Answer: a contest in which you are eliminated if you fail to spell a word correctly.
What does a roman numeral tell you when it is written after an element
Answer:
The correct answer is: Oxidation number
Explanation:
According to the IUPAC nomenclature for the inorganic compounds, roman numerals are used to denote the oxidation number of a positively charged ion, called cation.
This rule is only applicable to elements that can form more than one cation with different oxidation state or charge.
The oxidation state or charge of each cation of such an element is denoted or represented by a Roman numeral in the parentheses followed by the name of the element.
Therefore, the oxidation number of a cation is denoted by a roman numeral which is written after the name of the element.
Consider the two electron arrangements for neutral atoms A and B. What is the difference between atom A and atom B?
A - 1s22s22p63s1
B - 1s22s22p65s1
Atom B has lost some inner electrons. The outer electron of atom B has moved to a lower energy state. The outer electron of atom B has moved to a higher energy state.
Answer:
outer electron moved to a higher energy state
Explanation:
To know this, let's write again the electron arrangements:
[A] = 1s² 2s² 2p^6 3s^1
[B] = 1s² 2s² 2p^6 5s^1
Now, let's discart the options.
First option cannot be, because if you count the inner electrons, which are the numbers uppering (besides the s and p), in both cases, you have the same number of electrons, which is 11 in both cases.
Second option cannot be either, because electron configuration, always go from lower to higher state. In the case, that one electron move to a lower state, it should move fro 3s to the 2p level. In this case, it was not, mainly because the lower level of 2, has already maxed out it's capacity to hold electrons.
Therefore, option 3 it's the more accurate option, you can see that from 2p6, it jump to the energy level 5, skipping 3 and 4th level. Therefore this is the correct option.
Answer:
The outer electron of atom B has moved to a higher energy state.
Explanation: (Graded correct)
Mothballs are comprised primarily of naphthalene (C10H8). When 1.025 g of Naphthalene burns in a bomb calorimeter, the temperature rises from 24.25 degrees Celsius to 32.33 degrees Celsius. Find the change in energy for the combustion of a mole of naphthalene. The heat capacity of the bomb calorimeter is 5.11 kJ/degree Celsius.
Answer:
The change in energy for the combustion of a mole of naphthalene is 79 kJ
Explanation:
An excersise where you have to apply the Heat Capacity formula
C = Q . ΔT
where C is Heat Capacity and Q is Heat
ΔT means (T° final - T° initial)
5.11 kJ/°C = Q (32.33°C - 24.25°C)
5.11 kJ/°C = Q (32.33°C - 24.25°C)
5.11 kJ/°C = Q . 8.08°C
5.11 kJ / 8.08°C = Q
0.632 kJ = Q
This heat is released by 1.025 grams of naphtalene.
Molar mass Naphtalene : 128.17 g/m
1.025 g / 128.17 g/m = 0.008 moles
This are the moles, so we have to divide heat/moles to get the change in energy for one mole.
0.632 kJ/0.008 mol = 79 kJ/m
Identify each of the following energy exchanges as primarily heat or work and determine whether the sign of is positive or negative for the system. a. A rolling billiard ball collides with another billiard ball. The first billiard ball (defined as the system) stops rolling after the collision. b. A book is dropped to the floor (the book is the system). c. A father pushes his daughter on a swing (the daughter and the swing are the system
Answer:
A) Work (positive)
B) Heat (negative) and work (positive)
C) Work (negative) and heat (negative)
Explanation:
A) This energy change is work given that the first ball is using this energy to move the other one. The system (fist ball) is doing the work so it has + sign.
B) The book loses part of its energy by friction (heat) and part by applying a force to the floor (work). The heat is taken from the system so it has - sign and the work is done by the system and it has + sign.
C) The daughter is receiving the work done by the father. This work done to the system has - sign. Also, you can say that the system (daughter and swing) loses energy by friction (heat) and because of that it slows down.
Note: the sign of work and heat is defined by convention.
Mary, a sales representative for a small cleaning business, asks current customers for names of friends, customers and other businesses that might be interested in the company's cleaning services. Mary is relying on ____ to identify potential customers
Answer:
Personal Referral.
Explanation:
This technique of finding employees is called Personal Referral. In personal referral the existing employee of the company uses hi/her own connections to find employees for the company. He/she may ask his/her asks current customers for names of friends, customers and other businesses that might be interested in job provided by the company.
Final answer:
Mary is employing a 'Word of Mouth' marketing strategy by asking current customers for referrals, tapping into the informal network to reach potential customers with the trust built through personal recommendations.
Explanation:
Mary, a sales representative, is utilizing a marketing strategy known as Word of Mouth. By asking current customers for referrals, she is tapping into her business's informal network to identify potential customers who might be interested in the cleaning services her company provides.
This approach takes advantage of the trust and reliability found within personal networks, as individuals are more likely to believe and try services recommended by people they know and trust. Additionally, Mary's strategy could involve engaging with information hubs or influencers within her network who are more likely to share information about the cleaning services with a broader audience.
In the context of increasing sales, connecting with customers on a more personal level and leveraging their networks can be an effective method to gain new leads. This is particularly effective as it has been shown that a vast majority of jobs, similar to customer opportunities, are found through social networks, highlighting the power of personal contacts and referrals. Mary's approach can be seen as a grassroots method to increase customer outreach and foster organic growth for the business.
Consider the following equilibrium:
O2(g)+ 2F2(g) ↔ 2OF2(g); Kp = 2.3×10-15
Which of the following statements is true?
a. If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).
b. For this equilibrium, Kc=Kp.
c. If the reaction mixture initially contains only OF2(g), then the total pressure at equilibrium will be less than the total initial pressure.
d. If the reaciton mixture initially contains only O2(g) and F2(g), then at equilibrium, the reaction mixture will consist of essentially only OF2(g).
e. If the reaction mixture initially contains only O2(g) and F2(g), then the total pressure at equilibrium will be greater than the total initail pressure.
Answer:
a. If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g).
Explanation:
The answer is a) because the value for Kp is really close to zero (having x10⁻¹⁵), this means that at equilibrium O₂ and F₂ will be significantly more present than OF₂.
Final answer:
For the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g) with Kp = 2.3×10^-15, the reaction favors the reactants, making the correct answer that a mixture initially containing only OF2(g) will consist of essentially only O2(g) and F2(g) at equilibrium.
Explanation:
The question considers the equilibrium O2(g) + 2F2(g) ↔ 2OF2(g); Kp = 2.3×10-15 and asks which statement is true. Given the extremely low value of Kp, this indicates a strong preference for reactants at equilibrium. Therefore, the correct answer is: If the reaction mixture initially contains only OF2(g), then at equilibrium, the reaction mixture will consist of essentially only O2(g) and F2(g). This is because a very small Kp value means the equilibrium lies heavily on the side of the reactants, making statement (a) true.
Balance the following equation please
Answer:
2Fe + 3Br2 = 2FeBr3
Explanation:
A jar contains several different types of atoms. The proportions of these atoms can be changed slightly without changing any substance in the jar. This jar contains A. A molecule. B. A single element. C. A compound. D. A mixture.
The jar contains A Mixture
Explanation:
A mixture consists of different atoms that are not chemically bonded. It is of two types of the heterogeneous mixture and homogeneous mixture. The chemical substances can be differentiated visually in the heterogeneous mixture.
In a mixture, the proportion of atoms can be slightly changed without changing or modifying any substance. This is because the individual substances in mixture keep their properties when mixed together. Further, mixtures have variable compositions and substances. these mixtures can be are separated using physical methods.
The jar contains A Mixture
Explanation:
A mixture consists of different atoms that are not chemically bonded. It is of two types of the heterogeneous mixture and homogeneous mixture. The chemical substances can be differentiated visually in the heterogeneous mixture.
In a mixture, the proportion of atoms can be slightly changed without changing or modifying any substance. This is because the individual substances in mixture keep their properties when mixed together. Further, mixtures have variable compositions and substances. these mixtures can be are separated using physical methods.
A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 234 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.
The density of a metal that crystallizes in a face-centered cubic unit cell can be calculated by determining the number of atoms per unit cell, calculating the volume of a unit cell, and finally calculating the mass of the atoms in the unit cell using the given molar mass. The density is then found by dividing the mass by the volume.
Explanation:To answer this question, we need to know a couple of key pieces of information. First, a face-centered cubic unit cell consists of four atoms: one-eighth of an atom at each of the eight corners and half of an atom on each of the six faces (1/8*8+1/2*6=4 atoms).
Secondly, we need to find the volume of the atom that is engulfed in the cubic unit cell. Given that the radius of the metal is 234 picometers (or 234*10^-12 meters), the volume of the unit cell can be found by applying the formula for the volume of a cube (side^3) where the side equals 2*sqrt(2)*r.
We can then calculate the number of moles of atoms in the unit cell, convert them to grams using the molar mass, and finally calculate the density by dividing the calculated mass by the calculated volume.
We first convert the volume of the unit cell from cubic meters to cubic centimeters (1m^3 = 10^6 cm^3).
Then, we calculate the mass of the atoms in the unit cell using the molar mass (195.08 g/mol).
V=4/3*Pi*r^3Mass=No. of atoms per unit cell * Molar mass/Avogadro's numberFinally, we calculate the density. Density = Mass/Volume.
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To calculate the density of a metal in a face-centered cubic structure, you need to determine the volume of the unit cell and the molar mass of the metal.
Explanation:The density of a metal can be calculated using the formula:
Density = (Molar Mass ×Avogadro's Number) / (Volume of Unit Cell)
First, let's calculate the volume of the face-centered cubic (FCC) unit cell. The FCC unit cell consists of 4 atoms at the corners and 8 atoms at the face centers. The radius of the atom can be used to calculate the edge length of the unit cell using the formula:
Edge Length = 4×Radius / √2
Once we have the edge length, we can calculate the volume of the unit cell using the formula:
Volume of Unit Cell = Edge Length³
Finally, we can plug the values into the density formula and calculate the density of the metal.
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What is the entropy change for freezing 2.71 g of C2H5OH at 158.7 K? ∆H = −4600 J/mol. Answer in units of J/K.
Answer:
-1.71 J/K
Explanation:
To solve this problem we use the formula
ΔS = n*ΔH/T
Where n is mol, ΔH is enthalpy and T is temperature.
ΔH and T are already given by the problem, so now we calculate n:
Molar Mass C₂H₅OH = 46 g/mol
2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol
Now we calculate ΔS:
ΔS = 0.0589 mol * −4600 J/mol / 158.7 K
ΔS = -1.71 J/K
Separate samples of a solution of an unknown soluble ionic compound are treated with KCl, Na2SO4, and NaOH. A precipitate forms only when Na2SO4 is added. Which cations could be present in the unknown soluble ionic compound?
Common cations that can form insoluble sulfate salts that can form insoluble salts with the anion include calcium, barium, strontium, and lead.
A precipitate forms only when [tex]Na_2SO_4[/tex] is added to the solution of the unknown soluble ionic compound. This indicates that the anions in [tex]Na_2SO_4 (SO_4^2-)[/tex] react with certain cations in the unknown compound to form an insoluble compound.
To form an insoluble compound with sulfate ions [tex](SO_4^2-)[/tex], the cations need to form an insoluble sulfate salt. Common cations that can form insoluble sulfate salts include [tex]Ca^{2+}, Ba^{2+}, Sr^{2+}, Pb^{2+}[/tex].
A precipitate forms when [tex]Na_2SO_4[/tex] is added to the unknown solution, it suggests that the unknown compound could contain one or more of these cations. Other cations that do not form insoluble sulfate salts would not result in a precipitate with [tex]Na_2SO_4[/tex].
For the determination of the cation present in the compound, a confirmatory test is almost necessary.
Therefore, the common cations that can form insoluble sulfate salts that can form insoluble salts with the anion include calcium, barium, strontium and lead.
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The cation that can be present in the unknown soluble ionic compound can be Ba2+ since Barium sulfate(BaSO4) is known to be an insoluble sulfate salt which forms a precipitate.
Explanation:The unknown soluble ionic compound can have cations such as Ba2+ (barium). Here, the solubility is determined based on the interaction of ions in the solution. When a precipitate forms only with the addition of Na2SO4, it suggests that the sulfates form an insoluble compound with the cation present in the unknown solution. However, not with KCl or NaOH. According to solubility guidelines, sulfate salts (SO42-) are typically soluble, with exceptions for salts containing Pb2+, Sr2+, Ca2+, and Ba2+. So, one possible cation in the unknown soluble ionic compound is Ba2+.
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How many molecules of N2 are in a 400.0 mL container at 780 mm Hg and 135°C? Avogadro’s number = 6.022 x 1023A) 7.01 × 1021 moleculesB) 7.38 × 1021 moleculesC) 2.12 × 1022 moleculesD) 2.23 × 1022 molecules
Answer:
B
Explanation:
Firstly, we will need to calculate the number of moles. To do this, we make use of the ideal gas equation
PV = nRT
n = PV/RT
The parameters have the following values according to the question:
P = 780mmHg, we convert this to pascal.
760mHG = 101325pa
780mmHg = xpa
x = (780 * 101325)/760 = 103,991 Pa
V= 400ml = 0.4L
T = 135C = 135 + 273.15 = 408.15K
n = ?
R = 8314.463LPa/K.mol
Substituting these values into the equation yields the following:
n = (103991 * 0.4)/(8314.463 * 408.15)
= 0.012 moles
Now we know 1 mole contains 6.02 * 10^23 molecules, hence, 0.012moles will contain = 0.012 * 6.02 * 10^23 = 7.38 * 10^21 molecules
Answer: The number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]
Explanation:
To calculate the moles of gas, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 780 mmHg
V = Volume of the gas = 400.0 mL = 0.4 L (Conversion factor: 1 L = 1000 mL)
T = Temperature of the gas = [tex]135^oC=[135+273]K=408K[/tex]
R = Gas constant = [tex]62.364\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of nitrogen gas = ?
Putting values in above equation, we get:
[tex]780mmHg\times 0.4L=n\times 62.364\text{ L. mmHg }mol^{-1}K^{-1}\times 408K\\\\n=\frac{780\times 0.4}{62.364\times 408}=0.01226mol[/tex]
According to mole concept:
1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules
So, 0.01226 moles of nitrogen gas will contain = [tex](0.012\times 6.022\times 10^{23})=7.38\times 10^{21}[/tex] number of molecules
Hence, the number of nitrogen molecules in the container are [tex]7.38\times 10^{21}[/tex]
The balanced combustion reaction for C 6 H 6 is 2 C 6 H 6 ( l ) 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) 6 H 2 O ( l ) 6542 kJ If 7.700 g C 6 H 6 is burned and the heat produced from the burning is added to 5691 g of water at 21 ∘ C, what is the final temperature of the water
Answer: [tex]156.4^0C[/tex]
Explanation:
[tex]2C_6H_6(l)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)+6542kJ[/tex] [tex]\Delta H=-6542kJ[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.700g}{78.11g/mol}=0.9858moles[/tex]
According to stoichiometry :
2 moles of [tex]C_6H_6[/tex] releases = 6542 kJ of heat
0.9858 moles of [tex]C_6H_6[/tex] release =[tex]\frac{6542}{2}\times 0.9858=3224kJ[/tex] of heat
Thus heat given off by burning 7.700 g of [tex]C_6H_6[/tex] will be absorbed by water.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed= 3224kJ = 3224000J (1kJ=1000J)
m= mass of water = 5691 g
c = specific heat capacity = [tex]4.184J/g^0C[/tex]
Initial temperature of the water = [tex]T_i[/tex] = 21.0°C
Final temperature of the water = [tex]T_f[/tex] = ?
Putting in the values, we get:
[tex]3224000J =5691g\times 4.184J/g^0C\times (T_f-21)[/tex]
[tex]T_f=156.4^0C[/tex]
The final temperature of the water is [tex]156.4^0C[/tex]
A 10 g sample of ice, at -11 is mixed with 90 ml of water at 80. calculate the final temperature of the mixture assuming that no heat is lost to the surroundings. The heat capacities of H2O (s) and H2O(I) are 2.08 and 4.18 and the heat fusion for ice is 6.
Answer:
Final temperature is 71·31 °C
Explanation:
The amount of heat change in a body for a change in temperature of ΔT is m×s×ΔTwhere
m is the mass of the substance
s is the heat capacity of the substance
ΔT is the temperature difference
At constant temperature the amount of heat for fusion of ice is m×Lwhere
m is the mass of the substance
L is the latent heat of the substance
Let the final temperature be T
As there is no heat loss to the surrounding ∴ The heat gained by the ice = The heat lost by the waterAs the density of the water is 1 g/ml
∴ The mass of 90ml water will be 90 g
Heat gained by the ice when it is attained at 0°C and converted to water is
(10×2·08×11) + 10×6 = 288·8
But the heat loss by the water when water attains 0°C will be 90×4·18×80 = 30,096
As the heat loss is more therefore there will be further rise in temperature
Now the heat gain = 288·88 + 10×4·18×T
heat loss = 90×4·18×(80-T)
Here in the case of heat loss as we are already mentioning that the heat is lost so we are taking the magnitude of the heat change otherwise it would be 90×4·18×(T-80)
288·88 + 41·8×T = 9×41·8×(80-T)
41·8×9×80 - 288·8 = 10×41·8×T
∴T=71·31 °C
An unknown element is found to have three naturally occurring isotopes with atomic masses of 35.9675 (0.337%), 37.9627 (0.063%) and 39.9624 (99.600%). Which of the following is the unknown element? A) Ar B) K C) CI D) Ca E) None of the above could be the unknown element.
Answer:
The answer to your question is letter A, Argon
Explanation:
Isotope Atomic mass Percent (%)
1 35.9675 0.337
2 37.9627 0.063
3 39.9624 99.6
Average atomic mass = (Mass isotope 1)(percent 1) + (Mass isotope 2)(percent
2) + (Mass isotope 3)(percent 3)
Average atomic mass = (35.9675)(0.00337) + (37.9627)(0.00063) +
(39.9624)(0.996)
Average atomic mass = 0.1212 + 0.0239 + 39.8025
Average atomic mass = 39.9476
Theoretical Atomic mass
a) Ar 39.95
b) K 39.10
c) Cl 35.45
d) Ca 40.08
Glycerol (C3H8O3, 92.1 g/mol) is a nonvolatile nonelectrolyte substance. Consider that you have an aqueous solution that contains 34.4 % glycerol by mass. If the vapor pressure of pure water is 23.8 torr at 25oC, what is the vapor pressure of the solution at 25oC?
The vapor pressure of the solution can be calculated using Raoult's law. First, we need to calculate the mole fraction of glycerol and water in the solution. Then, we can use the mole fraction to calculate the vapor pressure of the solution.
Explanation:In order to find the vapor pressure of the solution, we can use Raoult's law. Raoult's law states that the vapor pressure of a solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent.
First, we need to calculate the mole fraction of glycerol in the solution. To do this, we need to convert the mass percent of glycerol to moles. Since the molar mass of glycerol is 92.1 g/mol and the solution contains 34.4% glycerol by mass, we can calculate the moles of glycerol:
Moles of glycerol = (34.4 g / 92.1 g/mol)
Next, we need to calculate the mole fraction of water in the solution. Since the solution is 34.4% glycerol, the mass percent of water is 100% - 34.4% = 65.6%. Using the molar mass of water (18.0 g/mol), we can calculate the moles of water:
Moles of water = (65.6 g / 18.0 g/mol)
Now we can calculate the mole fraction of glycerol and water:
Mole fraction of glycerol = (moles of glycerol) / ((moles of glycerol) + (moles of water))
Mole fraction of water = (moles of water) / ((moles of glycerol) + (moles of water))
Finally, we can calculate the vapor pressure of the solution using Raoult's law:
Vapor pressure of solution = (mole fraction of water) * (vapor pressure of water)
Oxide formation by heating element in oxygen atmosphere are best classified as which type of reaction?a. Synthesis b. Double displacement c. Decomposition d. None of the above
Answer: a. Synthesis
Explanation:
a. Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.
Example: [tex]Mg+O_2\rightarrow 2MgO[/tex]
Thus magnesium in its elemental form is combining with oxygen to form magnesium oxide.
b. Double displacement reaction is one in which exchange of ions take place.
Example: [tex]2NaOH(aq)+(NH_4)_2SO_4(aq)\rightarrow 2NH_4OH(aq)+Na_2SO_4(aq)[/tex]
c. Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.
Example: [tex]Li_2CO_3\rightarrow Li_2O+CO_2[/tex]
An oxide of nitrogen was found by elemental analysis to contain 30.4% nitrogen and 69.6% oxygen. If 23.0 g of this gas were found to occupy 5.6 L at STP, what are the empirical and molecular formulas for this oxide of nitrogen?
Express the empirical and molecular formulas for this oxide of nitrogen, separated by a comma.
Answer:
The answer is NO₂-NO₂, N₂O₄
Explanation:
If 23.0 g of this gas were found to occupy 5.6 L at STP we can apply the Ideal gas Law to determinate the quantity of mols of the oxide.
As I have the mass, I can find out the molar weight.
STP are 1 atm and 273, 1 K so:
P . V = n . R . T
1 atm . 5.6L = n . 0.082 L.atm/mol.K . 273.1K
(1 atm . 5.6L)/ 0.082 mol.K/L.atm . 273.1K = n
0.250 mol = n
Molar weight = mass/mols
23 g /0.250 mol = 92 g/m
The percentages are: 30.4% N and 69.6%O
100% ____ 92 g
30.4% _____ (30.4 .92)/100 = 28 g → 2 mols of N
100% _____ 92 g
69.6% ____ (69.6 . 92)/ 100 = 64 g → 4 mols of O
Molecular formula N₂O₄
Empirical formula: NO₂-NO₂
She dissolves a 10.0mg sample in enough water to make 30.0mL of solution. The osmotic pressure of the solution is 0.340torr at 25C. a). What is the molar mass of the gene fragment? b). If the solution density is 0.997g/mL, what is the freezing point for this aqueous solution?
Answer:
(a)The molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol
(b)The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C
Explanation:
The osmotic pressure (π) is given by the following equation:
[tex]\pi =cRT[/tex]
[tex]c[/tex]= Concentration of solution
R = universal gas constant = 62.364 [tex]\frac{L\times torr}{mol\times K}[/tex]
T = temperature
Weight of solute = w = 10.0 mg
Let the molecular weight of the solute be m g/mol.
Concentration = [tex]c=\frac{n}{V}\\ n=\frac{w}{m}\\ n=\frac{10\times10^{-3}}{m}\\c=\frac{10\times10^{-3}}{m\times30\times10^{-3}}M[/tex]
[tex]m=\frac{RT}{3\pi}[/tex]
m = 18220.071g/mol
Therefore, the molar mass of the gene fragment is 18220.071g/mol = 18.22 kg/mol
[tex]\Delta T_{f}=K_{f}m[/tex]
m is the molality of the solution.
m = [tex]1.835\times10^{-5}[/tex] mol/kg
[tex]\Delta T_{f}=1.86\times m[/tex]
[tex]\Delta T_{f}[/tex] = [tex]3.413\times10^{-5}[/tex] C
The freezing point for the aqueous solution is [tex]-3.413\times10^{-5}[/tex] C
The complete combustion of ethanol, C₂H₅OH (FW = 46.0 g/mol), proceeds as follows: [tex]C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)[/tex]; ΔH = −555 kJ What is the enthalpy change for combustion of 15.0 g of ethanol?
Answer: 181 kJ
Explanation:
The balanced chemical reaction is;
[tex]C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2Ol)[/tex] [tex]\Delta H=-555jJ[/tex]
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{15.0g}{46.0g/mol}=0.326moles[/tex]
According to stoichiometry:
1 mole of [tex]C_2H_5OH[/tex] on complete combustion give= 555 kJ
Thus 0.326 moles of [tex]C_2H_5OH[/tex] on complete combustion give=[tex]\frac{555}{1}\times 0.326=181kJ[/tex]
Thus the enthalpy change for combustion of 15.0 g of ethanol is 181 kJ
Metalloids have properties of what other families or elements? Group of answer choices
non-metals and halogens
metals and halogens
metals and non-metals
alkali and alkali earth metals
Answer:
Metalloids have properties of metals and non-metals.
Explanation:
A metalloid is a type of chemical element which has properties in between, or that are a mixture of, those of metals and nonmetals.
The six commonly recognized metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium.
Metalloids have a metallic appearance, but they are brittle (nonmetal property) and only fair conductors of electricity. Chemically, they behave mostly as nonmetals. They can form alloys with metal.
Most of their other physical properties and chemical properties are intermediate in nature.
That means Metalloids have properties of both metal and nonmetal.
The thermosphere is: 1. the layer of atmosphere closest to the Earth’s surface where weather occurs. 2. supports long distance communication because it reflects outgoing radio waves back to Earth without the use of satellites. 3. the layer where auroras form when electrically charged particles from the sun collide with gas molecules releasing energy visible as light of different colors
The thermosphere is a layer of Earth's atmosphere located 80-700km above sea level. It supports long-distance communication by reflecting radio waves back to Earth and is also where auroras form due to the collision of charged particles and gas molecules.
Explanation:The thermosphere is the fourth layer of Earth's atmosphere, located above the mesosphere and extend from about 80 to 700 kilometers above sea level. This layer holds a unique property due to the presence of electrically charged particles, or ions which allows it to support long-distance communication by reflecting radio waves back to Earth, bypassing the need for satellites. This is also the layer in which auroras form. Auroras occur when these charged particles from the sun collide with gas molecules in the Earth's atmosphere, releasing energy that manifests as visible, colorful light.
The Earth's atmosphere is divided into five main layers: The troposphere, the stratosphere, the mesosphere, the thermosphere and the exosphere. The Troposphere being the one closest to the Earth's surface and is where weather is generally observed. This layer expands to a height of roughly 12 km above the sea level and makes up around 80% of the atmosphere's mass. The Thermosphere is not the layer of atmosphere closest to the Earth’s surface where weather typically occurs, this property represents the troposphere.
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The third option is correct.
3. The thermosphere is the layer where auroras form when electrically charged particles from the sun collide with gas molecules releasing energy visible as light of different colors.
Scientific knowledge A. Never changes because scientists are never wrong. B. Changes as the public's opinion of a given topic changes. C. Can change as new research and experiments are done. D. Is always accurate because scientists know everything about the natural world.
Answer:can change as new researches and experiments are done
Explanation:
A 32.8 g iron rod, initially at 22.4 C, is submerged into an unknown mass of water at 63.1 C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.1 C.What is the mass of the water?
Answer:
mass water = 32.4 g
Explanation:
specific heat iron = 0.450 J/g°C
specific heat water = 4.18 J/g°C
32.8 x 0.450 ( 59.1 - 22.4) + mass water x 4.18 ( 59.1- 63.1)=0
541.7 - mass water x 16.7 = 0
mass water = 32.4 g
A sample of impure tin of mass 0.526 g is dissolved in strong acid to give a solution of Sn2 . The solution is then titrated with a 0.0448 M solution of NO3−, which is reduced to NO(g). The equivalence point is reached upon the addition of 3.67×10−2 L of the NO3− solution.
Find the percent by mass of tin in the original sample, assuming that it contains no other reducing agents.
Answer:
55.7%
Explanation:
The reaction that takes place is:
3Sn²⁺ + 2NO₃⁻ + 8H⁺ → 2NO + 3Sn⁺⁴ + 4H₂OWith the volume and concentration of NO₃⁻ solution, we can calculate the moles of Sn²⁺ that reacted:
3.67x10⁻² L * 0.0448 M = 1.64x10⁻³ mol NO₃⁻1.64x10⁻³ mol NO₃⁻ * [tex]\frac{3molSn^{2+}}{2molNO_{3}^{-}}[/tex] = 2.47x10⁻³mol Sn²⁺Now we convert moles of Sn to mass, using its atomic weight:
2.47x10⁻³mol Sn²⁺ * 118.71 g/mol = 0.293 g SnFinally we calculate the percent by mass of Sn:
0.293 g / 0.526 g * 100% = 55.7%27.0 L of HCl gas at STP is dissolved in water, giving 785 mL of solution. What is the molarity of the HCl solution? 9.46 M
Answer:
Molarity = 1.53 M
Explanation:
At STP,
Pressure = 1 atm
Temperature = 273.15 K
Given, Volume = 27.0 L
Using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
1 atm × 27.0 L = n × 0.0821 L.atm/K.mol × 273.15 K
⇒n = 1.20 moles
Given that volume = 785 mL
Also,
[tex]1\ mL=10^{-3}\ L[/tex]
So, Volume = 785 / 1000 L = 0.785 L
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{1.20}{0.785}[/tex]
Molarity = 1.53 M
2 A(g) + B(g) ⇄ 2 C(g)
When the concentration of substance B in the reaction above is doubled, all other factors being held constant, it is found that the rate of the reaction remains unchanged. The most probable explanation for this observation is that:
(A) the order of the reaction with respect to substance B is 1.
(B) substance B is not involved in any of the steps in the mechanism of the reaction.
(C) substance B is not involved in the rate-determining step of the mechanism but is involved in subsequent steps.
(D) substance B is probably a catalyst, and as such, its effect on the rate of the reaction does not depend on its concentration.
(E) the reactant with the smallest coefficient in the balanced equation generally has little or no effect on the rate of the reaction.
Answer:
(C) substance B is not involved in the rate-determining step of the mechanism but is involved in subsequent steps.
Explanation:
A. Is incorrect because if the order of the reaction with respect B was one then the rate would increase by the same multiple that B is increased by.
B. If B is reactant then it must be involved in the mechanism of the reaction and in the formation of the product
D. If B was a catalyst then increasing it's amount would affect the rate
E. That is just factually untrue. Effect of reactants on rates can only be found experimentally, not stoichiometrically.
The most probable explanation for this observation is (C) substance B is not involved in the rate-determining step of the mechanism but is involved in subsequent steps.
The key observation here is that doubling the concentration of substance B has no effect on the rate of the reaction, implying that substance B is not involved in the rate-determining step.In a typical multi-step reaction mechanism, the rate-determining step is the slowest step and primarily controls the overall rate of the reaction. As a result, even if other substances are involved in subsequent steps, they will not affect the initial rate if they do not appear in the rate-deciding step.
Our atmosphere is composed primarily of nitrogen and oxygen, which coexist at 25°C without reacting to any significant extent. However, the two gases can react to form nitrogen monoxide according to the reaction: N2(g) + O2(g) = 2 NO(g)
a. Calculate AGⓇ and K, for this reaction at 298 K. Is the reaction spontaneous?
b. Estimate AGⓇ at 2000 K. Does the reaction become more spontaneous as temperature increases?
To determine the spontaneity of the reaction between N2 and O2 to form 2NO and the equilibrium constant K, one would use the Gibbs Free Energy (ΔG) formula and the equilibrium constant (K) formula. While exact values can't be computed without data for the standard enthalpy and entropy changes, it can be inferred that this reaction is typically not spontaneous at room temperature (298 K), but becomes more spontaneous as temperature increases toward 2000 K.
Explanation:The given chemical reaction is between nitrogen and oxygen to form nitrogen monoxide: N2(g) + O2(g) = 2 NO(g). To answer your question, we'll need to use the Gibbs Free Energy (ΔG) formula and the equilibrium constant (K) formula which are linked by the equation ΔG = -RTlnK. These formulas are what we use to determine the spontaneity of a reaction and the equilibrium state at a given temperature.
Without the specific values of the standard enthalpy (ΔH) and entropy (ΔS) changes for this reaction, it's impossible to calculate ΔG and K exactly. However, at room temperature (298 K), the reaction is typically not spontaneous because the formation of NO requires high-energy conditions - usually above 2000 K. This is inferred by the presence of NO only in extreme conditions such as lightning strikes or in high-temperature combustion processes.
At 2000 K, although I can't give an exact numerical estimate without the ΔH and ΔS values, we can say that the reaction becomes more spontaneous as temperature increases since high energy and high temperature favor the formation of NO. It's always important to remember that whether a chemical reaction is spontaneous or not depends not only on the change in enthalpy (ΔH), but also on the change in entropy (ΔS) and the absolute temperature (T).
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The reaction N2(g) + O2(g) = 2NO(g) is not spontaneous at 298 K, with a very small equilibrium constant (K). However, at 2000 K, the reaction becomes more spontaneous as the free energy change (ΔG) is negative.
Explanation:The reaction N2(g) + O2(g) = 2NO(g) has a standard free energy change (ΔG°) of 173.4 kJ/mol at 298 K. To find the equilibrium constant (K) at this temperature, we use the equation ΔG° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. Rearranging the equation, we get K = e-ΔG°/(RT).
Substituting the values into the equation and solving, we find K to be 6.95 x 10-31, which is a very small value. Since K is less than 1, the reaction is not spontaneous at 298 K.
To estimate ΔG° at 2000 K, we can use the equation ΔG = ΔH - TΔS, where ΔH is the enthalpy change and ΔS is the entropy change. Assuming ΔH and ΔS do not change significantly with temperature, we can use the same values as at 298 K.
Substituting the values into the equation, we find ΔG to be -545.4 kJ/mol at 2000 K. Since ΔG is negative, the reaction becomes more spontaneous as temperature increases.
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When sodium metal is added to water, the following reaction occurs:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
What is the theoretical number of moles of hydrogen gas should be produced when 0.068 mol of sodium metal reacts?
a. 1.0 mol H2
b. 0.14 mol H2
c. 0.068 mol H2
d. 0.034 mol H2
Answer:
d. n H2(g) = 0.034 mol
Explanation:
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)2 - Na - 2
4 - H - 4
2 - O - 2
∴ n Na(s) = 0.068 mol
⇒ n H2(g) = ( 0.068 mol Na(s) )( mol H2(g) / 2 mol Na(s) )
⇒ n H2(g) = 0.034 mol
Based on the stoichiometric relationship between sodium and hydrogen in the given balanced chemical equation, 0.068 mol of sodium would yield 0.034 mol of hydrogen gas.
Explanation:The correct answer is (b) 0.034 mol H2. The balanced chemical equation given states that 2 moles of sodium metal (Na) react with 2 moles of water molecules (H2O) to produce 2 moles of sodium hydroxide (NaOH) and 1 mole of hydrogen gas (H2). This implies a stoichiometric ratio of 2:1 between sodium and hydrogen. Therefore, if we have 0.068 moles of sodium, we would obtain half of this number in moles of hydrogen because of this stoichiometric relationship. 0.068 mol ÷ 2 = 0.034 mol H2.
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Covalent network solids typically have blank melting points and blank boiling points
Answer:
Covalent network solids typically have high melting points and high boiling points
Explanation:
Covalent solids are the solids which have covalent bonds as intermolecular or interatomic interactions.
As covalent bonds are very strong, these solids are generally characterized by high melting points and high boiling points.
The examples of such solids are: Diamond and Graphite.
The other classes of solids are
i) ionic
ii) molecular
iii) metallic.