Answer:
Dissolve the impure solid in a minimal amount of boiling solvent, cool the solution to form crystals, vacuum filter the solution to collect the pure crystals.
Explanation:
Recrystallization is a process when a solid with impurities is purified. To do this a solvent of the compound we want must be used. We need to use only the quantity necessary to dissolve the compound of interest, otherwise, the solvent will dissolve the impurities or it will interfere in the crystallization.
For most of the solids, the solubility increases with the increase of the temperature, so to speed up the process, heat must be added at the system, or the solvent must be boiling. Then, the solution will be cooled to form the crystals of the compound purified, and then it must be filtered in a vacuum because the crystals can slow down the filtration.
Given a water solution that contains 1.704 kg of HNO:/kg H:O and has a specific gravity of 1.382 at 20 °C, express the composition in the following ways: (a) Weight percent HNO (b) Pounds HNO3 per cubic foot of solution at 20 °c (c) Molarity (gmoles of HNOs per liter of solution at 20 °C)
Answer:
a) 63,0%
b) 54,4 Pounds HNO₃ per cubic foot of solution
c) 13,8 M
Explanation:
a) Weight percent is the ratio solute:solution times 100:
[tex]\frac{1,704 kg HNO_3}{2,704 kg Solution}[/tex] = 63,0%
b) Pounds HNO₃ per cubic foot of solution at 20 °c
Pounds HNO₃:
1,704 kg [tex]\frac{2,20462 pounds}{1 kg}[/tex] = 3,7567 pounds
Cubic foot:
2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex]x[tex]\frac{1 cubic foot }{28,3168 L}[/tex] = 0,069 ft³
Thus:[tex]\frac{3,7567 pounds}{0,069ft^3} =[/tex] = 54,4 Pounds HNO₃ per cubic foot of solution
c) Moles of HNO₃:
1704 g HNO₃ [tex]\frac{1 mol HNO_3}{63,01 g }[/tex] = 27,04 moles
Liters of solution:
2,704 kg [tex]\frac{1 L}{1,382 kg}[/tex] = 1,96 L of solution
Molarity:
[tex]\frac{27,04 mol}{1,96 L}[/tex] = 13,8 M
Final answer:
The composition of the solution can be expressed as 63.02 wt% HNO3, 54.28 lb HNO3/ft3, and a molarity of 14.2 M at 20 °C.
Explanation:
To express the composition of a water solution with 1.704 kg of HNO3 per kg of H2O and a specific gravity of 1.382 at 20 °C, we can calculate the following:
Weight Percent HNO3
Weight percent (wt%) is calculated as the mass of the solute divided by the total mass of the solution, multiplied by 100. For 1.704 kg HNO3 in 1 kg of water, the total mass of the solution is 1.704 kg + 1 kg = 2.704 kg. The weight percent HNO3 is then ((1.704 kg) / (2.704 kg)) × 100 = 63.02 wt%.
Pounds HNO3 per Cubic Foot of Solution at 20 °C
Firstly, we need to convert the specific gravity to density in g/mL: 1.382 (Specific Gravity) × 1.000 g/mL (density of water) = 1.382 g/mL. To find the density in lb/ft3, we multiply by 62.43 lb/ft3 which is the conversion factor from g/mL to lb/ft3. Then, multiply the density of the solution by the weight percent of HNO3 to find pounds HNO3 per cubic foot of solution: (1.382 g/mL) × (62.43 lb/ft3) × (63.02%) = 54.28 lb HNO3/ft3.
Molarity of HNO3 at 20 °C
Using the density of the concentrated HNO3 solution, 1.42 g/mL, and the given formula for molarity which is Molarity = [(%) (d)/ (Mw)] × 10, we substitute the values: Molarity = [(63.02) (1.42 g/mL) / (63.01)] × 10 = 14.2 M.
Khoisan salts is the number 1 producer of salts in SA for both local and international markets. 500 kg of KCl is dissolved in sufficient water to make a saturated solution at 370 K. At 370 Kthe solubility of KCl is 42 mass %. The solution is cooled to 320 K and the solubility is 31,5 mass % It is assumed that no water is evaporated. 2.1. Determine the amount of water is added to the 500 kg of KCl to produce the required saturated solution at 370 K. (3) 2.2. Determine the mass of KCl crystals formed after the cooling process to a temperature of 320 K. (Use the formula method)
Answer:
690 kg of water.125 kg of KCl crystals are formed after cooling processExplanation:
1) A saturated solution is a chemical solution containing the maximum concentration of a solute dissolved in the solvent. The solubility of KCl at 370K is 42%mass -42kg of KCl in 100kg of water+KCl , thus, the amount of water added to 500kg of KCl is:
500 kg of KCl × [tex]\frac{100 kg Water+KCl}{42kgKCl}[/tex] = 1190 kg of water +KCl
1190 kg of water +KCl - 500 kg of KCl = 690 kg of water
2) The maximum amount that this solution could solubilize of KCl at 320K is:
1190 kg of water + KCl × [tex]\frac{31,5 kgKCl}{100kgWater+KCl}[/tex] = 375 kg
Thus, the mass of KCl crystals formed are:
500 kg of KCl - 375 Kg of KCl = 125 kg of KCl
I hope it helps!
Order the following elements based upon their electronegativity, highest to lowest: carbon, flourine, hydrogen, nitrogen, oxygen, and sodium.
Answer:
The correct order is: Fluorine; Oxygen; Nitrogen; Carbon; Hydrogen; Sodium.
Explanation:
The electronegativity in the periodic table increases to the right in a period and up in a group. We can figure out the electronegativity of each element according to its electron configuration:
Carbon: [He] [tex]2s^{2} 2p^{2}[/tex]
Fluorine: [He] [tex]2s^{2} 2p^{5}[/tex]
Hydrogen: [tex]1s^{1}[/tex]
Nitrogen: [He] [tex]2s^{2} 2p^{3}[/tex]
Oxygen: [He] [tex]2s^{2} 2p^{4}[/tex]
Sodium: [Ne] [tex]3s^{1} [/tex]
The period of a chemical element is given by the last energy level of electron configuration.
The group of a chemical element is given by the amount of electrons in the last energy level of electron configuration.
Therefore,
Carbon: Period 2 Group 4
Fluorine: Period 2 Group 7
Hydrogen: Period 1 Group 1
Nitrogen: Period 2 Group 5
Oxygen: Period 2 Group 6
Sodium: Period 3 Group 1
Electronegativity increases when the number of the group increases.
Electronegativity increases when the number of the period decreases.
In conclusion, Fluorine has the greatest number of group and Sodium has the lowest number of group (also it has a greater number of period than hydrogen, so it is less electronegative than hydrogen)
F>O>N>C>H>Na
How many grams of solid Ca(OH)2 (74.1 g/mol) are required to make 500 ml of a 3 M solution?
Answer:
111.15 g are required to prepare 500 ml of a 3 M solution
Explanation:
In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.
Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of
(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.
How are hydrogen bonds formed between water molecules? Use a drawing to illustrate.
Answer:
They hydrogen bonds exist at the covalents molecules, as water where the 0 is an atom electronegative. The hydrogen bonds are formed because the hydrogen gives its electron to oxygen (d-), that's why we say, there is a dipole at the molecule. As the dipole has been formed, the hydrogen who gave the electron to oxygen will try to get another electro from other molecules. That is how the hydrogen bonds are formed.
Explanation:
a reasonable pka for a weak base is:
10.3
0.9
5.8
7.4
13.1
Answer:
0.9
Explanation:
The pka represents the force by which the molecules need to dissociate for the acids ,
Hence , lower the pka stronger will be the acid and that therefore will dissociate completely and vice versa , for a weak acid higher the pka .
And in case of a base , it will be completely reversed , lower pKa , weaker base ,
and higher pKa , stronger base .
From the data of the question ,
0.9 is the lowest value of the pKa , hence , weakest base .
Which radioactive emission has the smallest (least) mass? Question options:
A) Positron
B) Alpha
C) All have equal mass
D) Gamma
E) Beta
Answer:
The answer is D. gamma rays
Explanation:
A radioactive atom can have three different types of emission:
alpha particles (α) = they have a mass of 4 amu and they have a very low penetrating power.
Beta particles (β) = they have 5x[tex]10^{-4}[/tex] amu and they have an intermediate penetrating power
Gamma rays (γ) = they are not particles basically just energy so its mass is ≈ 0 and its penetrating power is higher
For this reason Gamma emissions (γ) has the smallest mass value.
Among the options provided, the gamma photon in radioactive emissions has the smallest mass (zero mass), which means that it's the correct answer to the question. Gamma radiation is energy emission without a corresponding mass.
Explanation:In the context of radioactive emissions, each type has a different mass. The alpha particle is the heaviest and is comprised of two neutrons and two protons. Beta particles, which are electrons (or positrons in the case of beta-plus decay), have a smaller mass. Positrons have the same mass as electrons but with a positive charge.
However, the gamma photon, which is a type of electromagnetic radiation, has no rest mass at all. So, out of the given options, the gamma ray has the smallest (in this case, zero) mass. This is an example of energy being emitted without a corresponding mass in a radioactive process, which often occurs when the remaining nucleus is in an excited state post decay and moves to a lower energy level by emitting a gamma photon.
Learn more about Radioactive Emissions here:https://brainly.com/question/35544778
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When preparing 1-bromobutane from 1-butanol and HBr, why
mustit be washed with H2O, cold NaOH and NaCl?
Answer:
Answer has been given below
Explanation:
During reaction of 1-butanol and HBr, some unreacted amount of HBr might present in reaction mixture. Hence cold NaOH is used to neutralize unreacted HBr.Washing with [tex]H_{2}O[/tex] is done to remove the salt produced during neutralization i.e. NaBr from organic layer of 1-bromobutane.NaCl is added to remove water from organic layer. Because NaCl is more soluble in water than organic layer.Question: Consider the molecule cyclohexene, which
containsonly hydrogen and carbon atoms.
When a sample is burned in oxygen, 4.822g of CO(2) and1.650g
of H(2)O are obtained.
What is its simplest formula?
Answer:
The simplest formula is C₃H₅.
Explanation:
The moles of CO₂ are the moles of Carbon because oxygen is in excess. Thus, the double of moles of H₂O are the moles oh hydrogen (Double because H₂O has two hydrogens for each mole).
4,822g CO₂ × (1mol / 44,01 g) = 0,1096 moles of C
1,650g H₂O × (1mol / 18,01 g) × 2= 0,1832 moles of H
The next step is divide these numbers between the smaller one, thus:
0,1096 moles / 0,1096 moles = 1
0,1832 moles / 0,1096 moles = 1,672
The last step is to find a number to multiple that gives integers:
1 × 3 = 3 Carbon
1,672 × 3 = 5 hydrogen
Thus, the simplest formula is C₃H₅.
I hope it helps!
why diamond has a very high melting point than aluminium??
Calculate the mass of 12 mL of a substance whose density is 1.1 g/mL. Express your answer in grams using the correct number of significant figures. Do not enter your answer using scientific notation.
Final answer:
To find the mass of the substance, multiply the given density (1.1 g/mL) by the given volume (12 mL) to get a mass of 13.2 grams, with the answer rounded to two significant figures.
Explanation:
To calculate the mass of a substance, you can use the formula mass = density × volume. Given a density of 1.1 g/mL and a volume of 12 mL, you simply multiply these two values together to find the mass.
Mass = 1.1 g/mL × 12 mL = 13.2 g
The density has two significant figures and the volume has two as well, so our final answer will be reported with two significant figures, hence 13.2 grams is the mass.
A 0.1375 g sample of solid magnesium is burned in
aconstant-volume bomb calorimeter that has a heat capacity of3024
J/ oC. The temperature increases by 1.126oC.
Calculate the heat given off by burning Mg, inkJ/g and in
kJ/mol.
Answer:
24.76 kJ/g
844.8 kJ/mol
Explanation:
The heat produced by the burning of the solid magnesium is equal to the heat absorbed by the calorimeter, which causes its temperature to increase.
First, we calculate the heat absorbed by the calorimeter, where C is the heat capacity and Δt is the temperature change.
Q = CΔt = (3024 J·°C⁻¹)(1.126 °C) = 3405 J
This is the same amount of heat that was produced by burning the magnesium
Now we can calculate the heat produced per gram of magnesium:
(3405 J)(1 kJ/1000 J) / (0.1375 g) = 24.76 kJ/g
We can convert grams to moles using the atomic weight of Mg (24.305 g/mol).
(24.76 kJ/g)(24.305 g/mol) = 844.8 kJ/mol
A worker is exposed to the following airborne concentrations of ethylbenzene: 8:00 – 10:00 125 ppm 10:15 – 12:15 88 ppm 12:45 – 3:15 112 ppm 3:30 – 5:00 70 ppm What is the 8-hour TWA exposure for the employee (calculate the exposure)? Does it exceed the OSHA PEL?
Answer:
The 8-hour TWA exposure for the employee is 101 ppm and it exceeds the OSHA PEL of 100 ppm for ethylbenzene.
Explanation:
The TWA for 8 hours is calculated by the sum of airbone concentrations multiplied by the time it has been exposed to that period. The total is divided by 8 which refers to the 8 hours total the employee has been exposed.
TWA = (125x2+88x2+112x2.5+70x1.5)/8.
The OSHA PEL is a known number for every compound and it can be find in PEL tables. In the case of ethylbenezene, it is 100 ppm.
What is the overall (X-fold) dilution of 4 serial 2-fold dilutions?
Answer:
16-fold dilution.
Explanation:
A serial dilution is any dilution where the concentration decrease by the same quantity in each successive step. So, for a 2-fold dilution, the concentration decrease 1/2, it means that if we have a sample with 10 M of concentration, after a 2-fold dilution it will be 5 M. For the next step it will be 1/2 of 5= 2.5 M, and successively.
Then, we just multiply the factor for each dilution. After 4 serial dilutions:
1/2 x 1/2 x 1/2 x 1/2 = 1/16
So, it would be a 16-fold dilution in the end.
If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount of salt? OL 0 0.01 L the task is impossible, as manganese nitrate is insoluble in water O 0.10 L 398 ml
Answer: The volume of water required is 398 mL
Explanation:
To calculate the molarity of solution, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]
We are given:
Mass of solute (manganese (II) nitrate tetrahydrate) = 16 g
Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol
Molarity of solution = 0.16 M
Putting values in above equation, we get:
[tex]0.16M=\frac{16g\times 1000}{251g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=398mL[/tex]
Hence, the volume of water required is 398 mL
The unit conversion between rankine and kelvin is (linear, exponential, quadratic, none of the above)?
The unit conversion between degrees Celsius and degrees Fahrenheit is (linear, exponential, quadratic, none of the above)?
Answer:
Unit conversion between Rankine and Kelvin is linear.
Unit conversion between degree Celsius and degree Fahrenheit is linear.
Explanation:
Relation between rankine and Kelvin is
[tex]R\ =\ \dfrac{9}{5}\ K[/tex]
So, the plot between Rankine and Kelvin is a straight line with zero intercept and has a slope having value [tex]\dfrac{9}{5}[/tex].
Relation between degrees Celsius and degrees Fahrenheit is given by
[tex]^{\circ}C\ =\ \dfrac{5}{9}\ (^{\circ}F\ -\ 32)[/tex]
So, the plot between degrees Celsius and degrees Fahrenheit is a straight line with slope [tex]\dfrac{5}{9}[/tex] and negative intercept of [tex]\dfrac{160}{9}[/tex].
One kilogram of saturated steam at 373 K and 1.01325 bar is contained in a rigid walled vessel. It has a volume of 1.673 m3. It is cooled to a temperature at which the specific volume of water vapour is 1.789 m. The amount of water vapour condensed in kilograms is (a) 0.0 (b) 0.065 (c) 0.1 (d) 1.0
Answer: Option (b) is the correct answer.
Explanation:
The given data is as follows.
Initial volume [tex](v_{1})[/tex] = 1.673 [tex]m^{3}[/tex]
Final volume [tex](v_{2})[/tex] = 1.789 [tex]m^{3}[/tex]
As, the amount of water vapor condensed will be as follows.
[tex]\frac{(v_{2} - v_{1})}{v_{2}}[/tex]
= [tex]\frac{(1.789 m^{3} - 1.673 m^{3})}{1.789 m^{3}}[/tex]
= [tex]\frac{0.116 m^{3}}{1.789 m^{3}}[/tex]
= 0.065 kg
Hence, we can conclude that the amount of water vapour condensed in kilograms is 0.065 kg.
The density of a certain type of plastic is 0.81 g/cm3. If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick, what is its mass?
mass = _____ x 10_____g
g
Enter your answer in scientific notation.
Answer: The mass of plastic sheet is [tex]8.1\times 10^4g[/tex]
Explanation:
The plastic sheet is in the form of cuboid. To calculate the volume of cuboid, we use the equation:
[tex]V=lbh[/tex]
where,
V = volume of cuboid
l = length of cuboid = 10.0 m = 1000 cm (Conversion factor: 1 m = 100 cm)
b = breadth of cuboid = 1.0 m = 100 cm
h = height of cuboid = 1 cm
Putting values in above equation, we get:
[tex]V=1000\times 100\times 1=10^5cm^3[/tex]
To calculate mass of a substance, we use the equation:
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
We are given:
Density of plastic sheet = [tex]0.81g/cm^3[/tex]
Volume of plastic sheet = [tex]10^5cm^3[/tex]
Putting values in equation 1, we get:
[tex]0.81g/cm^3=\frac{\text{Mass of plastic sheet}}{10^5cm^3}\\\\\text{Mass of plastic sheet}=8.1\times 10^4g[/tex]
Hence, the mass of plastic sheet is [tex]8.1\times 10^4g[/tex]
The mass of the plastic sheet is found by multiplying the density (0.81 g/cm³) with its volume (100,000 cm³), resulting in a mass of 81,000 g, or 8.1 × 10⁴ g in scientific notation.
To calculate the mass of a plastic sheet with a given density, you need to multiply its density by its volume. The density of the plastic is given as 0.81 g/cm³. To find the volume, convert all dimensions into the same unit, in this case, centimeters, because the density is given in g/cm³. The sheet is 10.0 m long (1000 cm), 1.0 m wide (100 cm), and 1 cm thick. Therefore, the volume is 1000 cm × 100 cm × 1 cm = 100,000 cm³.
Now, calculate the mass using the formula: mass = density × volume, so mass = 0.81 g/cm³ × 100,000 cm³. The mass of the plastic sheet is thus 81,000 g. To express this in scientific notation, we write it as 8.1 × 10⁴ g.
The heat capacity of chloroform (trichloromethane,CHCl3)
in the range 240K to 330K is given
byCpm/(JK-1mol-1) = 91.47
+7.5x10-2(T/K). In a particular experiment,
1.0molCHCl3 is heated from 273K to 300K. Calculate the
changein molar entropy of the sample.
Answer : The change in molar entropy of the sample is 10.651 J/K.mol
Explanation :
To calculate the change in molar entropy we use the formula:
[tex]\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p,m}dT}{T}[/tex]
where,
[tex]\Delta S[/tex] = change in molar entropy
n = number of moles = 1.0 mol
[tex]T_f[/tex] = final temperature = 300 K
[tex]T_i[/tex] = initial temperature = 273 K
[tex]C_{p,m}[/tex] = heat capacity of chloroform = [tex]91.47+7.5\times 10^{-2}(T/K)[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=1.0\int\limits^{300}_{273}{\frac{(91.47+7.5\times 10^{-2}(T/K))dT}{T}[/tex]
[tex]\Delta S=1.0\times [91.47\ln T+7.5\times 10^{-2}T]^{300}_{273}[/tex]
[tex]\Delta S=1.0\times 91.47\ln (\frac{T_f}{T_i})+7.5\times 10^{-2}(T_f-T_i)[/tex]
[tex]\Delta S=1.0\times 91.47\ln (\frac{300}{273})+7.5\times 10^{-2}(300-273)[/tex]
[tex]\Delta S=8.626+2.025[/tex]
[tex]\Delta S=10.651J/K.mol[/tex]
Therefore, the change in molar entropy of the sample is 10.651 J/K.mol
1. Make a 100 mL volume of 100 mM acetic acid (HAC) HAC (glacial, 100%, 17.4 M, 1.05 g/ml, MW 60.05) ddh,0 ML add to 100 mL Measure 250 ml beaker. Add mL of ddH20 using a 100-ml graduated cylinder and transfer this volume to a ul of 100 mM HAC to the water. Mix with a stir bar.
Final answer:
To prepare a 100 mM acetic acid solution, calculate the moles needed, measure the glacial acetic acid based on its density and volume, and dilute with distilled water to the final volume. Mix well for homogeneity.
Explanation:
The question involves preparing a solution of acetic acid and relates to the subjects of molarity and solution preparation in chemistry. The solution is to be made to a specified concentration using a direct dilution from a more concentrated stock.
Steps to Prepare 100 mM Acetic Acid Solution
First, we calculate the number of moles of acetic acid required for 100 mL of a 100 mM solution using the molar mass of acetic acid (MW 60.05). Second, we use the density of glacial acetic acid (1.05 g/mL) and the volume needed (from the calculated moles and molarity) to determine how much of the glacial acetic acid to measure out. Finally, we add this measured amount of acid to a beaker and dilute to the mark with distilled water (ddH2O) to achieve the desired final concentration. Mixing with a stir bar ensures a homogeneous solution.
To calculate the mole fraction of acetic acid in a solution, we need the number of moles of acetic acid and the number of moles of water in the solution. This can be accomplished by converting the mass of each component to moles using their respective molar masses.
A liquid mixture contains water (H2O, MW = 18.0), ethanol (C2H5OH, MW = 46.0) and methanol (CH3OH, MW = 32.0). Using two different analytical techniques to analyze the mixture, it was determined that the water mole fraction was 0.250 while the water mass fraction was 0.134. Determine the mole fraction ethanol (C2H5OH) and the mole fraction methanol (CH3OH) in the solution. Report the values to the correct number of significant figures.
Answer:
Mole fraction of ethanol is 0.363.
Mole fraction of methanol is 0.387.
Explanation:
Mole fraction of water =[tex]\chi_1=0.250[/tex]
Mole fraction of ethanol =[tex]\chi_2[/tex]
Mole fraction of methanol = [tex]\chi_3[/tex]
[tex]\chi_1+\chi_2+\chi_3=1[/tex]
[tex]\chi_2+\chi_3=1-\chi_1=0.750[/tex]
[tex]\chi_2+\chi_3=0.750[/tex]
[tex]\chi_1=\frac{n_1}{n_1+n_2+n_3}[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}[/tex]
[tex]\chi_3=\frac{n_3}{n_1+n_2+n_3}[/tex]
[tex]n_1+n_2+n_3=1[/tex]
Moles of water = [tex]n_1=0.250 mol[/tex]
Moles of ethanol= [tex]n_2[/tex]
Moles of methanol= [tex]n_3[/tex]
[tex]n_2+n_3=0.750 mol[/tex]
Mass of the mixture = M
Mass of water, [tex]m_1[/tex]
[tex]=n_1\times 18.0 g/mol=0.250 mol\times 18.0 g/mol=4.5 g[/tex]
Fraction of water by mass = 0.134
[tex]\frac{n_1\times 18.0 g/mol}{M}=0.134[/tex]
[tex]M=\frac{0.250 mol\times 18.0 g/mol}{0.134}=33.58 g[/tex]
Mass of ethanol = [tex]m_2[/tex]
Mass of methanol = [tex]m_3[/tex]
[tex]m_1+m_2+m_3=M[/tex]
[tex]4.5 g+m_2+m_3=33.58 g[/tex]
[tex]m_2+m_3=29.08 g[/tex]..[1]
[tex]\frac{m_2}{46.0 g/mol}+\frac{m_3}{32.0 g/mol}=0.750 mol[/tex]
[tex]16m_1+23m_3=552[/tex]..[2]
On solving [1] and [2]:
[tex]m_2 = 16.70, m_3= 12.38 g[/tex]
Mole fraction of ethanol =[tex]chi_2[/tex]
[tex]\chi_2=\frac{n_2}{n_1+n_2+n_3}[/tex]
[tex]=\frac{\frac{16.70 g}{46.0 g/mol}}{1 mol}= 0.363[/tex]
Mole fraction of methanol = [tex]chi_3[/tex]
[tex]\chi_3=\frac{n_3}{n_1+n_2+n_3}[/tex]
[tex]=\frac{\frac{12.38 g}{32.0 g/mol}}{1 mol}= 0.387[/tex]
Calculate the concentration of A bottle of wine contains 12.9% ethanol by volume. The density of ethanol (CH3OH) is 0.789 g/cm ethanol in wine in terms of mass percent and molality Mass percent Molality =
Answer:
The mass percentage of the solution is 10.46%.
The molality of the solution is 2.5403 mol/kg.
Explanation:
A bottle of wine contains 12.9% ethanol by volume.
This means that in 100 mL of solution 12.9 L of alcohol is present.
Volume of alcohol = v = 12.9 L
Mass of the ethanol = m
Density of the ethanol ,d= [tex]0.789 g/cm^3=0.789 g/mL[/tex]
[tex]1 cm^3=1 mL[/tex]
[tex]m=d\times v=0.798 g/ml\times 12.9 mL = 10.1781 g[/tex]
Mass of water = M
Volume of water ,V= 100 mL - 12.9 mL = 87.1 mL
Density of water = D=1.00 g/mL
[tex]M=D\times V=1.00 g/ml\times 87.1 mL =87.1 g[/tex]
Mass percent
[tex](w/w)\%=\frac{m}{m+M}\times 100[/tex]
[tex]\frac{10.1781 g}{10.1781 g+87.1 g}\times 100=10.46\%[/tex]
Molality :
[tex]m=\frac{m}{\text{molar mass of ethanol}\times M(kg)}[/tex]
M = 87.1 g = 0.0871 kg (1 kg =1000 g)
[tex]=\frac{10.1781 g}{46 g/mol\times 0.0871 kg}[/tex]
[tex]m=2.5403 mol/kg[/tex]
To calculate the concentration of ethanol in wine, multiply the volume percent of ethanol by the density of ethanol. Then divide by the mass of the solution and multiply by 100 to get the mass percent.
Explanation:To calculate the concentration of ethanol in wine, we can use the mass percent formula. Mass percent is calculated by dividing the mass of the solute (ethanol) by the mass of the solution (wine), and multiplying by 100. The mass of ethanol can be found by multiplying the volume percent of ethanol (12.9%) by the density of ethanol (0.789 g/cm³). The density of wine is typically close to 1 g/cm³. So, to find the concentration of ethanol in terms of mass percent, we can follow these steps:
Assuming we have 100 mL of wine, we can calculate the volume of ethanol in the wine by multiplying 100 mL by 12.9% (0.129).Next, we can convert the volume of ethanol to mass by multiplying it by the density of ethanol. This gives us 0.129 mL * 0.789 g/cm³ = 0.101961 g.Since the total mass of the wine is equal to the mass of ethanol (since ethanol is the only solute), the concentration of ethanol in terms of mass percent is 0.101961 g / 100 mL * 100% = 0.101961%.Learn more about Concentration of Ethanol in Wine here:https://brainly.com/question/14641045
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Many elements combine with hydrogen(valence = 1) to give compounds called hydrides. Use the position of an element in the periodic table to deduce its valence. Write formulas for these hydrides without using subscripts, for example XH3. If no hydride forms, write "none". What is the formula of the hydride formed by sulfur? What is the formula of the hydride formed by potassium ? ||
Answer:
Explanation:
Use the position of an element in the periodic table to deduce its valence.
The periodic table is a table that groups elements based on their periodic functions. A group is a vertical arrangement of elements. The group number shows the number of elements in the outer shell of the atoms.
The groups runs from 1 to 8. On the periodic table, elements that has more than 4 electrons in their outermost shell will have a valency of 8 less than the number of outermost electrons.
Write formulas for these hydrides without using subscripts, for example XH3. If no hydride forms, write "none".
for group I : NaH, KH,
group II : MgH2, CaH2,
group III: BH3, AlH3
Group 8 do not form hydrides because of their inertness.
What is the formula of the hydride formed by sulfur?
the hydride of sulfur is H₂S
What is the formula of the hydride formed by potassium ?
the hydride of potassium is KH
Final answer:
The formula of the hydride formed by sulfur is H2S, while the formula for the hydride formed by potassium is KH, based on the valence of sulfur and potassium corresponding to their positions in the periodic table.
Explanation:
When determining the formulas for hydrides, the valence of the non-hydrogen element is crucial. Sulfur, which has a valence of 2, forms a hydride by combining with two hydrogen atoms, giving us the formula H2S. In contrast, potassium belongs to the alkali metals with a valence of 1, thus it combines with one hydrogen atom to form a hydride, resulting in the formula KH.
The periodic table helps us understand these valences due to an element's group number. Sulfur, located in group 16, typically forms compounds where it has two bonding sites, as seen with its hydride, hydrogen sulfide. Potassium, found in group 1, forms compounds by donating a single electron, resulting in a 1:1 ratio with hydrogen in potassium hydride.
. Does each of these describe a physical change or a chemical change? Explain a. The moth balls gradually vaporize in a closet. Hydrofluoric acid attacks class, and is used to etch calibrations marks on glass laboratory utensils A French chef making a sauce with brandy is able to burn of the alcohol from the brandy leaving just the brandy flavoring d. Chemistry majors sometimes et holes in the cotton jeans they wear to lab, because of acid spills A piece of egg boiled in water for 20 minutes
Answer:
a. The mothballs gradually vaporize in a closet. Physical change
b. Hydrofluoric acid attacks glass, and is used to etch calibrations marks on glass laboratory utensils. Chemical change
c. A French chef making a sauce with brandy is able to burn of the alcohol from the brandy leaving just the brandy flavoring. Chemical change
d. Chemistry majors sometimes et holes in the cotton jeans they wear to lab, because of acid spills. Chemical change
e. A piece of egg boiled in water for 20 minutes. Chemical change
Explanation:
In a physical change, there is no change in the chemical composition of the substance, the change is only on the physical properties. For example, state changes (a).
In a chemical change instead, usually there is a combination of two or more substances that combine to form a new one different from the originals. For example, chemical reactions like combustion (c) or protein denaturalization (e). The attacking of hydrofluoric acid to glass (b) or the acid pill attacking cotton are chemical reactions too (d).
Physical change or a chemical change are given as:
a. The moth balls gradually vaporize in a closet -physical change.
b. Hydrofluoric acid attacks glass and is used to etch calibration marks on glass laboratory utensils-chemical change.
c. A French chef making a sauce with brandy is able to burn off the alcohol from the brandy, leaving just the brandy flavoring- a chemical change.
d. Chemistry majors sometimes etch holes in the cotton jeans they wear to the lab because of acid spills- physical change.
A piece of egg boiled in water for 20 minutes- chemical change.
a. In a closet, moth balls progressively vaporise: This represents a physical transformation. The moth balls sublimate, going straight from a solid state to a gaseous state, undergoing a phase change from a solid to a gas.
b. Glass is attacked by hydrofluoric acid, which is also used to etch calibration marks on glass laboratory utensils: This refers to a chemical transformation. The glass and hydrofluoric acid interact, causing a chemical reaction that etches the surface of the glass.
c. A French cook who uses brandy in a sauce can burn out the alcohol, leaving only the brandy flavouring: This explains a chemistry alteration. Alcohol is burned off through a chemical reaction, specifically alcohol combustion, which transforms ethanol (alcohol) into carbon dioxide and water vapour.
d. Because of acid spills, chemistry majors occasionally etch holes in the cotton pants they wear to the lab: This sentence depicts a physical alteration. The fibres in the jeans melt or break down as a result of acid spills, creating holes.
e. A chemical change is described by a piece of egg that was boiled in water for 20 minutes. The denaturation of proteins and the coagulation of the egg white are two distinct chemical reactions that occur inside the egg as a result of boiling.
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Urea (CH4N2O) is a common fertilizer that can be synthesized by the reaction of ammonia (NH3) with carbon dioxide as follows: 2NH3(aq)+CO2(aq)→CH4N2O(aq)+H2O(l) In an industrial synthesis of urea, a chemist combines 132.0 kg of ammonia with 211.4 kg of carbon dioxide and obtains 172.7 kg of urea. determine the limiting reactant
Answer:
NH₃ is the limiting reagent
Explanation:
Mass of ammonia = 132.0kg
Mass of carbon dioxide = 211.4kg
Mass of urea = 172.7kg
Unknown:
Limiting reagent = ?
Solution
First, we write the balanced stoichiometeric equation:
2NH₃ + CO₂ → CH₄N₂O + H₂O
The reactant that is present in short supply determines the amount of product that is formed in a reaction. This reactant is called the limiting reagent.
To establish the limiting reagent, we need to go find out what is happening at the start of the reaction:
Convert the masses of the reactants to moles.
Number of moles of NH₃ = [tex]\frac{mass}{molar mass}[/tex]
Molar mass of NH₃ = 14 + (3x1) = 17g/mol
Number of moles of NH₃ = [tex]\frac{132}{17}[/tex] = 7.765mole
Number of moles of CO₂ = [tex]\frac{mass}{molar mass}[/tex]
Molar mass of CO₂ = 12 + (2 x 16) = 44g/mol
Number of moles of CO₂ = [tex]\frac{211.4}{44}[/tex] = 4.805mole
From the reaction equation:
2 moles of NH₃ reacted with 1 mole of CO₂
so 7.765 mole of NH₃ will require [tex]\frac{7.765}{2}[/tex]mole, 3.883 of CO₂
But we are given 4.805mole of CO₂.
Therefore, CO₂ gas is in excess and NH₃ is the limiting reagent.
The limiting reactant for the synthesis of CH₄N₂O given the data is ammonia, NH₃
Balanced equation2NH₃ + CO₂ —> CH₄N₂O + H₂O
Molar mass of NH₃ = 17 g/mole
Mass of NH₃ from the balanced equation = 2 × 17 = 34 g = 0.034 Kg
Molar mass of CO₂ = 44 g/mole
Mass of CO₂ from the balanced equation = 1 × 44 = 44 g = 0.044 Kg
SUMMARY
From the balanced equation above,
0.034 Kg of NH₃ reacted with 0.044 Kg of CO₂
How to determine the limiting reactantFrom the balanced equation above,
0.034 Kg of NH₃ reacted with 0.044 Kg of CO₂
Therefore,
132 Kg of NH₃ will react with = (132 × 0.044) / 0.034 = 170.8 of Kg of CO₂
From the calculation made above, we can see that only 170.8 of Kg of CO₂ out of 172.7 Kg given is needed to react completely with 132 Kg of NH₃.
Thus, NH₃ is the limiting reactant
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A 200 ml sample of 0.1015 Miric acid is mbred with 2300 ml of water. What is the molar concentration of nitric acid in the final solution 0 406 M 127M 325 x 10 M 5.08 x 10 M 8.12 x 10 M water Backspace lu 'o Pin
Answer: Molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]
Explanation:
According to the dilution law,
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of stock [tex]HNO_3[/tex] solution = 0.1015 M
[tex]V_1[/tex] = volume of stock [tex]HNO_3[/tex]solution = 200 ml
[tex]M_2[/tex] = molarity of dilute [tex]HNO_3[/tex] solution = ?
[tex]V_2[/tex] = volume of dilute [tex]HNO_3[/tex] solution = (2300 +200 )ml = 2500 ml
Putting in the values we get:
[tex]0.1015M\times 200=M_2\times 2500[/tex]
[tex]M_2=8.12\times 10^{-3}M[/tex]
Thus the molar concentration of nitric acid in the final solution is [tex]8.12\times 10^{-3}M[/tex]
What is the theoretical Van’t Hoff factor for the following substances: HCL & ethanol (CH3CH2OH)
Answer:
i=2 for HCl and i=1 for ethanol
Explanation:
Hello,
Since the hydrochloric acid is composed by sodium and chloride ions which are completely dissociated, two types of ions are present, that's why i becomes 2.
On the other hand, as long as the ethanol doesn't present dissociation in aqueous solution, the Van't Hoff factor becomes 1.
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a 50-g sample contains 32.1 g of oxygen 2 g of hydrogen and 15.9 g of carbon
Answer:
i need more to solve this
Explanation:
Options for full question:
(A) % oxygen is 32.1%
(B) % hydrogen is 4%
(C) total percent composition of all elements is approximately 100%
(D) % carbon is 15.9%
Answer:
Options B and C
Explanation:
Information Given;
Mass of sample - 50g
Mass of Oxygen - 32.1g
Mass of Hydrogen - 2g
Mass of Carbon - 15.9g
The percentage composition of an element in a compound is the mass percentage of the element present in the compound. It tells the mass percentage of each element present in a compound.
The formular is given as;
Percentage composition = (Mass of element / Mass of compound) * 100
For oxygen;
Percentage Composition = (32.1 / 50) * 100 = 0.642 * 100= 64.2%
For Hydrogen;
Percentage Composition = (2 / 50) * 100 = 0.04 * 100= 4%
For Carbon;
Percentage Composition = (15.9 / 50) * 100 = 0.318 * 100= 31.8%
Total percentage composition of all elements = 4% + 31.8% + 64.2% = 100%
Based on this,
Option A is incorrect
Option B is Correct.
Option C is Correct.
Option D is incorrrect
At 0.70 atmosphere pressure, a certain mass of a gas has a volume of 3.8 L. What will be the volume if the pressure is decreased to 0.409 atmosphere pressure?
Answer: The new volume of the gas will be 6.50 L
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.
The equation given by this law is:
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.
We are given:
[tex]P_1=0.70atm\\V_1=3.8L\\P_2=0.409atm\\V_2=?L[/tex]
Putting values in above equation, we get:
[tex]0.70atm\times 3.8L=0.409atm\times V_2\\\\V_2=6.50L[/tex]
Hence, the new volume of the gas will be 6.50 L
Give the names for each of the elements between atomic numbers 1 and 86 that has a symbol starting with "C".
Identify the family each element belongs to. If the element does not belong to a family, identify it as a metal or non-metal.
2. In your own words, define the following terms. a. Element b. Atom c. Molecule d. Compound 3. How many oxygen atoms are in one molecule (or formula unit) of each of the following compounds? NO3 Al(OH)3 Ca(NO3)2 Ba(OCN)2
Explanation:
1)
Carbon with atomic number 6. The symbol of carbon is C.It belongs to non metals.Chlorine with atomic number of 17. The symbol of chlorine is Cl.It belongs to non metals.Calcium with atomic number 20. The symbol of calcium is Ca. It belongs to metals.Chromium with atomic number 24. The symbol of chromium is Cr. It belongs to metals.Cobalt with atomic number of 27. The symbol of cobalt is Co. It belongs to metals.Copper with atomic number of 29. The symbol of copper is Cu. It belongs to metals.Cadmium with atomic number of 48. The symbol of cadmium is Cd. It belongs to metals.Cesium with atomic umber of 55.The symbol of cesium is Cs. It belongs to metals.Cerium with atomic number of 58. the symbol for cerium in Ce. It belongs to metals.2)
Element are those substance which can not be split into simpler substance.they are made up of single atom.
Atom is the structural unit of the matter and smallest component of an element.
Molecules are group of atoms bonded together. The atom boned can be of more than one type.
Compounds are group of atoms of different elements bonded together.
3) Oxygen atoms in one molecule of following compounds:
a) [tex]NO_3[/tex]
1 × 3 = 3
There 3 oxygen atoms in the 1 molecule of [tex]NO_3[/tex].
b) [tex]Al(OH)_3[/tex]
3 × 1 = 3
There 3 oxygen atoms in the 1 molecule of [tex]Al(OH)_3[/tex].
c) [tex]Ca(NO_3)_2[/tex]
2 × 3 = 6
There 6 oxygen atoms in the 1 molecule of [tex]Ca(NO_3)_2[/tex].
d) [tex]Ba(OCN)_2[/tex]
2 × 1 = 2
There 6 oxygen atoms in the 1 molecule of [tex]Ba(OCN)_2[/tex].