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
Given any two elements within a group, is the element with the larger atomic number likely to have a larger or smaller atomic radius than the other element
Answer:
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
Consider the example of sodium and potassium.
Sodium is present above the potassium with in same group i.e, group one.
The atomic number of sodium is 11 and potassium 19.
So potassium will have larger atomic radius as compared to sodium.
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
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)
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.
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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Identify the species oxidized, the species reduced, the oxidizing agent and the reducing agent in the following electron transfer reaction. 2Fe3+ + 2I-2Fe2+ + I2 species oxidized species reduced oxidizing agent reducing agent As the reaction proceeds, electrons are transferred from to_______.
Answer:
In the given chemical reaction:
Species Oxidized: I⁻
Species Reduced: Fe³⁺
Oxidizing agent: Fe³⁺
Reducing agent: I⁻
As the reaction proceeds, electrons are transferred from I⁻ to Fe³⁺
Explanation:
Redox reaction is a chemical reaction involving the simultaneous movement of electrons thereby causing oxidation of one species and reduction of the other species.
The chemical species that gets reduced by gaining electrons is called an oxidizing agent. Whereas, the chemical species that gets oxidized by losing electrons is called a reducing agent.
Given redox reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
Oxidation half-reaction: 2 I⁻ + → I₂ + 2 e⁻ ....(1)
Reduction half-reaction: [ Fe³⁺ + 1 e⁻ → Fe²⁺ ] × 2
⇒ 2 Fe³⁺ + 2 e⁻ → 2 Fe²⁺ ....(2)
In the given redox reaction, Fe³⁺ (oxidation state +3) accepts electrons and gets reduced to Fe²⁺ (oxidation state +2) and I⁻ (oxidation state -1) loses electrons and gets oxidized to I₂ (oxidation state 0).
Therefore, Fe³⁺ is the oxidizing agent and I⁻ is the reducing agent and the electrons are transferred from I⁻ to Fe³⁺.
In this electron transfer reaction, iodide ions (I-) are oxidized to iodine (I2) and act as the reducing agent, while iron(III) ions (Fe3+) are reduced to iron(II) ions (Fe2+) and act as the oxidizing agent.
Explanation:In the provided electron transfer reaction, a notable chemical transformation unfolds where the iodide ion (I-) undergoes oxidation to form iodine (I2), while concurrently, the iron(III) ion (Fe3+) experiences reduction to yield iron(II) ion (Fe2+). In terms of the key participants, it is crucial to discern the roles of reducing and oxidizing agents. The reducing agent is the entity that itself undergoes oxidation, and in this specific scenario, the iodide ion (I-) fulfills this role by ceding electrons during the reaction.
Conversely, the oxidizing agent is the species that undergoes reduction in the course of the reaction, and here, it's the iron(III) ion (Fe3+). As the reaction unfolds, electrons are effectively transferred from iodide ions to iron(III) ions, signifying a fundamental aspect of electron transfer reactions. This intricate interplay between the reducing and oxidizing agents, with the exchange of electrons, characterizes the essence of redox reactions in chemistry.
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What happens to the atoms of most nonmetals when they react with other elements
Final answer:
Nonmetals generally gain electrons to form anions in reactions with metals, creating ionic compounds, or they share electrons in reactions with each other to form molecular compounds.
Explanation:
When nonmetals react with other elements, they typically undergo a process where they gain electrons to form anions, or negative ions, as they combine with metals to produce ionic compounds. In reactions with each other, nonmetals often form molecular compounds with covalent bonding, where electrons are shared rather than transferred. The most reactive nonmetals, like fluorine, have a particularly high electron affinity and can take part in vigorous redox reactions, changing their oxidation states. Considering the diverse group of nonmetals, these elements can form a variety of different compounds, both ionic and covalent, depending on the elements they react with and the environmental conditions.
What is a metabolic pathway? What is a metabolic pathway?
A) A process that results in mutations in genes.
B) A series of chemical reactions where enzymes work one at a time to convert a reactant into intermediates and the intermediates into a final product.
C) A series of steps that convert a precursor to arginine.
D) A series of steps only found in Neurospora crassa.
Answer:
B) A series of chemical reactions where enzymes work one at a time to convert a reactant into intermediates and the intermediates into a final product.
Explanation:
In our cells there are multiple chemical reactions happening all the time to convert the elements we ingest (food, water, air) into the products the organism needs.
This processes can be very complex and offen involves various reactions (steps). Also, many of this reactions need to by catalized by enzymes in order to happen.
Answer: B
Explanation:
A metabolic pathway is series of chemical reactions that are well organized, well co-ordinated, and purposeful manner; all these reactions are collectively called as metabolism.
Metabolic Pathways this is a series of biochemical reaction where enzymes work one at a time to convert a reactant into intermediates and the intermediates into a final product. The product of reactions serve as an a reactants for the next enzymatic reaction.
The reactants, intermediates, products of an enzymatic reaction are known as metabolites.
Types of Metabolic pathway
1. Catabolic
2. Anabolic
3. Amphibolic
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
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)
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]
Balance the following equation please
Answer:
2Fe + 3Br2 = 2FeBr3
Explanation:
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
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
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]
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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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
gold(iii) chloride reacts with Antimony(III) oxide and water to produce Gold Antimony(V) oxide and Hydrochloric acid balanced equation
The reaction N2 + 3 H2 → 2 NH3 is used to produce ammonia. When 450. g of hydrogen was reacted with nitrogen, 1575 g of ammonia were actually produced. What is the percent yield of this reaction?
a. 20,7%
b. 41,5%
c. 62,1%
d. 30,8%
e. More information is needed to solve this problem
Answer:
61.76%
Explanation:
Percent yield = Actual yield/Theoeetical yield × 100%
We need to calculate the theoretical yield for this reaction.
Firstly, we need to know the number of moles of hydrogen reacted = mass of hydrogen ÷ molar mass = 450 ÷ 2 = 225 moles
From the reaction, we can see that 3 moles of hydrogen yielded 2 moles of ammonia. Hence 225 moles of hydrogen will yield:
(225 × 2) ÷ 3 = 150 moles
The actual mass of ammonia yielded = number of moles × molar mass = 150 × 17 = 2550g
Actual yield = 1575g
Theoretical yield = 2550g
Percentage yield = 1575/2550 × 100% = 61.76%
The correct option is (c). The percent yield of this reaction is 62.1%
To calculate the percent yield of the reaction, we need to compare the actual yield to the theoretical yield. The balanced chemical equation for the production of ammonia is:
[tex]\[ N_2 + 3H_2 \rightarrow 2NH_3 \][/tex]
From the equation, we see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. The molar masses of[tex]\( N_2 \), \( H_2 \), and \( NH_3 \)[/tex] are approximately 28.02 g/mol, 2.016 g/mol, and 17.031 g/mol, respectively.
First, we calculate the number of moles of hydrogen gas that reacted:
[tex]\[ \text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} = \frac{450 \text{ g}}{2.016 \text{ g/mol}} \approx 223.2 \text{ moles} \][/tex]
Since the stoichiometric ratio of[tex]\( H_2 \) to \( NH_3 \)[/tex] is 3:2, we can find the theoretical yield of ammonia:
[tex]\[ \text{moles of } NH_3 \text{ produced theoretically} = \frac{2}{3} \times 223.2 \text{ moles} \approx 148.8 \text{ moles} \][/tex]
Now, we calculate the theoretical mass of ammonia that could be produced:
[tex]\[ \text{mass of } NH_3 \text{ theoretically} \approx 148.8 \text{ moles} \times 17.031 \text{ g/mol} \approx 2533.5 \text{ g} \][/tex]
The actual yield is given as 1575 g of ammonia. The percent yield is calculated by:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% \][/tex]
[tex]\[ \text{Percent Yield} = \left( \frac{1575 \text{ g}}{2533.5 \text{ g}} \right) \times 100\% \approx 62.1\% \][/tex]
Therefore, the percent yield of this reaction is 62.1%.
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.
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.
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.
alculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035 kJ 2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJ H2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJ H2O(l)→H2O(g), ΔH∘D=+4
Answer:
-2312 kJ
Explanation:
By the Hess Law, when a reaction is formed by several steps, the enthalpy change (ΔH) of the global reaction can be calculated by the sum of the enthalpy changes of the steps reactions.
Besides that, when a reaction is multiplied by a constant, ΔH must be multiplied by the same constant, and when the reaction is inverted, the signal of ΔH must be inverted.
B₂O₃(s) + 3H₂O(g) → 3O₂(g) + B₂H₆(g) ΔH = +2035 kJ (multiply by 2 and inverted)
2B(s) + 3H₂(g) → B₂H₆(g) ΔH = +36 kJ (multiply by 2)
H₂(g) + 1/2O₂(g) → H₂O(l) ΔH = -285 kJ (multiply by 6 and inverted)
H₂O(l) → H₂O(g) ΔH = +4 kJ (multiply by 6 and inverted)
-------------------------------------------------------
6O₂(g) + 2B₂H₆(g) → 2B₂O₃(s) + 6H₂O(g) ΔH = -4070 kJ
4B(s) + 6H₂(g) → 2B₂H₆(g) ΔH = +72 kJ
6H₂O(l) → 6H₂(g) + 3O₂(g) ΔH = +1710 kJ
6H₂O(g) → 6H₂O(l) ΔH = -24 kJ
Summing the equation, simplifying the equal terms in both sides:
4B(s) + 3O₂(g) → 2B₂O₃(s)
ΔH = -4070 + 72 + 1710 - 24
ΔH = -2312 kJ
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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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
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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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.
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
What is the balanced equation for the combustion of magnesium?
Answer:
2Mg + O2 → 2MgO
Explanation:
In all conbustion you should know, that reactans are an specific compound and O2, so the products must be CO2 and H2O, or in this case, the corresponding oxide.
The balanced equation for the combustion of Magnesium is 2Mg + O2 -> 2MgO. It's a redox reaction where Magnesium is oxidised and Oxygen is reduced.
Explanation:The combustion of Magnesium in Oxygen is a well-known reaction in Chemistry. The balanced equation for this reaction is 2Mg + O2 -> 2MgO. In this reaction, two moles of Magnesium react with one mole of Oxygen gas to produce two moles of Magnesium Oxide. This is a redox reaction where Magnesium gets oxidised and Oxygen gets reduced.
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