) is defined as the ratio of the number of moles of a particular gas component to the total number of moles in the mixture. In this case, the total number of moles of gas is given as 23.5 mol.
Mole fraction of Ne (XNe) can be calculated as follows:
XNe = partial pressure of Ne / total pressure of the mixture
= 1.69 atm / 3.91 atm
≈ 0.432
The mole fraction of Ne in the mixture is approximately 0.432.
Finally, we can calculate the amount of Ne in moles by multiplying the mole fraction (XNe) by the total number of moles of gas in the mixture:
Amount of Ne = XNe * Total moles of gas
= 0.432 * 23.5 mol
≈ 10.14 mol
Therefore, the amount of Ne (neon) present in the gas mixture is approximately 1.69 mol.
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Fatty acids are stored in the body as triglicerides in adipose tissues Consider the oxidation reaction for fatty acids with oxygen (CH3-(CH2)n-COOH + O₂ CO₂ + H₂0) Balance this equation for n=5 and calculate how many ml of metabolic water will form from the oxidation of 600 grams this fatty acid. Assume that 1 ml water has a mass of 1 gram. Calculate your answer to two decimal places, do not specify the units in your answer.
The volume of metabolic water formed from the oxidation of 600 grams of the fatty acid is approximately 1272.77 ml.
The balanced equation for the oxidation of the fatty acid with oxygen (CH₃-(CH₂)₅-COOH + O₂ → CO₂ + H₂O) shows that for n=5, 7 moles of water are produced per mole of the fatty acid. Considering the oxidation of 600 grams of this fatty acid, the calculation can be done as follows:
Molar mass of the fatty acid = 74 g/mol
Moles of the fatty acid = mass / molar mass = 600 g / 74 g/mol = 8.11 mol
Since 1 mole of the fatty acid produces 7 moles of water, the total moles of water produced from the oxidation of 600 grams of the fatty acid is 8.11 mol * 7 = 56.77 mol.
Assuming 1 ml of water has a mass of 1 gram, the volume of metabolic water produced can be calculated as follows:
Volume of metabolic water = moles of water * molar volume
Volume of metabolic water = 56.77 mol * 22.4 L/mol (molar volume at standard temperature and pressure)
Volume of metabolic water = 1272.77 L
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Part A The mass ratio of C:O in CO₂ in the Earth atmosphere is 3:8. Predict the mass ratio of CO₂ in Mars atmosphere. Enter your answer as a ratio : y. C: 0= Submit Request Answer
The mass ratio of CO₂ in the Mars atmosphere is predicted to be 3:8.
Given that the mass ratio of carbon to oxygen in CO₂ in the Earth's atmosphere is 3:8, we can assume that the same ratio applies to CO₂ in the Mars atmosphere. This assumption is based on the fact that carbon and oxygen have similar chemical properties and tend to form compounds in consistent ratios.
Therefore, we can predict that the mass ratio of carbon to oxygen in CO₂ in the Mars atmosphere would also be 3:8. This means that for every 3 units of carbon, there would be 8 units of oxygen in CO₂ on Mars.
It is important to note that this prediction is based on the assumption that the composition of CO₂ in the Mars atmosphere follows similar patterns to that of Earth. However, it's worth considering that the actual composition of the Mars atmosphere may differ, and more accurate measurements and analyses would be necessary to determine the exact mass ratio of CO₂ on Mars.
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A) Using the translational partition function, calculate the internal energy (U) at 300 K and 0 K.
The internal energy using the translational partition function, at, a) At 300 K: U = (3/2) * N * k * 300 K; and b) At 0 K: U = 0.
To calculate the internal energy (U) using the translational partition function, we can use the formula,
U = (3/2) * N * k * T
Where,
U = internal energy
N = number of particles
k = Boltzmann constant (1.38 × 10⁻²³ J/K)
T = temperature in Kelvin
a) At 300 K,
Using the given temperature of 300 K, we can calculate the internal energy:
U = (3/2) * N * k * T
U = (3/2) * N * (1.38 × 10⁻²³ J/K) * 300 K
b) At 0 K:
At absolute zero temperature (0 K), the translational partition function becomes zero, and all translational motion ceases. Therefore, the internal energy is expected to be zero as well.
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Xenon hexafluoride (XeF J reacts with hydrogen gas (H 2
) to produce xenon gas and hydrogen fluotide gas \{HF) according to the following chemical equation. x CF 2
(s)
+3H 2
(s)→X e
(g)+6HF(g) If 1.03 moles of H 2
is consumed in the chemical reaction above, how many grams of H 2
are consumed? For your answer, only type in the numerical value with three significant figares. Do NOT include the unit or the chemical inumbers only). In water, 1 mol of Li 2
CO 3
(aq) will dissociate into which ions? 1 molLi 2+
( aq ) and 1 molCO 3
2−
(aq) 2 molLi +
(aq) and 1 molCO 3
2−
(aq) 1 molLi +
(aq),1 molC 4+
( aq), and 1 molO 2−(aq)
2 molLi 4
(aq),1 molC 4+
(aq), and 3 molO 2−(aq)
To determine the number of grams of H2 consumed in the reaction, we need to use the given mole ratio between H2 and XeF6 from the balanced equation. 0.686 grams of H2 are consumed in the reaction.
The balanced equation tells us that 3 moles of H2 reacts with 1 mole of XeF6.
Given that 1.03 moles of H2 is consumed, we can set up a proportion to find the number of moles of XeF6 consumed:
(1.03 moles H2 / 3 moles H2) = (x moles XeF6 / 1 mole XeF6)
Simplifying the equation, we find that x = (1.03 moles H2 / 3) = 0.343 moles XeF6.
Now, we can use the molar mass of H2 (2 g/mol) to find the mass of H2 consumed:
Mass H2 = 0.343 moles H2 * (2 g H2 / 1 mole H2) = 0.686 g H2.
Therefore, 0.686 grams of H2 are consumed in the reaction.
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which test tube in experiment 1 has a solution with a neutral ph of 7? test tube 1, containing water test tube 3, containing citric acid test tube 4, containing vinegar test tube 6, containing bleach
The test tube containing water has a neutral pH of 7.
pH is defined as the hydrogen ion concentration in a solution. pH below 7 is acidic, above 7 is basic and exactly 7 is neutral.
Strong acids which dissociate completely when dissolved in water and have high concentration of will have low pH and vice versa.
Citric acid is an acid having pH 4 and vinegar is also known as acetic acid which has pH 6. while examples of bases such as sodium hydroxide, sodium chloride, bleach etc have pH greater than 7.
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The heat of vaporization for ethanol is 0.826 kJ/g. Calculate the heat energy in joules required to boil 73.55 g of ethanol.
The heat of vaporization for ethanol is 0.826 kJ/g, then the heat energy in joules required to boil 73.55 g of ethanol is 60.7 joules, and to calculate the heat energy required to boil a given mass of ethanol, one need to multiply the heat of vaporization by the mass.
Mass of ethanol (m) = 73.55 g
Heat of vaporization (ΔHvap) = 0.826 kJ/g
First, one need to convert the mass from grams to kilograms to ensure consistent units throughout the calculation:
m = 73.55 g = 73.55 g / 1000 g/kg = 0.07355 kg
Now, one can calculate the heat energy required using the formula:
Heat energy (Q) = mass × heat of vaporization
Q = 0.07355 kg × 0.826 kJ/g = 0.0607 kJ
To convert the heat energy from kilojoules (kJ) to joules (J), one has to multiply by 1000:
Q = 0.0607 kJ × 1000 J/kJ = 60.7 J
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Guided
Learning
ar Inequalities in One Variable - Item 7601
Pre-Quiz
Practice
from a, it will be greater than b.
Post-Quiz
Finish
If the ordered pair (a, b) satisfies the inequality y> x-4, three of these statements are
true. Which statement is NOT true?
a and b may be equal to each other.
We
The statement about the inequality that is NOT true is (A), a and b may be equal to each other.
How to determine true statements?This is because the inequality y> x-4 means that y must be greater than x-4. If a and b are equal, then y = x-4, which means that the inequality is not satisfied.
The other three statements are true because:
If 4 is subtracted from a, it will be greater than b because y> x-4 means that y must be greater than x.
If 4 is subtracted from b, it will be less than a because y> x-4 means that y must be greater than x.
If 4 is subtracted from both a and b, the inequality will still be true because y> x-4 means that y must be greater than x, even if x and b are both decreased by 4.
Therefore, the answer to the question is the statement a and b may be equal to each other.
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Complete question:
Guided Learning ar Inequalities in One Variable - Item 7601
Pre-Quiz
Practice
from a, it will be greater than b.
Post-Quiz
Finish
If the ordered pair (a, b) satisfies the inequality y> x-4, three of these statements are true. Which statement is NOT true?
a and b may be equal to each other.
If you subtract 4 from a, it will be greater than b.
If you subtract 4 from b, it will be less than a.
If you subtract 4 from both a and b, the inequality will still be true.
A concentrated phosphoric acid solution is 85.5% H3PO4 by mass and has a density of 1.69 g/mL at 25 ∘
C. What is the molarity of H3PO4 ? What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 5.6 g of urea in 30.1 g of methanol, CH3OH ? How will an understanding of this concept help you in your healthcare career?
The molarity of H₃PO₄ is 14.615 M, and the mole fraction of urea is 0.318. Understanding molarity and mole fraction in healthcare enables accurate preparation of medications, dilution of solutions, and dosage calculations. It also helps understand substance interactions and their impact on processes like osmosis, benefiting patient care and treatment.
Molarity of H₃PO₄:
The molarity of H₃PO₄ can be calculated using the following formula:
Molarity = mass / molar mass * density
The mass of H₃PO₄ is 85.5% of the total mass of the solution, and the molar mass of H₃PO₄ is 98.008 g/mol. The density of the solution is 1.69 g/mL.
[tex]\text{Molarity} = \frac{0.855 \times 1.69 \times 1000}{98.008} = 14.615 \text{ M}[/tex]
Therefore, the molarity of H₃PO₄ is 14.615 M.
Mole fraction of urea:
The mole fraction of urea can be calculated using the following formula:
Mole fraction = moles of urea / total moles
[tex]\text{Moles} = \frac{10.0 \text{ g}}{50.0 \text{ g/mol}} = 0.200 \text{ moles}[/tex]
Mole fraction = 0.093 / 0.294 = 0.318
Therefore, the mole fraction of urea is 0.318.
How will an understanding of this concept help you in your healthcare career?
An understanding of molarity and mole fraction is important in healthcare because it allows you to calculate the concentration of solutions. This is important for many tasks, such as preparing medications, diluting solutions, and calculating dosages.
For example, if you are preparing a medication that requires a specific concentration of H₃PO₄, you can use the molarity to calculate the amount of H₃PO₄ that you need to add to the solution. Similarly, if you need to dilute a solution, you can use the mole fraction to calculate the amount of solvent that you need to add.
An understanding of molarity and mole fraction can also help you to understand how different substances interact with each other. For example, if you know the molarity of a solution, you can calculate the osmotic pressure of the solution. Osmotic pressure is important because it can affect the movement of water and other molecules across cell membranes.
Overall, an understanding of molarity and mole fraction is an essential skill for healthcare professionals. These concepts are used in many different tasks, and they can help you to provide better care for your patients.
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Know how to find the charge on ions in an ionic compound. Example: What is the charge on the manganese ion in Mn2O3?(+3) 36.) Know how to find the conjugate acid and the conjugate base of a species. Example: What is the conjugate base of phosphoric acid, H3PO4 ? (H2PO4−) 43.) Know how to write Lewis structures. Example: Draw the Lewis structure for ozone, O3. 44.) Know how to predict how many resonance forms a compound has. Example: How many resonance forms are there for the carbonate ion, CO32 ? (3) 45.) Know VSEPR theory (how to predict the shape of a molecule). Example: What is the shape of the SCl2 molecule?(bent)
1) The charge on the manganese ion is + 3
2) Conjugate base of phosphoric acid is H2PO4−
3) The Lewis structure of ozone is shown in the image attached
4) There are three resonance forms for carbonate ion
5) The molecule is bent
What is an ionic compound?
An ionic compound is a type of chemical compound composed of positively charged ions (cations) and negatively charged ions (anions) held together by electrostatic forces of attraction.
These compounds are typically formed when one or more electrons are transferred from a metal atom (which tends to lose electrons and form cations) to a non-metal atom (which tends to gain electrons and form anions).
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(a) Explain how temperature can affect the wavelengths of emission and intensity of a fluorescence spectrum.
[8 marks]
(b) Titanium dioxide is commonly used as a photocatalyst in the remediation of waste-water, self-cleaning windows and other surface-cleaning applications.
(i) Outline the mechanism of metal oxide photocatalysis for the degradation of organic matter on a surface that incorporates a metal oxide film such as titanium dioxide.
(ii) Name two materials that can be added to titanium dioxide to increase its effectiveness as a photocatalyst.
(a) Temperature affects fluorescence spectrum by decreasing the Stokes shift and increasing quenching at higher temperatures.
(b) (i) Metal oxide photocatalysis involves light absorption, charge separation, surface reactions, and degradation of organic matter. (ii) Dopants and co-catalysts can enhance TiO2's effectiveness as a photocatalyst.
(a) Temperature can affect the wavelengths of emission and intensity of a fluorescence spectrum through two main mechanisms: Stokes shift and temperature-dependent quenching.
Stokes Shift: Fluorescence occurs when a molecule absorbs energy (usually through light) and then emits the energy as light of longer wavelength.
This emitted light is known as fluorescence emission. The wavelength of fluorescence emission is usually red-shifted (i.e., longer wavelength) compared to the wavelength of the absorbed light. This shift is called the Stokes shift.
Temperature-dependent quenching: Fluorescence quenching refers to the reduction in the intensity of fluorescence emission. Temperature can influence the quenching process by affecting the rate of energy transfer between the excited fluorophore and surrounding molecules.
At higher temperatures, collisions between the excited fluorophore and other molecules occur more frequently, increasing the probability of non-radiative energy transfer processes such as collisional quenching. This leads to a decrease in fluorescence intensity.
(i) The mechanism of metal oxide photocatalysis, such as titanium dioxide ([tex]TiO_2[/tex]), for the degradation of organic matter on a surface involves the following steps:
Absorption of light: When [tex]TiO_2[/tex]is exposed to ultraviolet (UV) light, it absorbs photons, generating electron-hole pairs [tex](e^-/h^+[/tex]). The energy of the absorbed photons should be equal to or greater than the bandgap energy of [tex]TiO_2[/tex].
Charge separation: The generated electron [tex](e^-)[/tex] and hole [tex](h^+)[/tex]pairs get separated due to the internal electric field of the [tex]TiO_2[/tex] material. The electrons are usually excited from the valence band (VB) to the conduction band (CB) of [tex]TiO_2,[/tex] leaving behind positively charged holes.
Surface reactions: The separated charges can participate in redox reactions on the surface of the [tex]TiO_2[/tex] material. The electrons in the conduction band can reduce oxygen molecules [tex](O_2)[/tex] to form superoxide radicals ([tex]O_2^-[/tex]), while the holes in the valence band can oxidize water molecules ([tex]H_2O[/tex]) to form hydroxyl radicals (•OH).
Reactive species formation: The superoxide radicals ([tex]O_2^-[/tex]•) and hydroxyl radicals (•OH) produced in the surface reactions are highly reactive and can interact with organic pollutants adsorbed on the[tex]TiO_2[/tex] surface. These radicals can break down organic molecules through oxidation processes, leading to the degradation of organic matter.
Regeneration: After the degradation of organic matter, the electron and hole recombination can occur. To maintain the photocatalytic activity, it is important to ensure rapid charge separation and minimize electron-hole recombination.
(ii) Two materials that can be added to titanium dioxide ([tex]TiO_2[/tex]) to increase its effectiveness as a photocatalyst.
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The normal freezing point of a certain liquid X is −4.70∘C, but when 27.1 g of potassium bromide (KBr) are dissolved in 300 . g of X the solution freezes at -6.3. ∘C instead. Use this information to caiculate the molal freezing point depression constant Kf of X. Round your answer to 2 significant digits. Kf=πmol∘C⋅kg
The molal freezing point of the solution is calculated to be 3.5°C kg/mol.
Molal freezing point depression constant (Kf) is the number of degrees by which the freezing point of a solvent changes when 1.00 molar of a non-volatile, non-ionizing solute is dissolved in one kg of solvent.
Depression in freezing point Δt = -4.7°C -(- 6.3°C)
= 1.6°C
Mass of KBr added = 27.1g
The molar mass of KBr = 119g/mol
Moles of potassium bromide = 27.1g/119g/mol
Molality = number of solute moles/mass of solution in kg
M = 27.1g/119g/mol ÷ 0.3
KBr→ K⁺ + Br⁻
So i = 2
1.6°C =Kf × 27.1g/119g/mol × 2
Kf = 1.6°C × 119g/mol / 27.1g × 2
Kf = 3.513
So Kf = 3.5°C kg/mol
Here it is assumed that KBr is completely dissociated in water.
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