Answer: The concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] .....(1)
For KOH:
Initial molarity of KOH solution = 0.205 M
Volume of solution = 25.0 mL = 0.025 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]0.205M=\frac{\text{Moles of KOH}}{0.025L}\\\\\text{Moles of KOH}=(0.205mol/L\times 0.025L)=5.123\times 10^{-3}mol[/tex]
For phosphoric acid:
Initial molarity of phosphoric acid solution = 0.175 M
Volume of solution = 25.0 mL = 0.025 L
Putting values in equation 1, we get:
[tex]0.175M=\frac{\text{Moles of }H_3PO_4}{0.025L}\\\\\text{Moles of }H_3PO_4=(0.175mol/L\times 0.025L)=4.375\times 10^{-3}mol[/tex]
The chemical equation for the reaction of KOH and phosphoric acid follows:
[tex]3KOH+H_3PO_4\rightarrow K_3PO_4+3H_2O[/tex]
By Stoichiometry of the reaction:
3 moles of KOH reacts with 1 mole of phosphoric acid
So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}mol[/tex] of phosphoric acid
As, given amount of phosphoric acid is more than the required amount. So, it is considered as an excess reagent.
Thus, KOH is considered as a limiting reagent because it limits the formation of product.
Excess moles of phosphoric acid = [tex](4.375-1.708)\times 10^{-3}=2.667\times 10^{-3}mol[/tex]
By Stoichiometry of the reaction:
3 moles of KOH produces 1 mole of potassium phosphate
So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}moles[/tex] of potassium phosphate
For potassium phosphate:Moles of potassium phosphate = [tex]1.708\times 10^{-3}moles[/tex]
Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L
Putting values in equation 1, we get:
[tex]\text{Molarity of potassium phosphate}=\frac{1.708\times 10^{-3}}{0.050}=0.0342M[/tex]
For phosphoric acid:Moles of excess phosphoric acid = [tex]2.667\times 10^{-3}moles[/tex]
Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L
Putting values in equation 1, we get:
[tex]\text{Molarity of phosphoric acid}=\frac{2.667\times 10^{-3}}{0.050}=0.0533M[/tex]
For KOH:Moles of KOH remained = 0 moles
Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L
Putting values in equation 1, we get:
[tex]\text{Molarity of KOH}=\frac{0}{0.050}=0M[/tex]
Hence, the concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.
2. Assume that a sample of 10.00 g of a solid unknown is dissolved in 25.0 g of water. Assuming that pure water freezes at 0.0 oC and the solution freezes at -5.58 oC, what is the molal concentration of the solution
Answer:
m = 3 moles/kg
Explanation:
This is a problem of freezing point depression, and the formula or expression to use is the following:
ΔT = i*Kf¨*m (1)
Where:
ΔT: Change of temperature of the solution
i: Van't Hoff factor
m: molality of solution
Kf: molal freezing point depression of water (Kf = 1.86 °C kg/mol)
Now, the value of i is the number of moles of particles obtained when 1 mol of a solute dissolves. In this case, we do not know what kind of solution is, so, we can assume this is a non electrolyte solute, and the value of i = 1.
Let's calculate the value m, which is the molality solving for (1):
m = ΔT/Kf (2)
Finally, let's calculate ΔT:
ΔT = T2 - T1
ΔT = 0 - (-5.58)
ΔT = 5.58 °C
Now, let's replace in (2):
m = 5.58/1.86
m = 3 moles/kg
This is the molality of solution.
The other data of mass, can be used to calculate the molecular mass of this unknown solid, but it's not asked in the question.
In the simulation, open the Micro mode, then select the solutions indicated below from the dropdown list above the beaker. The beaker will fill up to the 0.50 L mark with the solution. Arrange the solutions in increasing order of basicity.
drain cleaner, hand soap, blood, milk, orange juice, soda pop
Answer:
In order of basicity we have
1. Soda pop (Least basic Normally called acidic)
2. Orange juice
3. Milk
4. Blood (slightly basic)
5. Hand soap
6. Drain cleaner (Highly basic)
Explanation:
Orange juice; the pH of orange juice is in the 3.3 to 4.2 range
Milk; the pH of milk about 6.5 to 6.7
Blood; the blood pH is around 7.35 to 7.45
Hand soap with contents such as ammonium hydroxide is basic, its pH is about 9-10
Drain cleaner contains baking soda or sodium bicarbonate which basic with a pH of 12 to 14
Soda pop pH of soda pop is in the range of 2.34 to 3.10. It contains carbonated water with a pH of 3–4, making it mildly acidic.
Arranging the above listed in order of increasing basicity, we have
1. Soda pop
2. Orange juice
3. Milk
4. Blood
5. Hand soap
6. Drain cleaner
Final answer:
The solutions in increasing order of basicity are: orange juice, soda pop, milk, blood, hand soap, and drain cleaner. Indicators like phenolphthalein are used in chemistry to assess the basicity of solutions.
Explanation:
The simulation activity mentioned involves organizing various solutions by their basicity. Basicity refers to the ability of a solution to act as a base, which is typically measured on the pH scale where a pH greater than 7 indicates a basic (alkaline) nature. In increasing order of basicity, the solutions you've mentioned can be generally expected to arrange as follows:
Orange juice - It is acidic due to the presence of citric acid and therefore has the lowest basicity.Soda pop - Carbonation makes soda acidic, but it's typically less acidic than orange juice.Milk - It is slightly acidic due to the presence of lactic acid.Blood - It has a pH close to neutral but can act as a buffer to maintain its pH, so it's considered less basic than soap and drain cleaner.Hand soap - Soaps are generally basic due to the presence of fatty acid salts.Drain cleaner - This is often very basic, with substances like sodium hydroxide being commonly used due to their capacity to unblock drains.To test basicity in the lab, indicators such as phenolphthalein may be used, which turn pink or fuschia in basic solutions. Basicity and acidity are fundamental concepts in chemistry, important in understanding the properties of substances and their reactions.
If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the equivalence point is 5.125 g, how many moles of hydrogen peroxide molecules are present?
Answer:
0.004522 moles of hydrogen peroxide molecules are present.
Explanation:
Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%
Mass of the solution , m= 5.125 g
Mass of the hydrogen peroxide = x
[tex]w/w\% = \frac{x}{m}\times 100[/tex]
[tex]3\%=\frac{x}{5.125 g}\times 100[/tex]
[tex]x=\frac{3\times 5.125 g}{100}=0.15375 g[/tex]
Mass of hydregn pervade in the solution = 0.15375 g
Moles of hydregn pervade in the solution :
[tex]=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol[/tex]
0.004522 moles of hydrogen peroxide molecules are present.
To determine the number of moles of hydrogen peroxide molecules present in the solution, multiply the mass of the solution by the mass percent of hydrogen peroxide. Then, convert the mass of hydrogen peroxide to moles using its molar mass. The number of moles of hydrogen peroxide molecules is approximately 0.00452.
Explanation:To determine the number of moles of hydrogen peroxide molecules present, we first need to calculate the mass of hydrogen peroxide in the solution. We can do this by multiplying the mass of the solution (5.125 g) by the mass percent of hydrogen peroxide (3.00% or 0.03).
Mass of hydrogen peroxide = 5.125 g × 0.03 = 0.15375 g
Next, we need to convert the mass of hydrogen peroxide to moles using its molar mass. The molar mass of hydrogen peroxide (H2O2) is 34.0146 g/mol.
Moles of hydrogen peroxide = 0.15375 g ÷ 34.0146 g/mol ≈ 0.00452 mol
Therefore, there are approximately 0.00452 moles of hydrogen peroxide molecules present.
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Calculate the concentration of OH-in a solution that has a concentration of H+ = 8.1 x 10^−6 M at 25°C. Multiply the answer you get by 1010 and enter that into the field to 2 decimal places.
Answer:
The answer is 12.35
Explanation:
From the question we are given that the concentration of [tex]H^{+}[/tex] is [tex]8.1 * 18^{-6}M[/tex]
Generally The rate equation is given as
[tex]K_{w} = [H^{+} ][OH^{-} ][/tex]
and [tex]K_{w}[/tex] the rate constant has a value [tex]1 * 10^{-14}[/tex]
Substituting and making [[tex]OH^{-}[/tex]] the subject we have
[tex][OH^{-} ] = \frac{1 * 10^{-14}}{[H^{+}]} = \frac{1 * 10^{-14}}{8.1 *10^{-6}} =1.235 * 10^{-9}[/tex]
[tex][OH ^ {-}] = 1.235 * 10^{-9}M[/tex]
Multiply the value by [tex]10^{10}[/tex] as instructed from the question we have
Answer = [tex]1.235 * 10 ^{-9} * 10^{10} = 12.35[/tex]
Hence the answer in 2 decimal places is 12.35
The concentration of OH¯ in the solution multiplied by 10¹⁰ is 12.35
Data obtained from the question:Concentration of Hydrogen ion [H⁺] = 8.1×10¯⁶ MConcentration of Hydroxide ion [OH¯] =? How to determine [OH¯]The concentration of hydroxide ion [OH¯] can be obtained as follow:
K = [H⁺][OH¯]
Equilibrium constant (K) = 1×10¯¹⁴
1×10¯¹⁴ = 8.1×10¯⁶ × [OH¯]
Divide both side by 8.1×10¯⁶
[OH¯] = 1×10¯¹⁴ / 8.1×10¯⁶
[OH¯] = 1.235×10¯⁹
Multiply by 10¹⁰
[OH¯] = 1.235×10¯⁹ × 10¹⁰
[OH¯] = 12.35
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The temperature of a 10.0 L sample of nitrogen in a sealed container is increased from 22°C to 202°C, while its pressure is increased from 1.00 atm to 3.00 atm. What is the new volume (in liters) of the nitrogen sample?
Answer:
The new volume is 5.37 L
Explanation:
Step 1: Data given
Initial volume = 10.0 L
Initial temperature = 22.0 °C
Initial pressure = 1.00 atm
Final temperature = 202 °C
Final pressure = 3.00 atm
Step 2: Calculate final volume
(P1*V1)/T1 = (P2*V2)/T2
⇒ with P1 = The initial pressure = 1.00 atm
⇒ with V1 = The initial volume = 10.0 L
⇒ with T1 = The initial temperature = 22 °C = 295 Kelvin
⇒ with P2 = The final pressure = 3.00 atm
⇒ with V2 = The final volume = TO BE DETERMINED
⇒ with T2 = The final temperature = 202 °C = 475 Kelvin
(1.00 * 10.0) / 295 = (3.00 * V2) / 475
10 / 295 = 3V2/ 475
3V2 = 4750/295
V2 = 5.37 L
The new volume is 5.37 L
What is the concentration of K+K+ in 0.15 MM of K2SK2S? Express your answer to one decimal place and include the appropriate units.
Answer : The concentration of [tex]K^+[/tex] ion in 0.15 M [tex]K_2S[/tex] is, 0.3 M
Explanation :
The given compound is, [tex]K_2S[/tex]
When [tex]K_2S[/tex] dissociates then it gives potassium ion and sulfide ion.
The balanced dissociation reaction is:
[tex]K_2S\rightarrow 2K^++S^{2-}[/tex]
By the stoichiometry we can say that, 1 mole of [tex]K_2S[/tex] dissociates to give 2 moles of [tex]K^+[/tex] ion and 1 mole of [tex]S^{2-}[/tex] ion.
Or, in terms of concentration we can say that:
0.15 M of [tex]K_2S[/tex] dissociates to give [tex]2\times 0.15M=0.3M[/tex] of [tex]K^+[/tex] ion and 0.15 M of [tex]S^{2-}[/tex] ion.
Thus, the concentration of [tex]K^+[/tex] ion in 0.15 M [tex]K_2S[/tex] is, 0.3 M
The concentration of K+ in a 0.15 M solution of K2S is 0.3 M, calculated by multiplying the concentration of K2S by two, since each unit of K2S produces two K+ ions.
Explanation:The concentration of K+ in a solution of 0.15 M K2S can be determined by recognizing that each formula unit of K2S produces two K+ ions.
Therefore, to find the concentration of K+, you multiply the concentration of K2S by two:
Concentration of K2S = 0.15 MConcentration of K+ = 2 × (Concentration of K2S)Concentration of K+ = 2 × 0.15 M = 0.30 MThe concentration of K+ in the solution is 0.3 M, where M stands for molarity, representing moles of solute per liter of solution.
In the mid-17th century, Isaac Newton proposed that light existed as a stream of particles, and the wave-particle debate continued for over 250 years until Planck and Einstein presented their revolutionary ideas. Give two pieces of evidence for the wave model and two for the particle model.
Answer:
Light as a wave
1. Young's Double Slit Experiment
2. Davisson-Germer Experiment
Light as a particle
1. Einsteins Photoelectric Effect Phenomenon
2. Diffraction Phenomenon of Particles
Evidence for light as a wave includes diffraction and interference patterns, while evidence for the particle model includes the photoelectric effect and emission spectra.
The mid-17th century debate on whether light is a wave or a particle extended well into the 20th century until revolutionary concepts by Planck and Einstein. There is evidence for both the wave model and the particle model of light. Two pieces of evidence for the wave nature of light are diffraction and interference patterns, as seen in Thomas Young's double-slit experiment.
Diffraction occurs when light encounters an obstacle, spreading out as a result, and interference patterns occur when waves overlap and combine in constructive or destructive ways. Conversely, evidence for the particle nature includes phenomena like the photoelectric effect, as explained by Einstein, where light knocks electrons from a material, and the behavior of emission spectra, where individual energy quanta or "photons" are emitted from atoms.
What is the wavelength (in nm) of the least energetic spectral line in the infrared series of the H atom?
Answer:
The least energetic spectral line in the infrared series of the H atom is 656.1 nm
Explanation:
Photon wavelength is inversely proportional to energy. To obtain the least energetic spectral line of the hydrogen atom (H), we determine the longest wavelength possible.
[tex]\frac{1}{\lambda} = R_H[\frac{1}{n_f^2} -\frac{1}{n^2}][/tex]
Where;
nf = 2
n = 3
RH is Rydberg constant = 1.09737 × 10⁷m⁻¹
λ is the wavelength of the least energetic spectral line
Substituting the above values into the equation, we will have
[tex]\frac{1}{\lambda} = 1.09737 X 10^7[\frac{1}{2^2} -\frac{1}{3^2}][/tex]
[tex]\frac{1}{\lambda} = 1.09737 X 10^7[\frac{1}{4} -\frac{1}{9}][/tex]
[tex]\frac{1}{\lambda} = 1.09737 X 10^7[0.25 -0.1111][/tex]
[tex]\frac{1}{\lambda} = 1.09737 X 10^7[0.1389][/tex]
[tex]\frac{1}{\lambda} = 1524246.93[/tex]
[tex]\lambda} = \frac{1}{1524246.93}[/tex]
[tex]\lambda} = 6.561 X10^{-7} m[/tex]
λ = 656.1 X10⁻⁹ m
In (nm): λ = 656.1 nm
Therefore, the least energetic spectral line in the infrared series of the H atom is 656.1 nm
The wavelength of the least energetic spectral line in the infrared series of the hydrogen atom is approximately 18,400 nanometers (nm).
To find the wavelength of the least energetic spectral line in the infrared series of the hydrogen atom, we can use the Rydberg formula for the hydrogen atom:
1 / λ = R_H * (1/n₁² - 1/n₂²)
Where:
λ is the wavelength of the spectral line.
R_H is the Rydberg constant for hydrogen, approximately 1.097 x 10^7 m⁻¹.
n₁ is the principal quantum number of the initial energy level.
n₂ is the principal quantum number of the final energy level.
For the least energetic line in the infrared series, we need to consider the transition where the electron moves from a higher energy level (n₂) to a lower energy level (n₁). In this case, n₂ > n₁.
Since we are interested in the infrared series, we'll consider transitions ending in the n₁ = 3 energy level. We want the least energetic line, so we'll choose the smallest value for n₂.
Let's take n₂ = 4 and calculate:
1 / λ = 1.097 x 10^7 m⁻¹ * (1/3² - 1/4²)
1 / λ = 1.097 x 10^7 m⁻¹ * (1/9 - 1/16)
1 / λ = 1.097 x 10^7 m⁻¹ * (7/144)
Now, solve for λ:
λ = 144 / (1.097 x 10^7 m⁻¹ * 7)
λ ≈ 0.0184 meters or 18.4 millimeters
To express the wavelength in nanometers (nm), we can convert millimeters to nanometers:
λ ≈ 18.4 mm * 1,000,000 nm/mm = 18,400 nm
So, the wavelength of the least energetic spectral line in the infrared series of the hydrogen atom is approximately 18,400 nm.
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To make the five standard solutions you will need for the Beer's Law Plot, you will first create a stock solution from ( NH 4 ) 3 Fe ( ox ) 3 ⋅ 3 H 2 O . The stock solution is made using 125 mg of ( NH 4 ) 3 Fe ( ox ) 3 ⋅ 3 H 2 O dissolved in deionized (DI) water in volumetric glassware for a total volume of 25 mL. Calculate the concentration of the stock solution.
Answer:
Concentration of stock solution in g/L = 5.0g/L
Concentration of stock solution in mol/L = 0.00117 mol/L or 0.00117 M
Explanation:
Concentration in mol/L = (Concentration in g/L)/(Molar mass)
Mass of (NH₄)₃Fe(ox)₃.3H₂O stock = 125mg = 0.125g
Volume of stock solution required = 25mL = 0.025 L
Concentration in g/L = 0.125/0.025 = 5.0 g/L
Molar Mass of (NH₄)₃Fe(ox)₃.3H₂O = (NH₄)₃Fe(C₂O₄)₃.3H₂O = 428.0632 g/mol
Concentration in mol/L = 5/428.0632 = 0.00117 mol/L = 0.00117 M
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Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
(a) n = 2 to n = 4
(b) n = 2 to n = 1
(c) n = 2 to n = 5
(d) n = 2 to n = 1
The question is incomplete , complete question is:
Arrange the following H atom electron transitions in order of increasing frequency of the photon absorbed or emitted:
(a) n = 2 to n = 4
(b) n = 2 to n = 1
(c) n = 2 to n = 5
(d) n = 4 to n = 3
Answer:
Hence the order of the transition will be : d < a < c < b
Explanation:
[tex]E_n=-13.6\times \frac{Z^2}{n^2}ev[/tex]
where,
[tex]E_n[/tex] = energy of [tex]n^{th}[/tex] orbit
n = number of orbit
Z = atomic number
Energy of n = 1 in an hydrogen atom:
[tex]E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV[/tex]
Energy of n = 2 in an hydrogen atom:
[tex]E_2=-13.6\times \frac{1^2}{2^2}eV=-3.40eV[/tex]
Energy of n = 3 in an hydrogen atom:
[tex]E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV[/tex]
Energy of n = 4 in an hydrogen atom:
[tex]E_4=-13.6\times \frac{1^2}{4^2}eV=-0.85 eV[/tex]
Energy of n = 5 in an hydrogen atom:
[tex]E_5=-13.6\times \frac{1^2}{5^2}eV=-0.544 eV[/tex]
a) n = 2 to n = 4 (absorption)
[tex]\Delta E_1= E_4-E_2=-0.85eV-(-3.40eV)=2.55 eV[/tex]
b) n = 2 to n = 1 (emission)
[tex]\Delta E_2= E_1-E_2=-13.6 eV-(-3.40eV)=-10.2 eV[/tex]
Negative sign indicates that emission will take place.
c) n = 2 to n = 5 (absorption)
[tex]\Delta E_3= E_5-E_2=-0.544 eV-(-3.40eV)=2.856 eV[/tex]
d) n = 4 to n = 3 (emission)
[tex]\Delta E_4= E_3-E_4=-1.51 eV-(-0.85 eV)=-0.66 eV[/tex]
Negative sign indicates that emission will take place.
According to Planck's equation, higher the frequency of the wave higher will be the energy:
[tex]E=h\nu [/tex]
h = Planck's constant
[tex]\nu [/tex] frequency of the wave
So, the increasing order of magnitude of the energy difference :
[tex]E_4<E_1<E_3<E_2[/tex]
And so will be the increasing order of the frequency of the of the photon absorbed or emitted. Hence the order of the transition will be :
: d < a < c < b
Final answer:
The order of increasing frequency of photon absorbed or emitted for the H atom electron transitions is (a) n = 2 to n = 4, (c) n = 2 to n = 5, (b) n = 2 to n = 1, and (d) n = 2 to n = 1.
Explanation:
The frequency of a photon absorbed or emitted by a hydrogen atom is related to the energy difference between the initial and final energy levels. The energy difference decreases as the value of n increases, so (a) n = 2 to n = 4 has the lowest frequency, followed by (c) n = 2 to n = 5, (b) n = 2 to n = 1, and finally, (d) n = 2 to n = 1 has the highest frequency.
You wish to extract an organic compound from an aqueous phase into an organic layer (three to six extractions on a marco scale). With regard to minimizing the number of transfer steps, would it be better to use an organic solvent that is heavier or lighter than water
Answer:
Explanation:
It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.
When conducting liquid-liquid extractions, it is better to use an organic solvent that is lighter than water. Using a lighter organic solvent allows for easier phase separation and more efficient extraction of the organic compound from the aqueous phase.
Explanation:When conducting liquid-liquid extractions, it is better to use an organic solvent that is lighter than water. This preference is based on the fact that organic solvents that are lighter than water form a separate layer on top of the aqueous phase, allowing for easier phase separation. This means that the organic compound can be more efficiently extracted from the aqueous phase into the organic layer, minimizing the number of transfer steps required.
For example, if you are extracting an organic compound from water, using an organic solvent like diethyl ether (density ≈ 0.71 g/mL) would be more effective compared to using an organic solvent like chloroform (density ≈ 1.478 g/mL) that is heavier than water.
Question 7 of an unknown protein are dissolved in enough solvent to make of solution. The osmotic pressure of this solution is measured to be at . Calculate the molar mass of the protein. Round your answer to significant digits.
The question is incomplete, here is the complete question:
371. mg of an unknown protein are dissolved in enough solvent to make 5.00 mL of solution. The osmotic pressure of this solution is measured to be 0.118 atm at 25°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.
Answer: The molar mass of the unknown protein is [tex]1.54\times 10^3g/mol[/tex]
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 0.118 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of protein = 371. mg = 0.371 g (Conversion factor: 1 g = 1000 mg)
Molar mass of protein = ?
Volume of solution = 5.00 mL
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]25^oC=[25+273]K=298K[/tex]
Putting values in above equation, we get:
[tex]0.118atm=1\times \frac{0.371\times 1000}{\text{Molar mass of protein}\times 5}\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 298K\\\\\pi=\frac{1\times 0.371\times 1000\times 0.0821\times 298}{0.118\times 5}=1538.4g/mol=1.54\times 10^3g/mol[/tex]
Hence, the molar mass of the unknown protein is [tex]1.54\times 10^3g/mol[/tex]
At what temperature (in degrees Celsius) will xenon atoms have the same average speed that Cl2Cl2 molecules have at 41 ∘C∘C?
Answer:
Xenon atoms will have a temperature of 308.25 °C
Explanation:
Step 1: Data given
Molar mass of Cl2 = 70.9 g/mol
Molar mass of Xenon = 131.29 g/mol
Temperatue of Cl2 molecules = 41 °C = 314 K
Step 2: Calculate temperature
The average speed of a gas particle is given by v = √(8RT/πM).
We can simplify this to:
T1/M1 = T2/M2
⇒ with T1 = The temperature of Cl2 molecules = 314 K
⇒ with M1 = the molar mass of Cl2 = 70.9 g/mol
⇒ with T2= The temperature of Xenon = TO BE DETERMINED
⇒ with M2 = The molar mass of Xenon = 131.29 g/mol
314/70.9 = T2/131.29
T2 = 581.4 Kelvin
581.4 Kelvin = 308.25 °C
Xenon atoms will have a temperature of 308.25 °C
Final answer:
To find the temperature at which xenon atoms have the same average speed as Cl2 molecules at 41 degrees Celsius, use the equation for the average speed of gas molecules. Substitute the molar masses of Cl2 and xenon into the equation and solve for temperature. The temperature is approximately 191.89 degrees Celsius.
Explanation:
To determine the temperature at which xenon atoms have the same average speed as Cl2 molecules at 41 °C, we need to use the equation for the average speed of gas molecules:
average speed = sqrt((3 * kB * T) / (molar mass))
where kB is the Boltzmann constant and T is the temperature in Kelvin.
Since the gas molecules we are comparing are different, we need to find the molar mass of Cl2 molecules and xenon atoms. The molar mass of Cl2 is 70.90 g/mol and the molar mass of xenon is 131.29 g/mol.
Substituting the values into the equation, we have:
sqrt((3 * (1.380649 × 10-23) * T) / (70.90)) = sqrt((3 * (1.380649 × 10-23) * (41 + 273.15)) / (131.29))
Simplifying and solving for T, we find that the temperature at which xenon atoms have the same average speed as Cl2 molecules at 41 °C is approximately 191.89 °Celsius.
Write the expected ground-state electron configuration for the following. (Express your answers as a series of orbitals using noble gas notation, in order of increasing orbital energy. For example, the electron configuration of Li would be entered as [He]2s1.) a The element with one unpaired electron that forms a covalent compound with fluorine. Configuration: b The (as yet undiscovered) alkaline earth metal after radium.
Answer:
For a: The element forming covalent compound with fluorine is Hydrogen.
For b: The electronic configuration of the element X is [tex][Uuo]8s^2[/tex]
Explanation:
Electronic configuration is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom is determined by the atomic number of that atom.
To write the electronic configuration of the elements in the noble gas notation, we first count the total number of electrons and then write the noble gas which lies before the same element.
For the given options:
For a:Covalent compound is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound. These are usually formed when two non-metals react.
Hydrogen is the 1st element of the periodic table and is considered as a non-metal. It has 1 unpaired electron.
Its electronic configuration is [tex]1s^1[/tex]
Fluorine is the 9th element of the periodic table. It is also a non-metal.
Its electronic configuration is [tex][He]2s^22p^5[/tex]
The two elements will share 1 electron to attain stable configuration and form HF compound.
For b:Radium is the 88th element of the periodic table having electronic configuration of [tex][Rn]7s^2[/tex]
Let the alkaline earth metal that is not yet discovered be X
The electronic configuration of the element X is [tex][Uuo]8s^2[/tex]
The expected ground-state electron configuration for the element that forms a covalent compound with fluorine is [He]2s2p4. The electron configuration of the undiscovered alkaline earth metal after radium cannot be determined.
Explanation:a) The element with one unpaired electron that forms a covalent compound with fluorine has the electron configuration of [He]2s22p4.
b) The alkaline earth metal after radium has not been discovered, so its electron configuration cannot be determined at this time.
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Why is the water solubility of benzoic acid significantly less than the water solubility of the benzoate ion?
Answer:
It is because of difference in polarization
Explanation: Benzoic acid is a non-polar strong acid with low solubility and no internal stabilizing structures to favor the carboxylic acid . Benzoate ion is polar which has high solubility due to being polar.
The water solubility of benzoic acid is much lower than that of the benzoate ion because of the difference in polarization.
PolarizationThe water solubility of benzoic acid is much lower than that of the benzoate ion because of the difference in polarization.Polarization is a feature of certain electromagnetic radiations that describes how the direction and magnitude of the vibrating electric field are associated in a specific way.Benzoic acid is a non-polar strong acid with low solubility and no internal stabilizing structures that favor carboxylic acid. Because benzoate ion is polar, it has a high solubility.Find out more information about the polarization here:
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How many electrons in an atom can have each of the following quantum number or sublevel designations?
(a) 2s
(b) n = 3, l = 2
(c) 6d
Answer :
(a) Number of electrons in an atoms is, 2
(b) Number of electrons in an atoms is, 10
(c) Number of electrons in an atoms is, 10
Explanation :
There are 4 quantum numbers :
Principle Quantum Numbers : It describes the size of the orbital. It is represented by n. n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
Magnetic Quantum Number : It describes the orientation of the orbitals. It is represented as . The value of this quantum number ranges from . When l = 2, the value of
Spin Quantum number : It describes the direction of electron spin. This is represented as . The value of this is for upward spin and for downward spin.
Number of electrons in a sublevel = 2(2l+1)
(a) 2s
n = 2
Value of 'l' for 's' orbital : l = 0
Number of electrons in an atoms = 2(2l+1) = 2(2×0+1) = 2
(b) n = 3, l = 2
Number of electrons in an atoms = 2(2l+1) = 2(2×2+1) = 10
(c) 6d
n = 2
Value of 'l' for 'd' orbital : l = 2
Number of electrons in an atoms = 2(2l+1) = 2(2×2+1) = 10
In quantum chemistry, the 2s sublevel can hold 2 electrons, the n = 3, l = 2 sublevel (3d) can hold 10 electrons, and the 6d sublevel can also hold 10 electrons.
Explanation:In quantum chemistry, specific sublevels or orbitals within an atom can hold certain numbers of electrons.
(a) The 2s sublevel can hold a maximum of 2 electrons. This is because 's' orbitals can contain up to 2 electrons.
(b) For n = 3, l = 2, this refers to the 3d orbital. 'd' orbitals can hold a maximum of 10 electrons, so 10 electrons can exist in this energy level.
(c) For the 6d orbital, it can also hold up to 10 electrons, as it is also a 'd' orbital.
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A solution contains 0.25 M Ni(NO3)2 and 0.25 M Cu(NO3)2. Can the metal ions be separated by slowly adding Na2CO3? Assume that for successful separation 99% of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of Na2CO3.
Explanation:
Ksp of NiCO3 = 1.4 x 10^-7
Ksp of CuCO3 = 2.5 x 10^-10
Ionic equations:
NiCO3 --> Ni2+ + CO3^2-
CuCO3 --> Cu2+ + CO3^2-
[Cu2+][CO3^2-]/[Ni2+][CO3^2-]
= (2.5* 10^-10)/(1.4* 10^-7)
= 0.00179.
[Cu2+]/[Ni2+]
= 0.00179
= 0.00179*[Ni2+]
If all of Cu2+ is precipitated before Na2CO3 is added.
= 0.00179 * (0.25)
The amount of Cu2+ not precipitated = 0.000448 M
The percent of Cu2+ precipitated before the NiCO3 precipitates = concentration of Cu2+ unprecipitated/initial concentration of Cu2+ * 100
= 0.000448/0.25 * 100
= 0.18%
Therefore, percentage precipitated = 100 - 0.18
= 99.8%
The two metal ions can be separated by slowly adding Na2CO3. Thus that is the unpptd Cu2+.
The metal ions can be separated by slowly adding Na2CO3 based on the relative solubilities of their carbonates. Nickel carbonate (NiCO3) is less soluble than copper carbonate (CuCO3), allowing the selective precipitation of nickel ions before copper ions.
Explanation:To determine if the metal ions can be separated by slowly adding Na2CO3, we need to consider the solubility of the metal carbonates. Nickel carbonate (NiCO3) and copper carbonate (CuCO3) both have low solubilities, but it is crucial to examine their relative solubilities. If one carbonate is significantly less soluble than the other, it can be selectively precipitated first.
In this case, NiCO3 is less soluble than CuCO3. Therefore, by slowly adding Na2CO3 to the solution, we can precipitate the majority of the Ni2+ ions as NiCO3 before CuCO3 begins to precipitate. This satisfies the condition that 99% of the metal ion must be precipitated before the other metal ion begins to precipitate.
Therefore, it is possible to separate the nickel and copper ions in the solution by slowly adding Na2CO3.
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The addition of a ____________ group to butane increases the solubility of the resulting molecule in water.
methyl
ethyl
double bond
hydroxyl group
propyl
Answer: hydroxyl group
Explanation:
Which of the following should most favor the solubility of an ionic solid in the water? Note: high and low refers to the magnitudes (i.e. the absolute value) of lattice and hydration energies.
Select one:
A) a low lattice energy for the solid and a low hydration energy for its ions
B) low lattice energy for the solid and a high hydration energy for its ions
C) a high lattice energy for the solid and a low hydration energy for its ions
D) a high lattice energy for the solid and a high hydration energy for its ions
Answer:low lattice energy for the solid and a high hydration energy for its ions
Explanation:
The lattice energy is the energy that holds the ions in the crystal lattice together. This energy must also be supplied for the lattice to disintegrate into its component ions. When this energy is lower than the energy released during the hydration of ions, the lattice easily breaks apart releasing the ions which are now strongly hydrated and the ionic solid is said to be soluble in water. Hence, solubility is favoured by lower lattice energy and higher hydration energy.
Final answer:
The option that favors the solubility of an ionic solid in water is a low lattice energy for the solid and a high hydration energy for its ions.
Explanation:
The correct choice that would most favor the solubility of an ionic solid in water is:
a low lattice energy for the solid and a high hydration energy for its ions
This is because low lattice energy allows for easier breakup of the ionic lattice, while high hydration energy promotes the attachment of water molecules to the ions, increasing solubility.
A compound that would be most soluble in water would be: a. glucose b. cholesterol. c. a large protein. d. a triglyceride.
Answer:
a. glucose
Explanation:
Glucose is the compound that would be most soluble in water because it has regions of hydrogen-oxygen polar bonds, making it hydrophilic. Cholesterol, large proteins, and triglycerides are nonpolar and not very soluble in water.
Explanation:Water is considered a polar solvent, and as such, substances that dissolve in water are usually polar or ionic compounds. Glucose, a, would be the most soluble in water because it has regions of hydrogen-oxygen polar bonds, making it hydrophilic. Cholesterol, b, is a nonpolar molecule and therefore not very soluble in water. Large proteins, c, can have both polar and nonpolar regions, so their solubility in water depends on the specific protein. Triglycerides, d, are nonpolar and not soluble in water.
For the Bradford assay, the instructor will make a Bradford reagent dye by mixing 50 ml of 95% v/v ethanol with 100 mg of Coomassie Blue followed by the addition of 100 ml of 85% v/v phosphoric acid. This entire mixture is then diluted to 1 liter with water. What is the final concentration of phosphoric acid?
Answer:
4.25% is the final concentration of phosphoric acid.
Explanation:
Initial concentration of phosphoric acid = [tex]C_1=85\%=0.85[/tex]
Initial volume of phosphoric acid = [tex]V_1=50 mL[/tex]
Final concentration of phosphoric acid = [tex]C_2=?[/tex]
Final volume of phosphoric acid = [tex]V_2=1 L=1000 mL[/tex]
( 1L = 1000 mL)
[tex]C_1V_1=C_2V_2[/tex]
[tex]C_2=\frac{C_1times V_1}{V_2}[/tex]
[tex]=\frac{0.85\times 50 mL}{1000 mL}=0.0425=4.25%[/tex]
4.25% is the final concentration of phosphoric acid.
An unsaturated hydrocarbon is A. a hydrocarbon that contains oxygen. B. a compound in which all carbon atoms have four single bonds. C. a compound in which one or more carbon atoms have double or triple bonds. D. a hydrocarbon that is dissolved in water. E. a cycloalkane with five or more carbons.
Answer:
. C. a compound in which one or more carbon atoms have double or triple bonds.
Explanation:
An unsaturated hydrocarbon -
It refers to the organic compound , where the carbon atom remains unsaturated with the number of hydrogen atoms , i.e. , maximum number of hydrogen atom a carbon atom can hold is four , and less than four number of hydrogen atom attached to the carbon atom , makes the hydrocarbon unsaturated in nature .
Hence ,
It the carbon atom have one or more number of double or triple bond , then the compound is said to be unsaturated in nature .
Hence , from the question,
The correct option is c.
A radio wave has a frequency of 3.8 x 10¹⁰ Hz. What is the energy (in J) of one photon of this radiation?
The energy of a photon can be calculated using Planck's equation (E = hf). Given the radio wave frequency of 3.8 x 10^10 Hz and using Planck's constant (6.626 x 10^-34 J·s), the energy of one photon of this radiation is 2.52 x 10^-23 Joules.
Explanation:To find the energy of one photon of radiation, we need to use Planck's formula: E = hf. Here, 'E' represents energy, 'h' is Planck’s constant (6.626 x 10^-34 J·s), and 'f' is frequency. Given that the frequency (f) of the wave is 3.8 x 10^10 Hz, we substitute these values into Planck's equation:
E = hf = (6.626 x 10^-34 J·s) (3.8 x 10^10 Hz) = 2.52 x 10^-23 J
So, the energy of one photon of this radiation is 2.52 x 10^-23 Joules.
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Rank the elements in each of the following sets in order of increasing IE₁:
(a) Sb, Sn, I
(b) Sr, Ca, Ba
Explanation:
Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.
When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period.
But when we move from top to bottom in a group then there occurs an increase in size of the atoms. Hence, ionization energy decreases along a group.
(a) As Sb, Sn and I are all period 5 elements. Hence, these elements are arranged in order of increasing [tex]IE_{1}[/tex] as follows.
Sn < Sb < I
(b) As Sr, Ca, and Ba are all elements of group 2a. Hence, these elements are arranged in order of increasing [tex]IE_{1}[/tex] as follows.
Ba < Sr < Ca
During the GFP purification experiment, the instructor will have to make breaking buffer for the students to use. This buffer contains 150mM NaCl. Given a bottle of crystalline NaCl (M.W. = 40g/mol), describe how you would make 500ml of 150mM NaCl. Include the type and sizes of any measuring vessels that you use to make the solution.
Answer:
Explanation:
150 mM NaCl = 0.15 M NaCl
500 ml of 0.15 M NaCl = 500 x 0.15 ml of M NaCl
= 75 ml of M NaCl
M.W = 40 g
1000 ml of M NaCl requires 40 g
75 ml of M NaCl requires (40/1000) x75
= 3 g
We shall take 500 ml of water in a graduated cylinder . Then we weigh 3 g of salt and add it to 500 ml of water and mix it . We shall get 500 ml of 150mM NaCl.
Final answer:
To make 500mL of 150mM NaCl solution, calculate the required mass of NaCl (4.383 grams), dissolve it in less than 500mL of DI water, and then adjust the volume to 500mL using a graduated cylinder for precise measurement.
Explanation:
To make 500 mL of 150mM NaCl solution, you first need to calculate the amount of NaCl required using the molarity formula, which is Molarity (M) = moles of solute / liters of solution. Given that the molar mass of NaCl is 58.44g/mol (not 40g/mol as incorrectly stated in the question), to prepare a 150mM (0.150M) NaCl solution, the calculation is as follows:
Calculate the moles of NaCl needed: moles = 0.150M * 0.500L = 0.075 moles.Convert moles to grams: grams = moles * molar mass = 0.075 mol * 58.44 g/mol = 4.383 grams.Measure out 4.383 grams of NaCl using an electronic scale.Dissolve the measured NaCl in less than 500 mL of di-ionized water (DI water) in a beaker.Transfer the solution to a 500 mL graduated cylinder and add DI water until the volume reaches 500 mL to ensure the correct concentration.
It is important to dissolve the NaCl in less than the final volume before adjusting up to the final desired volume. This approach ensures accuracy in the concentration of the resulting solution.
Theistic religion is centered in a belief in Select one: a. spirits or animals that control our lives. b. such totems as a rabbit's foot or four-leaf clover. c. gods who are thought to be powerful, who have an interest in human affairs, and who merit worship. d. abstract ideals only.
Answer: C. gods who are thought to be powerful, who have an interest in human affairs, and who merit worship.
Explanation: Theistic religion is a religious belief in the existence of a supreme being in the form of God or gods who control the affairs of men. Theism is classified into different categories from Monotheism ( the belief in the existence of only one God), Polytheism ( the belief in the presence of different supreme beings), Pantheism (where it is believed that the universe in itself is a God),Autotheism( a believe that each person has a divine nature,it means that divinity exists in us), etc.
Theism is derived from the Greek word known as "Theo" which means God.
Write two different resonance forms for triphenylmethyl cation. Write the structure showing the positive charge at an ortho position.
Final answer:
Two resonance forms of the triphenylmethyl cation were provided, with the positive charge located at the central carbon atom and one of the ortho positions.
Explanation:
The triphenylmethyl cation is a carbocation formed by removing an electron from the triphenylmethyl radical. It has the formula C19H16+. To write two resonance forms, we need to show the positive charge at different positions in the structure. One possible resonance form is where the positive charge is at the central carbon atom, and the other form is where the positive charge is at one of the ortho positions. The resonance structures can be represented as follows:
C6H5--C+--C6H5
: :
C6H4--C6H5 C6H6--C6H4
A 279.6 mL sample of an aqueous solution at 25°C contains 91.6 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 torr, what is the molar mass (in g/mol) of the unknown compound?
Answer:
The molar mass of the compound is 720.8 g/mol
Explanation:
Let's apply the colligative property of Osmotic pressure to solve this.
Formula is π = M . R . T
where π is pressure (atm)
M is molarity (mol/L)
R, Universal Constant Gases
T, Absolute T° ( T° in K = T° in C + 273)
Let's replace the data:
8.44 Torr = M . 0.082 L.atm/mol.K . 298K
As we have the pressure in Torr, we must convert to atm, to work properly.
8.44 Torr . 1 atm/ 760 Torr = 0.0111 atm
0.0111 atm = M . 0.082 L.atm/mol.K . 298K
0.0111 atm / (0.082 L.atm/mol.K . 298K) = M → 4.54×10⁻⁴ mol/L
So molarity is the moles of solute (mass (g) / molar mass) / volume (L)
Let's convert the volume to L → 279.6 mL . 1L / 1000 mL = 0.2796 L
4.54×10⁻⁴ mol/L . 02796 L = 1.27×10⁻⁴ moles
This moles are represented by the 91.6 mg, so let's convert the mass of solute from mg to g
91.6 mg . 1 g / 1000 mg = 0.0916 g
Molar mass → g/mol → 0.0916 g / 1.27×10⁻⁴ moles → 720.8 g/mol
At high temperatures phosphine (PH_3) dissociates into phosphorus and hydrogen by the following reaction: 4PH3 rightarrow P_4 + 6H_2 At 800 degree C the rate at which phosphine dissociates is dC_PH_3/dt = -3.715 times 10^-6 C_PH_3. for t in seconds. The reaction occurs in a constant-volume, 2-L vessel, and the initial concentration of phosphine is 5kmol/m^3
a. If 3mol of the phosphine reacts, how much phosphorus and hydrogen is produced?
b. Develop expressions for the number of moles of phosphine, phosphorus, and hydrogen present at any time, and determine how long it would take for 3 mol of phosphine to have reacted.
When 3 mol of phosphine reacts, (3/4) mol of phosphorus and 4.5 mol of hydrogen are produced. The number of moles of phosphine, phosphorus, and hydrogen at any time can be determined using the stoichiometry of the reaction. The time needed for 3 mol of phosphine to react can be found by integrating the given rate equation.
Explanation:To determine the amount of phosphorus and hydrogen produced when 3 mol of phosphine reacts, we can use the stoichiometry of the reaction. From the balanced equation, we see that for every 4 mol of PH3 that reacts, we get 1 mol of P4 and 6 mol of H2. Therefore, when 3 mol of PH3 reacts, we will produce (3/4) mol of P4 and (3/4) x 6 = 4.5 mol of H2.
Let's denote the number of moles of phosphine as nPH3, the number of moles of phosphorus as nP4, and the number of moles of hydrogen as nH2. At any time t, when 3 mol of phosphine has reacted, the corresponding number of moles of phosphorus and hydrogen can be determined using the stoichiometry of the reaction as mentioned above.
To determine how long it would take for 3 mol of phosphine to have reacted, we can use the given rate equation: dCPH3/dt = -3.715 × 10-6 CPH3. Since we have the initial concentration of phosphine (CPH3) as 5 kmol/m3, we can integrate the rate equation to find the time needed for 3 mol of phosphine to react.
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Calculate the molalities of some commerical reagents from thefollowing data:
HCl has a formula weight (amu) of 36.465, Density of thesolution(g/mL) of 1.19, Weight % of 37.2, and Molarity of12.1.
HC2H3O2 has a formula weightof 60.05, Density of 1.05, Weight % of 99.8, and Molarity of17.4.
NH3(aq) has a formula weight of 17.03, Denisty of0.90, Weight % of 28.0, and Molarity of 14.8
Answer:
The molality of HCl solution is 16.24 mol/kg.
The molality of [tex]HC_2H_3O_2[/tex] solution is 82,500 mol/kg.
The molality of [tex]NH_3[/tex] solution is 27.78 mol/kg.
Explanation:
formula used:
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
1) Mass percentage of the HCl solution = 37.2%
This means that in 100 grams of solution 37.2 grams of HCl is present.
Mass of HCl (solute)= 37.2 g
Mass of water(solvent) = 100 g - 37.2 g = 62.8 g = 0.0628 kg (1g = 0.001 kg)
Mole of HCl = [tex]\frac{37.2 g}{36.465 g/mol}=1.020 mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{1.020 mol}{0.0628 kg}=16.24 mol/kg[/tex]
The molality of HCl solution is 16.24 mol/kg.
2) Mass percentage of the [tex]HC_2H_3O_2[/tex] solution = 99,8%
This means that in 100 grams of solution 99.8 grams of [tex]HC_2H_3O_2[/tex] is present.
Mass of [tex]HC_2H_3O_2[/tex](solute)= 99.8 g
Mass of water(solvent) = 100 g - 99.8 g = 0.2 g = 0.0002 kg (1g = 0.001 kg)
Mole of [tex]HC_2H_3O_2[/tex] = [tex]\frac{99.8 g}{60.05g/mol}=16.50 mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{16.50 mol}{0.0002 kg}=82,500 mol/kg[/tex]
The molality of [tex]HC_2H_3O_2[/tex] solution is 82,500 mol/kg.
3) Mass percentage of the [tex]NH_3[/tex] solution = 28.0%
This means that in 100 grams of solution 28.0 grams of [tex]NH_3[/tex] is present.
Mass of [tex]NH_3[/tex](solute)= 37.2 g
Mass of water(solvent) = 100 g - 28.0 g = 72.0 g = 0.072 kg (1g = 0.001 kg)
Mole of [tex]NH_3[/tex]= [tex]\frac{28.0g}{17 g/mol}=2mol[/tex]
[tex]Molality=\frac{Moles}{\text{Mass of solvent(kg)}}[/tex]
[tex]m=\frac{2 mol}{0.072kg}=27.78 mol/kg[/tex]
The molality of [tex]NH_3[/tex] solution is 27.78 mol/kg.
Final answer:
To calculate the molalities of HCl, HC2H3O2, and NH3, convert the weight percentage to mass of solute, then convert that to moles using the formula weight, and divide by the mass of the solvent in kilograms. Use the provided densities to infer the mass of water solvent and proceed with calculations respecting significant figures.
Explanation:
Calculating Molalities of Commercial Reagents
To calculate the molalities of commercial reagents, you'll first need to determine the moles of solute and then divide that by the mass of the solvent in kilograms. Below is the process for each reagent provided:
For HCl (hydrochloric acid), we use the weight % to find the mass of HCl per 100g of solution, convert that mass to moles using the formula weight, and then divide by the solvent mass (water) in kilograms to find the molality.
In the case of HC2H3O2 (acetic acid), we apply a similar approach, starting with the weight % to calculate the mass of acetic acid in 100g of solution, convert to moles using its formula weight, and then divide by the mass of solvent to find molality.
For NH3 (aqueous ammonia), we begin by taking the 28.0 weight % value to get the mass of NH3 in 100g of solution, converting this mass to moles using NH3's formula weight, and finding the molality by dividing moles over kilograms of solvent.
Molality (°m) is defined as the number of moles of solute per kilogram of solvent and is especially useful because it doesn’t change with temperature. Using the provided densities and weights, we can assume the solvent used is predominantly water and calculate the weight of the water to use in the molality equation. The molar mass information provided, such as the formula weight of HCl (36.465 g/mol), acts as a conversion factor between grams and moles of solute.
At each step, remember to account for significant figures based on the precision of the given data.