Answer: False
Explanation: the masses of reactants are always the same as the products in order to conform to the law of conservation of mass.
Answer: False
Explanation:
In a chemical reaction atoms are not created or destroyed, only rearranged.
No change in mass
According to observations, the overall chemical composition of our solar system and other similar star systems is approximately (a) 98% hydrogen and helium, 2% all other elements combined; (b) 98% ice, 2% metal and rock; (c) 100% hydrogen and helium.
Answer:A
Explanation:
The solar system consist of the sun, the planets, stars and other objects. The chemical composition of the Sun consist mainly of Hydrogen and helium.
The sun is the largest object in the Solar system, it comprises nearly all the matter in the Solar System, Also the largest planet after the Sun are Jupiter and Saturn are giant planets forming almost the remaining matter of the solar system.
Like the Sun, the mass of Jupiter and Saturn are composed of roughly 98% hydrogen and helium with 2% of all the other elements combined.
In nature, the element X consists of two naturally occurring isotopes. 107X with abundance 53.84% and isotopic mass 106.9051 amu and 109X with isotopic mass 108.9048 amu. Use the given information to calculate the atomic mass of the element X to an accuracy of .001% (Report your answer like this yyy.yyyy)
The atomic mass of element X is approximately 107.8682 amu with an accuracy of .001%.
To calculate the atomic mass of element X, we need to use the given information about the isotopes of element X.
We are given that element X consists of two naturally occurring isotopes: 107X with an abundance of 53.84% and an isotopic mass of 106.9051 amu, and 109X with an isotopic mass of 108.9048 amu.
To calculate the atomic mass of element X, we can use the formula:
Atomic mass = (abundance of isotope 1 x mass of isotope 1) + (abundance of isotope 2 x mass of isotope 2)
Plugging in the values for element X, we get:
Atomic mass = (0.5384 x 106.9051 amu) + (0.4616 x 108.9048 amu)
Atomic mass ≈ 107.8682 amu
Therefore, the atomic mass of element X is approximately 107.8682 amu to an accuracy of .001%.
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To calculate the atomic mass of element X, multiply the mass of each isotope by its abundance and sum them up. The atomic mass of element X is 106.22288 amu.
Explanation:To calculate the atomic mass of element X, we need to consider the abundance and isotopic mass of its two naturally occurring isotopes. The atomic mass is calculated by multiplying the mass of each isotope by its abundance and summing them up.
For isotope 107X with an abundance of 53.84%, we multiply its mass (106.9051 amu) by its abundance (0.5384). For isotope 109X with an abundance of (100% - 53.84% = 46.16%), we multiply its mass (108.9048 amu) by its abundance (0.4616).
Finally, we add these two values together to get the atomic mass of element X to an accuracy of .001%. The calculation is as follows:
(106.9051 amu * 0.5384) + (108.9048 amu * 0.4616) = 106.22288 amu
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For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does, will it raise or lower the actual yield:__________
Washing the crystals with hot water
Answer:
Lowers the actual yield
Explanation:
If x is a string, then x = new String("OH"); and x = "OH"; will accomplish the same thing. Group of answer choices True False
Answer:
True is the correct answer to the above question.
Explanation:
If x is a string then it can be assigned by the help of two ways in java:By the help of constructor:- When we write " x = new String("OH");", then it will create a pass a string "OH" into the constructor. It is because the String is a class in java and x is an object created by the constructor of the String class.With the help of assigning: The "x= OH", which assigns the value of x which is an object of String class it can also use the constructor to initialize the "OH" string on the class.The above question states the two scenarios which are defined above. Hence the question statement is true.Answer:
"True" is the correct answer to this question.
Explanation:
The program to the given question as follows:
Program:
public class data //defining class
{
public static void main (String [] aw)//defining the main method
{
String x="OH"; //defining string variable x and assign value
System.out.println("assign value: "+x); //print value
x = new String("OH"); //defining instance variable and assign value
System.out.println("assign value by creating instance: "+x); //print value
}
}
Output:
assign value: OH
assign value by creating instance: OH
Explanation of the program:
In the above java program, a class data is defined, inside the class the main method is declared, In this main method a string variable "x" is defined that holds a value "OH", then we the print function to print this variable value.
In the next line, An instance of variable x is created, which holds a value "OH" in its parameter. In this question, both are correct because both hold the same value.
In the construction industry, why are I beams generally used as support as opposed to solid rectangular beams? What might be an advantage of using a rectangular beam?
Answer:I beams can withstand greater Moment compared to other beams like rectangular beam when given the same stress.
Rectangular beams are better for casting concrete beams, and for making beams of high and low moment of inertia and its centriod is easy to understand.
Explanation: I beams are beams usually used in metal work and in making beams of high moments of inertia.
Rectangular beams are beams whose centroid ( the geometric center of a regular rectangular beam) are easily identifiable and they can be built to give high and low moment of inertia. This is one of the advantage of using rectangular beams.
The advantage of use of I beam over rectangular beam has been the wide spread and high moment of inertia.
In the construction industry, for the support to the buildings and the structure, beams of different shapes has been used. I beam are the steel or metal beams that have high functionality.
I beam in construction industryI beam has been made with the shape of rolled joist. They have been installed in the building because of their high moment of inertia.
I beam has been working with the spread of moment of inertia in the structure, that has been able to bend and withstand the pressure more than compared to the rectangular cross-section with concentrated inertia.
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if you take a dried out white rose and place the stem in food coloring, will the rose turn the color you put the stem in
Answer:
Yes, this is true. The reason is that the flower transpires and sucks the water in and distributes it as much as it can. You can also flip it upside down and hang it with petals down , allowing the liquid to enter the flower and then retaining color for longer periods of time and having more color.
Explanation:
Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 26°C and 0.80 atm, the density of this gas mixture is 2.2 g·L−1. What is the partial pressure of each gas?
Answer: The partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm
Explanation:
Assuming ideal gas behavior, the equation follows:
PV = nRT
We know that:
[tex]\text{Density}=\frac{\text{Mass}}{\text{Volume}}[/tex]
Rearranging the above equation:
[tex]M=\frac{dRT}{P}[/tex]
where,
d = density of gas mixture = 2.2 g/L
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]26^oC=[26+273]K=299K[/tex]
P = pressure of the mixture = 0.80 atm
M = average molar mass of mixture
Putting values in above equation, we get:
[tex]M_{avg}=\frac{2.2g/L\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 299}{0.80atm}\\\\M_{avg}=67.5g/mol[/tex]
We know that:
Molar mass of [tex]NO_2[/tex] = 46 g/mol
Molar mass of [tex]N_2O_4[/tex] = 92 g/mol
Let the mole fraction of [tex]NO_2[/tex] be 'x' and that of [tex]N_2O_4[/tex] be '(1-x)'
For average molar mass calculation:
[tex]M_{avg}=M_{NO_2}\chi_{NO_2}+M_{N_2O_4}\chi_{N_2O_4}[/tex]
Putting values in above equation:
[tex]67.5=46x+92(1-x)\\\\x=0.533[/tex]
Mole fraction of [tex]N_2O_4[/tex] = (1 - x) = (1 - 0.533) = 0.467
To calculate the partial pressure, we use the equation given by Raoult's law, which is:
[tex]p_{A}=p_T\times \chi_{A}[/tex]
For [tex]NO_2[/tex] :We are given:
[tex]p_T=0.80atm\\\chi_{NO_2}=0.533[/tex]
Putting values in above equation, we get:
[tex]p_{NO_2}=0.80atm\times 0.533\\\\p_{NO_2}=0.426atm[/tex]
For [tex]N_2O_4[/tex] :We are given:
[tex]p_T=0.80atm\\\chi_{N_2O_4}=0.467[/tex]
Putting values in above equation, we get:
[tex]p_{N_2O_4}=0.80atm\times 0.467\\\\p_{N_2O_4}=0.374atm[/tex]
Hence, the partial pressure of [tex]NO_2[/tex] is 0.426 atm and that of [tex]N_2O_4[/tex] is 0.374 atm
A reducing agent gets oxidized as it reacts. A reducing agent gets oxidized as it reacts. true false
A reducing agent does get oxidized as it reacts in a process known as a redox reaction. During this reaction, the reducing agent loses electrons, effectively donating them to another substance. Disproportion reactions are also possible, where the same substance gets oxidized and reduced.
Explanation:Yes, the statement is true: A reducing agent gets oxidized as it reacts. In redox reactions, the substance that is oxidized loses electrons and is therefore referred to as the reducing agent. For example, in the reaction, 'Al(s) + NiO(s) --> Al2O3(s) + Ni(s)', aluminum (Al) is the reducing agent as it gets oxidized from 0 to +3 oxidation state while reducing nickel oxide (NiO).
Redox reactions involve the transferring of electrons from one atom (which gets oxidized) to another (which gets reduced). In the process, the reducing agent is oxidized because it essentially donates its own electron to the other substance.
There are also cases of disproportion reactions where the same substance gets oxidized and reduced. These reactions are very interesting in the context of oxidation-reduction chemistry.
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What happens when you add more solute to a saturated solution
Answer:
If you add more solute in a saturated solution, it will have no effect. Additional solute does not dissolve in a saturated solution.
Explanation:
If more solute is added and it does not dissolve, then the original solution was saturated. If the added solute dissolves, then the original solution was unsaturated. A solution that has been allowed to reach equilibrium but which has extra undissolved solute at the bottom of the container must be saturated.
If the rate of evaporation is equal to the rate of condensation, the system is in a state of dynamic equilibrium, which cannot be disturbed. TRUE FALSE
Answer: False
Explanation:
From definition dynamic equilibrium is a state of balance between continuing processes. When the rate of evaporation equals the rate of condensation the system has reached a dynamic equilibrium.
However, it is possible, to disturb a system that is in dynamics equilibrium by changing conditions of the system. In fact anything that changes the thermodynamic state of the system will disturb the system and it will no longer be at equilibrium.
An increase in temperature to the system will favour evaporation more than condensation. Molecules with higher kinetic energy will escape the system.
Silica, sio2, is formed on silicon as an electrically insulating layer for microelectronic devices. silica is formed when silicon is exposed to o2 gas at an elevated temperature. at 900˚c, it takes 90 minutes for the oxygen to diffuse from the surface to form a 0.06 micron (0.06 x 10-6 m) thick layer of sio2 on
Final answer:
Silicon dioxide, or silica, is integral to microelectronics as an electrical insulator and is used in applications such as isolation, gate insulation, and dopant diffusion. Chemical Vapor Deposition is commonly used to deposit silica thin films. The material's tetrahedral structure ensures stability under rapid temperature changes, vital for semiconductor industries.
Explanation:
Silicon dioxide (SiO2), often referred to as silica, plays a crucial role in the world of microelectronics. Its excellent electrical insulating properties make it essential for various applications within semiconductor devices. Silicon dioxide is employed for the isolation of conductive layers as well as for its dielectric properties which are used in gate insulation. Moreover, it serves to facilitate the diffusion of dopants from oxides and as a means to prevent the loss of dopants when capping films.
The Chemical Vapor Deposition (CVD) method is widely used for depositing thin layers of SiO2 during semiconductor processing. This is due to the unique challenges associated with its application in creating insulating thin films. Additionally, silica's unique diamond-like network structure allows for rapid temperature changes, making it invaluable in the steel, electronic, and semiconductor industries.
Silica's three-dimensional tetrahedral structure, where silicon atoms are bonded to oxygen, confers stability and resilience that is crucial for the high temperature processes involved in microelectronic device fabrication. Different forms of silicon dioxide, such as quartz and fused silica, contribute to the diversity of silica's properties and applications. Its naturally abundant presence and the numerous crystalline forms it can take, underscore the material's importance to technology and industry.
Given a unsorted list of 1024 elements, what is the runtime for linear search if the search key is less than all elements in the list?
Answer:
10
Explanation:
Binary search's runtime is proportional to log (base two) of the number of list elements.
The density of air under ordinary conditions at 25 degrees * C is 1.19g / L . How many kilograms of air are in a room that measures 9.0ft * 11.0ft and has a 10.0 ft ceiling?
Answer:
33.3 kg of air
Explanation:
This is a problem of conversion unit.
Density is mass / volume
Therefore we have to calculate the volume in the room, to be multiply by density. That answer will be the mass of air.
Volume of the room → 9 ft . 11 ft . 10 ft = 990 ft³
Density is in g/L, therefore we have to convert the ft³ to dm³ (1 dm³ = 1L)
990 ft³ . 28.3 dm³ / 1ft³ = 28017 dm³ → 28017 L
This is the volume of the room, if we replace it in the density formula we can know the mass of air in g.
1.19 g/L = Mass of air / 28017 L
Mass of air = 28017 L . 1.19 g/L → 33340 g of air
Finally, let's convert the mass in g to kg → 33340 g . 1kg / 1000 g = 33.3 kg
Sodium tends lose a single electron in natural settings. Based on what you know, what are two other elements that tend to do the same thing?
Answer:
Lithium and Sodium
Explanation:
Losing and electron in natural setting is characterizes of elements in group one. These are elements known as the alkaline earth metals. They are the most electropositive elements on the periodic table.
These elements ionize by losing an electron yin their outermost shell to attain the configuration of the nearest noble gas. These elements are usually found in combined and rarely seen in uncombined state principally due to their very reactive nature.
Sodium naturally would ionize by losing one electron. Other elements capable of this even at a better rate because they are more electropositive are potassium and lithium. Both are also group one alkaline metals
Which of the following would be expected to have the lowest freezing point? a. 0.1 M NaCl b. 0.1 M MgCl2 c. 0.1 M AlCl3
Answer:
option c, 0.1 M [tex]AlCl_3[/tex]
Explanation:
Addition of non-volatile solute to a solvent decreases its vapour pressure which results in decrease in melting point.
Decrease in melting point is known as depression in freezing point. the depression in freezing point is related with molality and no. of ions as follows:
[tex]\Delta T_f = imK_f[/tex]
Where, i is von't Hoff factor, m is molatilty and ΔTf is depression in freezing point.
As, in the given case concentration of all the solution is same, therefore, depression in freezing point will depend upon the no. of ions produced by the ionization of the salts in the aqueous solution.
In case of NaCl, the no. of ions produced will be 2.
Therefore, value of i will be 2
In case of [tex]MgCl_2[/tex], the no. of ions produced will be 3.
Therefore, value of i will be 3
In case of [tex]AlCl_3[/tex], the no. of ions produced will be 4.
Therefore, value of i will be 4.
More, the value of van't Hoff factor, more will be depression in freezing point.
Therefore, assuming the given solution to be aqueous, the solution expected to have lowest freezing point is 0.1 M [tex]AlCl_3[/tex].
The correct option is b. 0.1 M MgCl2 would be expected to have the lowest freezing point.
To understand why 0.1 M MgCl2 has the lowest freezing point, we need to consider the colligative properties of solutions, specifically the freezing point depression. The freezing point depression is directly proportional to the number of particles (ions or molecules) in the solution. When a solute is dissolved in a solvent, it dissociates into ions, and the more ions it dissociates into, the greater the freezing point depression.
Let's analyze each of the given solutions:
a. 0.1 M NaCl: Sodium chloride (NaCl) dissociates into two ions in solution, Na+ and Cl-. Therefore, the total number of particles in solution is 2 times the concentration of NaCl, which is 0.2 moles of particles per liter.
b. 0.1 M MgCl2: Magnesium chloride (MgCl2) dissociates into three ions in solution, Mg2+ and 2Cl-. Therefore, the total number of particles in solution is 3 times the concentration of MgCl2, which is 0.3 moles of particles per liter.
c. 0.1 M AlCl3: Aluminum chloride (AlCl3) can dissociate into four ions in solution, Al3+ and 3Cl-. However, AlCl3 is not fully dissociated in aqueous solution due to its Lewis acid behavior and the formation of complex ions such as AlCl4-. Assuming complete dissociation for simplicity, which is not the case in reality, the total number of particles in solution would be 4 times the concentration of AlCl3, which is 0.4 moles of particles per liter.
Since MgCl2 dissociates into the most number of ions compared to NaCl and AlCl3 (assuming complete dissociation for AlCl3), it will have the greatest effect on freezing point depression. Therefore, 0.1 M MgCl2 is expected to have the lowest freezing point among the given options.
It is important to note that in reality, the actual freezing point depression of AlCl3 would be less than calculated here due to its incomplete dissociation and complex ion formation. This further supports the conclusion that MgCl2 would have the lowest freezing point, as AlCl3 would not reach the expected freezing point depression based on complete dissociation.
A. After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A 25.0-mL portion of the liquid had a mass of 21.95 gg. A chemistry handbook lists the density of benzene at 15∘C∘C as 0.878 g/mg/mL. Is the calculated density in agreement with the tabulated value?
B. An experiment requires 15.0 g of cyclohexane, whose density at 25oC is 0.7781 g/mL. What volume of cyclohexane should be used?
C. A spherical ball of lead has a diameter of 5.0 cm. What is the mass of the sphere if lead has a density of 11.34 g/cm3?
Answer:
A. Yes, the calculated density in agreement with the tabulated value.
B. 19.28 mL of volume of cyclohexane should be used.
C. 742.20 is the mass of the sphere of lead.
Explanation:
A.
Volume of the liquid = V = 25.0 mL
Mass of the liquid = m = 21.95 g
Density of the liquid = d
[tex]d=\frac{m}{V}[/tex]
[tex]=\frac{21.95 g}{25.0 mL}=0.878 g/mL[/tex]
Density mentioned in the report book = d' = 0.878 g/mL
d' = d = 0.878 g/mL
Yes, the calculated density in agreement with the tabulated value.
B.
Volume of the liquid cyclohexane= V = ?
Mass of the liquid cyclohexane= m = 15.0 g
Density of the liquid cyclohexane = d = 0.7781 g/mL
[tex]d=\frac{m}{V}[/tex]
[tex]V=\frac{15.0 g}{0.7781 g/mL}=19.28 mL[/tex]
19.28 mL of volume of cyclohexane should be used.
C.
Diameter of the ball = d = 5.0 cm
Radius of the ball = r = 0.5 × d = 2.5 cm
Volume of sphere ,V= [tex]\frac{4}{3}\pi r^3[/tex]
[tex]V = \frac{4}{3}\times 3.14\times (0.25 cm)^3=65.45 cm^3[/tex]
Volume of the spherical lead ball = V
Mass of the spherical lead ball= m = ?
Density of the spherical lead ball = d = [tex]11.34 g/cm^3[/tex]
[tex]d=\frac{m}{V}[/tex]
[tex]m=d\times V=11.34 g/cm^3\times 65.45 cm^3=742.20 g[/tex]
742.20 is the mass of the sphere of lead.
A. The tabulated density of benzene [tex]15^0 C[/tex] is 0.878 g/mL. B. 19.28 mL of volume of cyclohexane should be used. C. The mass of the lead sphere is approximately 743.5 g.
A. To determine if the calculated density of the liquid matches the tabulated value for benzene, we need to calculate the density using the provided data and then compare it to the handbook value. The density is calculated by dividing the mass of the liquid by its volume.
Given:
Mass of the liquid = 21.95 g
The volume of the liquid = 25.0 mL
Calculated density = [tex]Mass / Volume = 21.95 g / 25.0 mL = 0.878 g/mL[/tex]
The tabulated density of benzene [tex]15^0 C[/tex] is 0.878 g/mL. Since the calculated density matches the tabulated value, the liquid in the bottle is likely benzene.
B. To find the volume of cyclohexane required for the experiment, we can use the formula: Volume = Mass / Density
Given:
Mass of cyclohexane needed = 15.0 g
Density of cyclohexane at [tex]25^0C[/tex] = 0.7781 g/mL
Volume of cyclohexane = 15.0 g / 0.7781 g/mL = 19.28 mL
Therefore, 19.28 mL of cyclohexane should be used for the experiment.
C. The mass of the lead sphere can be calculated using the formula for the volume of a sphere and the density of lead:
The volume of a sphere = [tex](4/3)\pi r^3[/tex]
Density = Mass / Volume
Given:
The diameter of the lead sphere = 5.0 cm
Radius of the lead sphere = Diameter / 2 = 5.0 cm / 2 = 2.5 cm
Density of lead = [tex]11.34 g/cm^3[/tex]
The volume of the lead sphere =[tex](4/3)\pi (2.5 cm)^3 =20.94375\pi cm^3[/tex]
Mass of the lead sphere = Density — Volume = [tex]11.34 g/cm^3 - 20.94375\pi cm^3= 11.34 g/cm^3- 65.45902 cm^3= 743.5 g[/tex]
Therefore, the mass of the lead sphere is approximately 743.5 g.
Analysis of a volatile liquid showed that it is 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen by mass. A separate 0.345-gram sample of its vapor occupied 120. mL at 100.°C and 1.00 atm. What is the molecular formula for the compound?
Answer: The molecular formula for the given organic compound is [tex]C_4H_8O_2[/tex]
Explanation:
We are given:
Percentage of C = 54.5 %
Percentage of H = 9.1 %
Percentage of O = 36.4 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 54.5 g
Mass of H = 9.1 g
Mass of O = 36.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{54.5g}{12g/mole}=4.54moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{9.1g}{1g/mole}=9.1moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{36.4g}{16g/mole}=2.28moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.28 moles.
For Carbon = [tex]\frac{4.54}{2.28}=1.99\approx 2[/tex]
For Hydrogen = [tex]\frac{9.1}{2.28}=3.99\approx 4[/tex]
For Oxygen = [tex]\frac{2.28}{2.28}=1[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : O = 2 : 4 : 1
Hence, the empirical formula for the given compound is [tex]C_2H_{4}O_1=C_2H_4O[/tex]
Mass of empirical formula = [tex]C_2H_4O[/tex] = 2(12) + 4(1) + 16 = 44 g/eq.
Now we have to determine the molar mass of compound by using ideal gas equation.
[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]
where,
P = pressure of gas = 1.00 atm
V = volume of gas = 120 mL = 0.120 L
T = temperature of gas = [tex]100^oC=273+100=373K[/tex]
w = mass of gas = 0.345 g
M = molar mass of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the above formula, we get:
[tex]PV=\frac{w}{M}RT[/tex]
[tex](1.00atm)\times (0.120L)=\frac{0.345g}{M}\times (0.0821L.atm/mol.K)\times (373K)[/tex]
[tex]M=88.04g/mol[/tex]
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
We are given:
Mass of molecular formula = 88.04 g/mol
Mass of empirical formula = 44 g/mol
Putting values in above equation, we get:
[tex]n=\frac{88.04g/mol}{44g/mol}=2[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_2H_4O=(C_2H_4O)_n=(C_2H_4O)_2=C_4H_8O_2[/tex]
Thus, the molecular formula for the given compound is [tex]C_4H_8O_2[/tex]
The empirical formula of the given compound is [tex]\bold {C_2H_4O}[/tex]. The empirical formula is the smallest whole-number ratio of the compound.
Assume the mass of the compound is 100 g. So, the percentages given are taken as mass.
Mass of C = 54.5 g = 4.54 moles
Mass of H = 9.1 g = 9.1 moles
Mass of O = 36.4 g = 2.28 moles
To formulate the empirical formula, Calculate the molar ratio by dividing moles by the smallest number,
For the mole ratio, we divide each value of the moles by the smallest number of moles, we get.
For Carbon = 2
For Hydrogen =4
For Oxygen = 1
Therefore, the empirical formula of the given compound is [tex]\bold {C_2H_4O}[/tex].
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The best mixture of antifreeze and water is 50% antifreeze, 50% water. The cooling system in your car has a mixture of 6.00L water and 6.00 L ethylene glycol (antifreeze). The molality of the solution is 17.9m. The chemical formula of antifreeze is C2H6O2 and its density is 1.1132 g/cm3.
If the summer temperatures rise and the coolant reaches a temperature of 108°C, will it boil?
No, it would boil at 109.13°C
No, it would boil at 123.09°C
No, it would boil at 119.13°C
Yes, it would boil at 99.0°C
Yes, it would boil at 100.13°C
Answer:
The correct answer is No, it would boil at 109.13°C
Explanation:
This question can be solved by knowing the boiling point elevation formula and the fact that ethylene glycol dissolves in water without dissociation
The boiling point elevation formula is given by
ΔT = i × [tex]K_{b}[/tex] ×[tex]m_{solute}[/tex]
Where [tex]K_{b}[/tex] = 0.51 °C/mole
i = Vant't Hoff factor
m = molality of the solution
When ethylene glycol, C2H6O2, (antifreeze) enters into solution in water it disociates into
C2H6O2 (aq) ---> 2OH(-1)(aq) + C2H4(+2)(aq)
Thus one mole of C2H6O2 dissociates into two moles of hydroxyl ions and one mole of C2H4(+2) ion
Hence the Van't Hoff factor, i, = 3
Therefore the mass of the mole
Therefore ΔT = 3 × 0.51 × 17.9 = 27.387 K
However Ethylene formula = (CH2OH)2 it dissolves in water without dissociation
Therefore i = 1
and ΔT = 1 × 0.51 × 17.9 = 9.129 ≅ 9.13
Hence at the boiling point of the water with antifreeze dissolved in it it
Boiling point of water + Boiling point elevation = 100 + 9.13 = 109.13 °C
The water will not boil until it reaches 109.13 °C
The coolant mixture will not boil at 108°C. The calculated boiling point elevation suggests it will boil at approximately 109.16°C, based on the molality and the boiling point elevation constant for water.
Explanation:To determine whether the coolant mixture in your car will boil at 108°C, we can use the concept of boiling point elevation. The boiling point of a solution increases when a solute is added to a solvent due to the colligative properties of the solution. Using the molality provided (17.9m), and knowing the boiling point elevation constant (Kb) for water is approximately 0.512 °C/m, we can calculate the boiling point elevation.
ΔTb = i * Kb * m
Where ΔTb is the boiling point elevation, i is the van't Hoff factor (i = 1 for ethylene glycol as it does not dissociate in solution), Kb is the ebullioscopic constant for water, and m is the molality of the solution.
ΔTb = 1 * 0.512 °C/m * 17.9m = 9.16 °C
The normal boiling point of water is 100°C, so adding the elevation to the normal boiling point gives us:
100°C + 9.16°C = 109.16°C
Therefore, the solution will not boil at 108°C because it would boil at approximately 109.16°C.
An element has four naturally occurring isotopes with the masses and natural abundances given in the table below. Find the atomic mass of the element (express your answer to four significant figures and include the appropriate units).
Isotope Mass (amu) Abundance (%)
1 203.97304 1.390
2 205.97447 24.11
3 206.97590 22.09
4 207.97665 52.41
Identify the element, by spelling out its full name.
Answer:
Lead (Pb)
Explanation:
To find the atomic mass of the element, we need to take into consideration all the naturally occurring isotopes. We then find the atomic mass by multiplying the natural abundance of the isotopes by their mass for all of the isotopes and summing them together.
1. 1.390/100 * 203.97304 = 2.835225256
2. 24.11/100 * 205.97447 = 49.660444717
3. 22.09/100 * 206.97590 = 45.72097631
4. 52.41/100 * 207.97665 = 109.000562265
We then add all of these masses together:
109.000562265 + 45.72097631+ 49.660444717 + 2.835225256 = 207.217208548
To 4 sf = 207.2 amu
Element is Lead (Pb)
Which of the following is true about a municipal bond with a put option? (A) An investor will exercise the option to put the bond if yields rise significantly. (B) An investor must have the issuer's permission to put the bond. (C) The market price of bonds is never affected by a put option feature. (D) Yields are usually higher for a new issue bond with a put option than for a new issue bond without a put option.
Answer:
(A) An investor will exercise the option to put the bond if yields rise significantly
Explanation:
A put option on the bond is a mechanism to allow the buyer of the bond the ability to compel the lender to repay the principal on the bond. The put option offers the buyer of the bond the ability to collect the principal of the bond anytime they choose until maturity for any purpose.
Recall that once the price drops (that is, the yield increases), put options are exercised. If the yield significantly increased, the put choice on a municipal bond is executed.
Ice Station Bravo near the North Pole launched a helium-filled balloon to check atmospheric conditions. At sea level (1.0 atm) where the balloon was launched, it had a volume of 0.93m^3 . It rose to an altitude of 18000m where the atmospheric pressure dropped to 0.072atm.
What is the volume of the balloon at that altitude assuming that the temperature was the same at sea level?
Answer:
12.9 m³ is the new volume
Explanation:
As the temperature keeps on constant, and the moles of the gas remains constant too, if we decrease the pressure, the volume will increase. If the volume is decreased, pressure will be higher.
The relation is this: P₁ . V₁ = P₂ . V₂
1 atm . 0.93m³ = 0.072 atm . V₂
0.93m³ .atm / 0.072 atm = V₂
V₂ = 12.9 m³
In conclusion and as we said, pressure has highly decreased so volume has highly increased.
Crime scene investigators keep a wide variety of compounds on hand to help with identifying unknown substances they find in the course of their duties. One such investigator, while reorganizing their shelves, has mixed up several small vials and is unsure about the identity of a certain powder. Elemental analysis of the compound reveals that it is 63.57% carbon, 6.000% hydrogen, 9.267% nitrogen, 21.17% oxygen by mass. Which of the following compounds could the powder be?
a.) C11H15NO2 = 3,4-methylenedioxymethamphetamine (MDMA), illicit drug
b.) C3H6NO3 = hexamethylene triperoxide diamine (HMTD), commonly used explosive
c.) C21H23NO5 = heroin, illicit drug
d.) C8H9NO2 = acetaminophen, analgesic
e.) C7H5N3O6 = 2,4,6-trinitrotoluene (TNT), common used explosive
f.) C17H19NO3 = morphine, analgesic
g.)C10H15N = methamphetamine, stimulant
h.) C4H5N2O = caffeine, stimulant
Answer:
Option d: C₈H₉NO₂ = acetaminophen, analgesic
Explanation:
% composition of compound is:
63.57 g of C
6 g of H
9.267 g of N
21.17 g of O
First of all we divide each by the molar mass of the element
63.57 g / 12 gmol = 5.29 mol of C
6 g of H / 1 g/mol = 6 mol H
9.267 g of N / 14 g/mol = 0.662 mol of N
21.17 g of O / 16 g/mol = 1.32 mol of O
We divide each by the lowest value, in this case 0.662
5.29 / 0.662 = 8
6 / 0.662 = 9
0.662 / 0.662 = 1
1.32 / 0.662 = 2
Molecular formula of the compound is C₈H₉NO₂
A solution containing CaCl2 is mixed with a solution of Li2SO4 to form a solution that is 2.1 × 10-5 M in calcium ion and 4.75 × 10-5 M in sulfate ion. What will happen once these solutions are mixed? The Ksp for CaSO4 is 2.4 x 10-5.
Answer : The precipitate will not be formed when these solutions are mixed.
Explanation :
The chemical equation for the reaction of calcium chloride and lithium sulfate follows:
[tex]CaCl_2(aq)+Li_2SO_4(aq)\rightarrow 2LiCl(aq)+CaSO_4(s)[/tex]
We are given:
Concentration of calcium ion = [tex]2.1\times 10^{-5}M[/tex]
Concentration of sulfate ion = [tex]4.75\times 10^{-5}M[/tex]
[tex]K_{sp}=2.4\times 10^{-5}[/tex]
The salt produced is calcium sulfate.
The equation follows:
[tex]CaSO_4(s)\rightleftharpoons Ca^{2+}(aq)+SO_4^{2-}(aq)[/tex]
The expression of [tex]Q_{sp}[/tex] (ionic product) for above equation follows:
[tex]Q_{sp}=[Ca^{2+}]\times [SO_4^{2-}][/tex]
Putting values of the concentrations in above expression, we get:
[tex]Q_{sp}=(2.1\times 10^{-5})\times (4.75\times 10^{-5})\\\\Q_{sp}=9.9\times 10^{-10}[/tex]
There are 3 conditions:
When [tex]K_{sp}>Q_{sp}[/tex]; the reaction is product favored. (No precipitation)When [tex]K_{sp}<Q_{sp}[/tex]; the reaction is reactant favored. (Precipitation)When [tex]K_{sp}=Q_{sp}[/tex]; the reaction is in equilibrium. (Sparingly soluble)As, the [tex]K_{sp}>Q_{sp}[/tex]. The above reaction is product favored. This means that no salt or precipitate will be formed.
Hence, the precipitate will not be formed when these solutions are mixed.
A(n) _______ solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
Answer : A hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
Explanation :
Solution : It is made up of the combination of amount solute and solvent.
Isotonic solutions : It is defined as the solutions in which the concentration of solute inside the cell and outside the cell is same.
Hypotonic solutions : It is defined as the solutions in which the concentration of solute inside the cell is lower than outside the cell.
For example : Diluted sugar syrup
Hypertonic solutions : It is defined as the solutions in which the concentration of solute inside the cell is higher than outside the cell.
For example : Concentrated sugar syrup
Hence, a hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution.
The term for a solution that has a higher concentration of water and a lower concentration of solute than a cell is 'hypotonic'. In this scenario, water moves into the cell via osmosis.
Explanation:A(n) hypotonic solution has a higher concentration of water and lower concentration of solute than the cell placed in the solution. In biology, we often talk about the relationship between cells and their surrounding environment in terms of tonicity. In a hypotonic environment, there is less solute (like salt or sugar) outside the cell compared to inside the cell. This causes water to move into the cell by osmosis, because water moves from areas of high concentration to areas of low concentration until equilibrium is reached.
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If 75 gm's of codeine phosphate is dissolved in 1.5 liters of sterile water, what is the resultant percentage strength of the solution?
Answer:
w/v% = gm per 100 ml
75gm X gm 7500
------- = ------ == --------
1500ml 100ml 1500X
X=5%
A sample of N2 gas is collected over water at 20o C and pressure of 1 atm. The volume collected is 250 Liters. What mass of N2 is collected?
Answer : The mass of nitrogen gas collected is, 290.9 grams
Explanation :
To calculate the mass of nitrogen gas we are using ideal gas equation:
[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]
where,
P = pressure of nitrogen gas = 1 atm
V = volume of nitrogen gas = 250 L
T = temperature of nitrogen gas = [tex]20^oC=273+20=293K[/tex]
R = gas constant = 0.0821 L.atm/mole.K
w = mass of nitrogen gas = ?
M = molar mass of nitrogen gas = 28 g/mole
Now put all the given values in the ideal gas equation, we get:
[tex](1atm)\times (250L)=\frac{w}{28g/mole}\times (0.0821L.atm/mole.K)\times (293K)[/tex]
[tex]w=290.9g[/tex]
Therefore, the mass of nitrogen gas collected is, 290.9 grams.
The observation that 4.0 g of hydrogen reacts with 32.0 g of oxygen to form a product with O:H mass ratio-8:1, and 6.0 g of hydrogen reacts with 48.0 g of oxygen to form the same product with O/H mass ratio = 8:1 is evidence for the law of 1. multiple proportions 2. erergy conservation. 3. mass conservation 4. definite proportion
Answer:
Law of definite proportion
Explanation:
As per law of definite proportion, ratio of elements present in a compound is always fixed irrespective of the source, amount and method of preparation.
In the given case, the hydrogen and oxygen react with each other to form a compound. The ratio of oxygen and hydrogen is fixed which is 8 : 1 and this ratio does not change upon changing amount of oxygen and hydrogen.
So, the this experiment support the law of definite proportion.
Therefore, the correct option is option 4
The following peptides are subjected to normal electrophoretic analysis at pH 6.0. State whether the peptides will migrate towards the cathode or anode and predict the relative rate of migration of each peptide. a.GlyArg Phe.b.Gly.Arg Phe.c.Glu.Glu Phe.d.GIy.Glu
Answer:
The peptide will definitely migrated towards cathode (negative terminal)
Explanation:
The positively and negatively charged side chains of proteins cause them to behave like amino acids in an electrical field; that is, they migrate during electrophoresis at low pH values to the cathode (negative terminal) and at high pH values to the anode (positive terminal). The isoelectric point, the pH value at which the protein molecule does not migrate, is in the range of pH 5 to 7 for many proteins.
if you begin with 210.3 g of Cl2, how many grams of HCl will you end up with? CH4 + 3Cl2 → CHCl3 + 3HCl *
Answer:
108.04g
Explanation:
Looking at the balanced chemical equation theoretically, we can see that 3 moles of chlorine yielded 3 moles of HCl. This means that they have equal mole ratio.
Now let’s get the actual reactive moles. The number of moles is obtained by dividing the mass by the molar mass.
The molecular mass of chlorine gas is (2 * 35.5) = 72g/mol
The number of moles is thus 210.3/71 = 2.96 moles
Since the number of moles are equal, the number of moles of HCl produced too is 2.96 moles.
Now to get the mass of HCl produced, we multiply the number of moles by the molar mass of HCl. The molar mass of HCl is 36.5g/mol
The mass is thus 2.96 * 36.5 = 108.04g
By what factor does the rate change in each of the following cases (assuming constant temperature)? (a) A reaction is first order in reactant A, and [A] is doubled. (b) A reaction is second order in reactant B, and [B] is halved. (c) A reaction is second order in reactant C, and [C] is tripled.
Answer:
a) 2
b) 1/4
c) 9
Explanation:
a) for a first order reaction in reactant A
r initial = k*[A initial]
then if the concentration is doubled [A final ]= 2*[A initial] , then
r final = k*[A final ] = 2* k*[A initial] = 2*r initial
then the velocity changes by a factor of 2
b) for a second order reaction in reactant B
r initial = k*[B initial]²
then if the concentration is halved: [B final ]= [B initial]/2 , then
r final = k*[B final ]² = k*( [B initial]/2 )² =k* [B initial]² /4 = r initial /4
then the velocity changes by a factor of 1/4
c) for a second order reaction in reactant C
r initial = k*[C initial]²
then if the concentration is tripled : [C final ]= 3* [C initial] , then
r final = k*[C final ]² = k*( 3*[C initial] )² =k* [C initial]² *9 = 9 * r initial
then the velocity changes by a factor of 9
For first-order reactions, doubling the concentration of the reactant doubles the reaction rate. For second-order reactions, halving the concentration reduces the rate by a factor of four, while tripling the concentration increases the rate by a factor of nine.
Explanation:(a) For a reaction that is first order in reactant A, if [A] is doubled, the rate of the reaction will also double. This is because the rate of a first-order reaction is directly proportional to the concentration of the reactant. When [A] is doubled, the rate is doubled as well.
(b) For a reaction that is second order in reactant B, if [B] is halved, the rate of the reaction will decrease by a factor of four. This is because the rate of a second-order reaction is directly proportional to the square of the concentration of the reactant. When [B] is halved, the concentration is squared and the rate is decreased by a factor of four.
(c) For a reaction that is second order in reactant C, if [C] is tripled, the rate of the reaction will increase by a factor of nine. Similar to the previous case, the rate of a second-order reaction is directly proportional to the square of the concentration of the reactant. When [C] is tripled, the concentration is squared and the rate is increased by a factor of nine.
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