Answer:
Wt Avg Atomic Wt = 28.0577 amu
Explanation:
Atomic Mass = ∑Wt Avg contributions of isotopes of an element
Si-28 => 92.21% => 0.9221 decimal fraction => Wt Avg Si-28 = 0.9221(27.9769) = 25.7695 amu
Si-29 => 4.69% => 0.0469 decimal fraction => Wt Avg Si-29 = 0.0469(28.9765) = 1.3590 amu
Si-30 => 3.10% => 0.0310 decimal fraction => Wt Avg Si-30 = 0.0310(29.9737) = 0.9292
Wt Avg Atomic Wt = ∑Wt Avg Contributions = (25.7695 + 1.3590 + 0.9292)amu = 28.0577 amu
Answer : The average atomic mass of silicon is, 28.08 amu
Explanation :
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]
For isotope Si-28 :
Mass of isotope Si-28 = 27.9769 amu
Fractional abundance of isotope Si-28 = 92.21 % = 0.9221
For isotope Si-29 :
Mass of isotope Si-29 = 28.9765 amu
Fractional abundance of isotope Si-29 = 4.69 % = 0.0469
For isotope Si-30 :
Mass of isotope Si-30 = 29.9737 amu
Fractional abundance of isotope Si-30 = 3.10 % = 0.0310
Putting values in equation 1, we get:
[tex]\text{Average atomic mass }=[(27.9769\times 0.9221)+(28.9765\times 0.0469)+(29.9737\times 0.0310)][/tex]
[tex]\text{Average atomic mass }=28.08amu[/tex]
Thus, the average atomic mass of silicon is, 28.08 amu
How many moles of Cu are needed to react with 5.8 moles of AgNO3?
Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
Answer:
moles of Cu° needed = ½ (5.8 moles) = 2.9 moles Cu° needed.
Explanation:
Cu + 2AgNO₃ => Cu(NO₃)₂ + 2Ag
? moles Cu + 5.8 moles AgNO₃ => Cu(NO₃)₂ + 2Ag
Note coefficient of Cu° vs coefficient of AgNO₃
=> 1 mole Cu° < 2 moles AgNO₃ ...
=> Since moles of Cu° are smaller than moles of AgNO₃ in the given equation, the moles of Cu° needed to react with 5.8 moles AgNO₃ will be smaller than the 5.8 by the ratio of coefficients that will make 5.8 smaller. That is ...
moles of Cu° needed = ½ (5.8 moles) = 2.9 moles Cu° needed.
Note: Using 2/1(5.8) will make value greater than 5.8 and incorrect.
One can also set up a ratio relationship as follows...
1 mole Cu° <=> 2 moles AgNO₃ (fm equation)
? mole Cu° <=> 5.8 moles AgNO₃ (problem)
=> (1 mole Cu°)/X=(2 mole AgNO₃)/(5.8 moles AgNO₃)
=> X = (1 mole Cu°)(5.8 mole AgNO₃)/2 mole AgNO₃
= ½(5.8) mole Cu° = 2.9 mole Cu°
Final answer:
2.9 moles of Cu are needed to react with 5.8 moles of AgNO3 according to the stoichiometry of the balanced chemical reaction.
Explanation:
To determine how many moles of Cu (copper) are required to react with 5.8 moles of AgNO3 (silver nitrate), we use the stoichiometry of the balanced chemical reaction:
Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
This balanced equation tells us that one mole of Cu reacts with two moles of AgNO3. If we have 5.8 moles of AgNO3, then the amount of Cu required is calculated by dividing the moles of AgNO3 by 2:
(5.8 moles AgNO3) / (2 moles AgNO3/moles Cu) = 2.9 moles of Cu
Therefore, 2.9 moles of Cu are needed to react with 5.8 moles of AgNO3.
Some hydrogen gas is enclosed within a chamber being held at 200∘C with a volume of 0.0250 m3. The chamber is fitted with a movable piston. Initially, the pressure in the gas is 1.50×106Pa (14.8 atm). The piston is slowly extracted until the pressure in the gas falls to 0.950×106Pa. What is the final volume V2 of the container
Answer:
0.04 m³.
Explanation:
We can use the general law of ideal gas: PV = nRT.where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and T are constant, and have different values of P and V:(P₁V₁) = (P₂V₂).
V₁ = 0.025 m³, P₁ = 1.5 x 10⁶ Pa,
V₂ = ??? m³, P₂ = 0.95 x 10⁶ Pa.
∴ V₂ = (P₁V₁)/(P₂) = (0.025 m³)(1.5 x 10⁶ Pa)/(0.95 x 10⁶ Pa) = 0.04 m³.
Answer: The final volume of the container is [tex]0.040m^3[/tex]
Explanation:
To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature.
The equation given by this law is:
[tex]P_1V_1=P_2V_2[/tex]
where,
[tex]P_1\text{ and }V_1[/tex] are initial pressure and volume.
[tex]P_2\text{ and }V_2[/tex] are final pressure and volume.
We are given:
[tex]P_1=1.50\times 10^6Pa\\V_1=0.0250m^3\\P_2=0.950\times 10^6\\V_2=?m^3[/tex]
Putting values in above equation, we get:
[tex]1.50\times 10^6Pa\times 0.0250m^3=0.950\times 10^6Pa\times V_2\\\\V_2=\frac{1.50\times 10^6\times 0.0250}{0.950\times 10^6}=0.040m^3[/tex]
Hence, the final volume of container is [tex]0.040m^3[/tex]
For the reaction shown, calculate how many moles of NO2 form when each of the following completely reacts. 2N2O5(g)→4NO2(g)+O2(g) Part A 1.0 mol N2O5 Express your answer using two significant figures. nothing mol m o l Request Answer Part B 5.4 mol N2O5 Express your answer using two significant figures.
Answer:
For part A: The number of moles of [tex]NO_2[/tex] is 2.0 moles.
For part B: The number of moles of [tex]NO_2[/tex] is [tex]1.1\times 10^1[/tex] moles.
Explanation:
For the given chemical equation:
[tex]2N_2O_5(g)\rightarrow 4NO_2(g)+O_2(g)[/tex]
For Part A: When 1.0 mole of [tex]N_2O_5[/tex] is reactedBy Stoichiometry of the reaction:
2 moles of [tex]N_2O_5[/tex] produces 4 moles of [tex]NO_2[/tex]
So, 1.0 moles of [tex]N_2O_5[/tex] will produce = [tex]\frac{4}{2}\times 1.0=2.0moles[/tex] of [tex]NO_2[/tex]
Hence, the number of moles of [tex]NO_2[/tex] expressed in two significant figures are 2.0 moles.
For Part B: When 5.4 moles of [tex]N_2O_5[/tex] is reactedBy Stoichiometry of the reaction:
2 moles of [tex]N_2O_5[/tex] produces 4 moles of [tex]NO_2[/tex]
So, 5.4 moles of [tex]N_2O_5[/tex] will produce = [tex]\frac{4}{2}\times 5.4=10.8moles[/tex] of [tex]NO_2[/tex]
To express it in two significant figures, we round this value of 10.8 mol to 11 mole and then express it in scientific notation.
Scientific notation is defined as the way of representing a number which have very large value or very small value and is written in the decimal form.
Hence, the number of moles of [tex]NO_2[/tex] expressed in two significant figures are [tex]1.1\times 10^1[/tex] moles.
In the given reaction, 2 moles of NO2 are produced for every mole of N2O5. So, in part A, 1 mole of N2O5 will produce 2 moles of NO2, and in part B, 5.4 moles of N2O5 will produce around 11 moles of NO2.
Explanation:The question is asking about the stoichiometry of a chemical reaction - specifically, how many moles of nitrogen dioxide (NO2) are formed from a given number of moles of dinitrogen pentoxide (N2O5). The balanced chemical equation is 2N2O5(g)→4NO2(g)+O2(g).
Part A: With 1.0 mol of N2O5, according to the stoichiometric ratio in the balanced equation (2:4), you would get 2 mol of NO2 for every 1 mol of N2O5, so you would have 2.0 moles of NO2.
Part B: With 5.4 mol of N2O5, again using the stoichiometric ratio, you would get 2 * 5.4 = 10.8 ≈ 11 moles of NO2 (rounded to two significant figures).
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The polymerization of amino acids produces: DNA, RNA, carbohydrates, or polypeptides?
Answer:
The polymerization of amino acids produces polypeptides.
Explanation:
The polymerization of amino acids producing polypeptide is known as translation. mRNA produces amino acids.
The polymerization of amino acids produces polypeptides. Hence, option D is correct.
What is polymerization?Polymerization is defined as "A chemical process that combines several monomers to form a polymer or polymeric compound."
Polymerization of amino acids is the formation of proteins. Amino acids are small molecules consisting of an amino group, a carboxyl group, a hydrogen atom, and an ‘R’ group.
In polymerization, two amino acids make a bond with each other with a peptide bond, which is formed between the amino group of one amino acid and the carboxyl group of another amino acid.
Hence, the polymerization of amino acids produces polypeptides which are also known as translation. mRNA produces amino acids.
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If we find that there is a linear correlation between the concentration of carbon dioxide in our atmosphere and the global temperature, does that indicate that changes in the concentration of carbon dioxide cause changes in the global temperature? Choose the correct answer below. A. Yes. The presence of a linear correlation between two variables implies that one of the variables is the cause of the other variable. B. No. The presence of a linear correlation between two variables does not imply that one of the variables is the cause of the other variable.
Answer:
A
Explanation:
A linear correlation means the increase in one variable cause an increase in the other variable. In a graph, the linear correlation can be demonstrated by a right-slanted straight diagonal line. Therefore if an increase in carbon dioxide causes a directly proportional increase in global temperatures then the two are correlated.
A linear correlation between two variables, such as the concentration of carbon dioxide in our atmosphere and the global temperature, does not necessarily imply that changes in one cause changes in the other. Correlation does not imply causation.
Explanation:The correct answer to your question is B: No. The presence of a linear correlation between two variables does not imply that one of the variables is the cause of the other variable. In statistics, correlation describes the degree to which two variables move in relation to each other, but it does not imply causation. In other words, just because the concentration of carbon dioxide in our atmosphere and the global temperature move together (e.g., they both increase or decrease at the same time), it does not necessarily mean that changes in the concentration of carbon dioxide cause changes in the global temperature. Other factors might be influencing both, or the relationship might be coincidental.
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An aqueous solution of potassium hydroxide is standardized by titration with a 0.194 M solution of hydrobromic acid. If 25.2 mL of base are required to neutralize 24.2 mL of the acid, what is the molarity of the potassium hydroxide solution? M potassium hydroxide
Answer: The molarity of potassium hydroxide is 0.186 M
Explanation:
To calculate the molarity of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HBr
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=1\\M_1=0.194M\\V_1=24.2mL\\n_2=1\\M_2=?M\\V_2=25.2mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.194\times 24.2=1\times M_2\times 25.2\\\\M_2=0.186M[/tex]
Hence, the concentration of potassium hydroxide is 0.186 M.
For the decomposition of calcium carbonate, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): ΔH∘rxn 178.5kJ/mol ΔS∘rxn 161.0J/(mol⋅K) Calculate the temperature in kelvins above which this reaction is spontaneous. Express your answer to four significant figures and include the appropriate units.
Answer: [tex]1.109\times 10^3[/tex] with four significant digits.
Explanation:
Given,
[tex]\Delta H[/tex] = 178.5 KJ/mole = 178500 J/mole (1kJ=1000J)
[tex]\Delta S[/tex] = 161.0 J/mole.K
According to Gibbs–Helmholtz equation:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
[tex]\Delta G[/tex] = Gibbs free energy
[tex]\Delta H[/tex] = enthalpy change
[tex]\Delta S[/tex] = entropy change
T = temperature in Kelvin
[tex]\Delta G[/tex]= +ve, reaction is non spontaneous
[tex]\Delta G[/tex]= -ve, reaction is spontaneous
[tex]\Delta G[/tex]= 0, reaction is in equilibrium
As per question the reaction is spontaneous that means the value of [tex]\Delta G[/tex] is negative or we can say that the value is less than zero.
Thus
[tex]T\Delta S>\Delta H[/tex]
[tex]T\times 161J/Kmol> 178500J/mol[/tex]
[tex]T>1109K[/tex]
Significant figures are the figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Thus the temperature is [tex]1.109\times 10^3[/tex] in kelvins above which this reaction is spontaneous.
22. A flask containing 450 mL of 0.50 M H2SO4 was accidentally knocked to the floor. How many grams of NaHCO, do you need to put on the spill to neutralize the acid according to the following equation: H2SO4(aq)+2 NaHCOs(aq) Na,SO(aq) +2 H20()+2 CO2(g) D) 38 g A) 2.3 g B) 9.5 g C) 19 g
To neutralize the 0.50 M H2SO4, calculate the moles of NaHCO3 required using the mole ratio from the balanced equation and then convert the moles to grams.
Explanation:To neutralize the 0.50 M H2SO4, we need to understand the stoichiometry of the reaction.
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaHCO3.
Using the molarity and volume of the H2SO4 solution, we can calculate the moles of H2SO4, and then use the mole ratio to find the moles of NaHCO3 required to neutralize it. Finally, we can convert the moles of NaHCO3 to grams by using the molar mass of NaHCO3.
The correct answer is C) 19 g.
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What is the mole fraction of potassium dichromate, K2Cr2O7, in a solution prepared from 24.42 g of potassium dichromate and 240. g of water? Report your answer in scientific notation with three significant figures
Answer:
[tex]6.19\times 10^{-3}[/tex] is the mole fraction of potassium dichromate.
Explanation:
Mass of potassium dichromate = 24.42 g
Moles of potassium dichromate =[tex]n_1=\frac{24.42 g}{294.185 g/mol}=0.0830 mol[/tex]
Mass of water = 240.0 g
Moles of water =[tex]n_2=\frac{240.0 g}{18.015 g/mol}=13.3222 mol[/tex]
Mole fraction is calculated by:
[tex]\chi_1=\frac{n_1}{n_1+n_2}[/tex]
[tex]\chi_1=\frac{0.0830 mol}{0.0830 mol+13.3222 mol}=0.00619=6.19\times 10^{-3}[/tex]
[tex]6.19\times 10^{-3}[/tex] is the mole fraction of potassium dichromate.
In a popular classroom demonstration, solid sodium is added to liquid water and reacts to produce hydrogen gas and aqueous sodium hydroxide. Part A Write a balanced chemical equation for this reaction. Express your answer as a chemical equation. Identify all of the phases in your answer.
Answer: The chemical reaction is given below.
Explanation:
When solid sodium metal reacts with water molecule to produce aqueous sodium hydroxide and hydrogen gas. The equation for this follows:
[tex]2Na(s)+2H_2O(l)\rightarrow 2NaOH(aq.)+H_2(g)[/tex]
By Stoichiometry of the reaction:
2 moles of solid sodium metal reacts with 2 moles of water molecule to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas.
Sodium metal is present in solid state, Water molecule is present in liquid state, Sodium hydroxide is present in aqueous state and hydrogen is present in gaseous state.
A chemical reaction between sodium (Na) and water (H₂O) yields hydrogen gas (H₂) and aqueous sodium hydroxide (NaOH). The reaction is highly exothermic as sodium hydroxide reacts vigorously with water, generating a lot of heat. The balanced chemical equation is '2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g)'.
Explanation:To answer your question, we need to first understand the reaction occurring in this process. This is a classic example of a reaction between a metal and water, specifically it's an alkali metal (sodium) reacting with water. Upon addition of solid sodium to liquid water, the sodium displaces hydrogen from the water, leading to the production of hydrogen gas and aqueous sodium hydroxide.
The balanced chemical equation is thus written as: 2Na(s) + 2H₂O(l) -> 2NaOH(aq) + H₂(g). In this equation, (s) represents solid, (l) represents liquid, (aq) represents aqueous or a material dissolved in water, and (g) represents gas.
Sodium hydroxide NaOH is a strong base and it reacts vigorously with water, generating a great deal of heat and forming very basic solutions. In fact, 40 grams of sodium hydroxide can dissolve in only 60 grams of water at 25 °C.
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A metal initially at 450.℃ and mass 2.00 g is dropped into a liquid with specific heat capacity 4.00 Jg^-1℃^-1, mass 10.0 g and initial temperature 40.0℃. The final temperature of the liquid is 50.0℃. What is the specific heat capacity of the metal. Express your answer in Jg^-1℃^-1.
Answer:
0.5
Explanation:
for more info see the attachment.
The specific heat capacity of the unknown metal is 0.5Jg-¹°C-¹
SPECIFIC HEAT CAPACITY:
The specific heat capacity of a metal can be calculated by using the following formula:Q = mc∆T
Where;
Q = quantity of heat absorbed or released (J)m = mass of substance (g)c = specific heat capacity (Jg°C)∆T = temperature (°C) However, in a calorimeter, the following expression applies:mc∆T (liquid) = - (mc∆T) metalMETAL:
m = 2.00g∆T = 50°C - 450°C = -400°Cc = ?LIQUID:
m = 10.0g∆T = 50°C - 40°C = 10°Cc = 4.00 Jg-¹℃-¹10 × 4 × 10 = -{2 × -400 × c}400 = -{-800c}400 = 800cc = 400 ÷ 800c = 0.5Jg-¹°C-¹Therefore, the specific heat capacity of the unknown metal is 0.5Jg-¹°C-¹.
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2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initial temperature at 1336K. The solid and liquid other. When the two reach thermal equilibrium will the mixture be entirely solid, or will they be in a mixed solid/liquid phase? Explain how you know. Draw two separate temp. vs. energy added diagrams to help you answer this question. can only exchange heat with each
Answer:
liquid phase
Explanation:
see the non-graph solution (used only formulas of the thermodynamic); the final temperature of the mixture is ~1139.701 (°K). The details are in the attachment.
Note:
[tex]c_l \ is \ for \ liquid \ gold;\\ c_s \ is \ for \ solid \ gold.[/tex]
Water's surface tension and heat storage capacity are accounted for by its ______. A) orbitals B) weight C) hydrogen bonds D) mass E) size
Answer:
The correct answer is C: hydrogen bonds
Explanation:
Water's surface tension and heat storage capacity are accounted for by its hydrogen bonds. Molecules of water are very strongly attracted to each other through hydrogen bonding.
Hydrogen bonds are constantly formed and broken in water molecules. Surface tension results from this hydrogen bonding which means that Water has a higher storage capacity for heat.
That is why at night, the Earth gets colder must faster than water.
Water takes time to slowly release heat to keep the atmospheric temperature moderate at night.
Final answer:
Hydrogen bonds are responsible for water's high surface tension and heat storage capacity. They result in strong cohesion between water molecules and a significant amount of energy is required to change water's state due to these bonds.
Explanation:
Water's surface tension and heat storage capacity are accounted for by its hydrogen bonds. The correct answer to the student's question is C) hydrogen bonds. These bonds occur because the hydrogen atoms in a water molecule have a slight positive charge, while the oxygen atom has a slight negative charge. Each water molecule can form four hydrogen bonds with surrounding molecules, leading to a high degree of cohesion. This cohesion is the main reason behind water's significant surface tension.
Additionally, because of the strength of these bonds, water has a high heat capacity and high heat of vaporization. It requires a large amount of heat energy to break the hydrogen bonds for water to change state, which explains why water is able to absorb and store a lot of heat without undergoing a significant increase in temperature.
Calculate the pH values of the following solutions: (Hint: See Special Topic I in the Study Guide and Solutions Manual.) a 1.0 M solution of acetic acid (pKa=4.76) a 0.1 M solution of protonated methylamine (pKa=10.7) a solution containing 0.3 M HCOOH and 0.1 M HCOO− (pKa of HCOOH=3.76)
Answer:
See Explanation
Explanation:
a. pH of 1M HOAc(aq)
HOAc ⇄ H⁺ + OAcˉ
C(eq) 1.0M x x
Ka = [H⁺][OAc⁻]/[HOAc] = x²/1.0M = 1.85x10⁻⁵
=> x = [H⁺] = SqrRt([HOAc]Ka) = SqrRt[(1M)(1.85x10ˉ⁵)] = 4.30x10ˉ³M
=> pH = -log[H⁺] = -log(4.30x10ˉ³) = 2.37
b. pH of 0.10M CH₃NH₃OH(aq)
CH₃NH₃OH => CH₃NH₃⁺ + OHˉ; Kb = 4.4x10ˉ⁴
C(eq) 0.10M x x
=> Kb = [CH₃NH₃⁺][OH⁻]/[CH₃NH₃] = x²/0.10M
=> x = [OHˉ] = SqrRt([CH₃NH₃OH]Kb) = SqrRt[(0.10M)(4.4x10ˉ⁴)] = 6.63x10ˉ³M
=> pOH = -log[OHˉ] = -log(6.63x10⁻³) = 2.18
=> pH = 14 – pOH = 14 – 2.18 = 11.82
c. pH of 0.30M HOAc/0.10M OAcˉ(aq)
HOAc ⇄ H⁺ + OAcˉ
C(eq) 0.30M x 0.10M
=> Ka = [H⁺][OAcˉ]/[HOAc] => [H⁺] = Ka[HOAc]/[OAcˉ]
= 1.85X10ˉ⁵(0.30M)/(0.10M) = 5.55X10ˉ⁵M
=> pH = -log[H⁺] = -log(5.55x10ˉ⁵) = 4.26
To calculate pH for acetic acid, protonated methylamine, and a formic acid/formate mixture, we utilize an ICE table for the weak acids and the Henderson-Hasselbalch equation for the buffer solution, considering their respective pKa values.
Explanation:To calculate the pH values of the given solutions, we apply the pH formula and utilize ICE (Initial, Change, Equilibrium) tables for weak acids and bases. Using the provided pKa values, we can determine the pH for each solution:
These calculations showcase the principles of acid-base equilibrium and buffer systems in solution chemistry.
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Consider the reaction. FeCl2(aq)+(NH4)2 SO4(aq)⟶FeSO4+2NH4Cl Identify the precipitate, or lack thereof, for the reaction. (A) FeSO4 (B) no precipitate (C) NH4Cl
Answer: The correct answer is Option B.
Explanation:
Precipitate is defined as the insoluble solid substance which is formed when two different aqueous solutions are mixed. It settles down at the bottom of the solution.
For the given chemical equation:
[tex]FeCl_2(aq.)+(NH_4)_2SO_4(aq.)\rightarrow FeSO_4(aq.)+2NH_4Cl(aq.)[/tex]
The products formed in the reaction are ferrous sulfate and ammonium chloride. Both the products are soluble in aqueous solutions. Thus, no precipitate will be formed in the reaction.
Hence, the correct answer is Option B.
How many molecules of XeF6 are formed from 12.9 L of F2 (at 298 K and 2.6 atm) according to 11) the following reaction? Assume that there is excess Xe. Xe(g) +3F2(g)→XeF6(g) A) 8.25 × 1023 molecules XeF6 B) 1.21 × 1023 molecules XeF6 C) 1.37 × 1023 molecules XeF6 D) 7.29 × 1023 molecules XeF6 E) 2.75 × 1023 molecules XeF6
Answer:
#Molecules XeF₆ = 2.75 x 10²³ molecules XeF₆.
Explanation:
Given … Excess Xe + 12.9L F₂ @298K & 2.6Atm => ? molecules XeF₆
1. Convert 12.9L 298K & 2.6Atm to STP conditions so 22.4L/mole can be used to determine moles of F₂ used.
=> V(F₂ @ STP) = 12.6L(273K/298K)(2.6Atm/1.0Atm) = 30.7L F₂ @ STP
2. Calculate moles of F₂ used
=> moles F₂ = 30.7L/22.4L/mole = 1.372 mole F₂ used
3. Calculate moles of XeF₆ produced from reaction ratios …
Xe + 3F₂ => XeF₆ => moles of XeF₆ = ⅓(moles F₂) = ⅓(1.372) moles XeF₆ = 0.4572 mole XeF₆
4. Calculate number molecules XeF₆ by multiplying by Avogadro’s Number (6.02 x 10²³ molecules/mole)
=> #Molecules XeF₆ = 0.4572mole(6.02 x 10²³ molecules/mole)
= 2.75 x 10²³ molecules XeF₆.
To determine the number of molecules of XeF₆,
First, we will determine the number of moles of F₂ present in the 12.9L of F₂
From the ideal gas equation
PV = nRT
Where P is the pressure
V is the volume
n is the amount of substance ( number of moles)
R is ideal gas constant
T is the temperature
From the question,
P = 2.6 atm
V = 12.9 L
R = 0.082057 L atm mol⁻¹ K⁻¹
T = 298 K
Putting these values into the equation
PV = nRT
2.6 × 12.9 = n × 0.082057 × 298
33.54 = n × 24.452986
∴ n = 33.54 ÷ 24.452986
n = 1.3716116 moles
The number of moles of F₂ present is 1.3716116 moles
From the given equation of reaction
Xe(g) +3F₂(g)→XeF₆(g)
1 mole of Xe reacts with 3 moles of F₂ to produce 1 mole of XeF₆
∴ 1.3716116 moles of F₂ will give (1.3716116/3) moles of XeF₆
Hence, the number of moles of XeF₆ produced = 0.457204 moles
To determine the number of molecules formed,
From the formula
Number of molecules = number of moles × Avogadro's number
∴ Number of molecules of XeF₆ formed = 0.457204 × 6.02214 ×10²³
Number of molecules of XeF₆ formed = 2.7533 × 10²³ molecules
Number of molecules of XeF₆ formed ≅ 2.75 × 10²³ molecules
Hence, the number of molecules of XeF₆ that are formed from 12.9 L of F₂ (at 298 K and 2.6 atm) is 2.75 × 10²³ molecules.
The correct option is E) 2.75 × 1023 molecules XeF6
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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic (As).(As). Therefore, prior to starting construction, the group decides to measure the current level of arsenic in the lake. If a 14.3 cm314.3 cm3 sample of lake water is found to have 159.5 ng As,159.5 ng As, what is the concentration of arsenic in the sample in parts per billion (ppb), assuming that the density of the lake water is 1.00 g/cm3?1.00 g/cm3?
Final answer:
To find the arsenic concentration in ppb, we convert the volume of the sample to liters and then use the formula Concentration (ppb) = (Mass of As (ng) / Volume of water (L)) x 1 billion. For a 14.3 cm3 sample containing 159.5 ng As, the arsenic concentration is 11,153.85 ppb.
Explanation:
To find the concentration of arsenic (As) in the lake water sample in parts per billion (ppb), we need to use the given information about the mass of arsenic and the volume of the water sample. Since we are provided with a 14.3 cm3 sample containing 159.5 ng As and assuming that the density of the lake water is 1.00 g/cm3, we can calculate the concentration.
First, we need to convert the volume of the lake water from cm3 to liters since 1 cm3 is equal to 1 mL and there are 1,000 mL in a liter. For the 14.3 cm3 sample, this converts to:
14.3 cm3 × (1 mL/1 cm3) = 14.3 mL
14.3 mL × (1 L/1,000 mL) = 0.0143 L
Next, we can calculate the concentration of arsenic in the water sample in ppb using the following equation:
Concentration (ppb) = (Mass of As (ng) / Volume of water (L)) × 1 billion
Filling in the values we have:
Concentration (ppb) = (159.5 ng / 0.0143 L) × 1,000,000,000 ng/billion
Concentration (ppb) = 11,153.85 ppb
The arsenic concentration in the sample is therefore 11,153.85 ppb.
A 21.82 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 21.33 grams of CO2 and 4.366 grams of H2O are produced. In a separate experiment, the molar mass is found to be 90.04 g/mol. Determine the empirical formula and the molecular formula of the organic compound.
Answer: The molecular formula for the given organic compound is [tex]C_2H_2O_4[/tex]
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of [tex]CO_2=21.33g[/tex]
Mass of [tex]H_2O=4.366g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 21.33 g of carbon dioxide, [tex]\frac{12}{44}\times 21.33=5.82g[/tex] of carbon will be contained.
For calculating the mass of hydrogen:In 18g of water, 2 g of hydrogen is contained.
So, in 4.366 g of water, [tex]\frac{2}{18}\times 4.366=0.485g[/tex] of hydrogen will be contained.
Mass of oxygen in the compound = (21.82) - (5.82 + 0.485) = 15.515 gTo 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{5.82g}{12g/mole}=0.485moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.485g}{1g/mole}=0.485moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{15.515g}{16g/mole}=0.969moles[/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 0.485 moles.
For Carbon = [tex]\frac{0.485}{0.485}=1[/tex]
For Hydrogen = [tex]\frac{0.485}{0.485}=1[/tex]
For Oxygen = [tex]\frac{0.969}{0.485}=1.99\approx 2[/tex]
Step 3: Taking the mole ratio as their subscripts.The ratio of C : H : O = 1 : 1 : 2
Hence, the empirical formula for the given compound is [tex]C_1H_{1}O_2=CHO_2[/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 = 90.04 g/mol
Mass of empirical formula = 45 g/mol
Putting values in above equation, we get:
[tex]n=\frac{90.04g/mol}{45g/mol}=2[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4[/tex]
Thus, the molecular formula for the given organic compound is [tex]C_2H_2O_4[/tex]
The friction loss term is always a positive value O True O False No
What will happen to the pH of a buffer if acid/base ratio is increased by 100?
plz respond within an hour :)
Answer:
It will decrease by 2 units.
Explanation:
The Henderson-Hasselbalch equation for a buffer is
pH = pKa + log(base/acid)
Let's assume your acid has pKa = 5.
(a) If the base: acid ratio is 1:1,
pH(1) = 5 + log(1/1) = 5 + log(1) = 5 + 0 = 5
(b) If the base: acid ratio is 1:100,
pH(2) = 5 + log(1/100) = 5 + log(0.01) = 5 - 2 = 3
(c) Difference
ΔpH = pH(2) - pH(1) = 5 - 3 = -2
If you increase the acid:base ratio to 100:1, the pH will decrease by two units.
At a certain temperature, Kc equals 1.4 × 102 for the reaction: 2 CO(g) + O2(g) ⇌ 2 CO2(g). If a 3.50-L flask contains 0.400 mol of CO2 and 0.100 mol of O2 at equilibrium, how many moles of CO are also present in the flask?
Answer:
[tex]\boxed{\text{0.200 mol}}[/tex]
Explanation:
The balanced equation is
2CO + O₂ ⇌ 2CO₂
Step 1. Calculate the known concentrations.
[tex]\text{[O$_{2}$]} = \dfrac{\text{0.100 mol}}{\text{3.50 L}} = \text{0.02857 mol/L}\\\\\text{[CO$_{2}$]} = \dfrac{\text{0.400 mol}}{\text{3.50 L}} = \text{0.1143 mol/L}[/tex]
Step 2. Calculate the concentration of CO
[tex]K_{\text{eq}} = \dfrac{\text{[CO$_{2}$]$^{2}$}}{\text{[CO]$^{2}$[O$_{2}$]}} = 1.4 \times10^{2}[/tex]
[tex]\begin{array}{rcl}\\\\\dfrac{0.1143^{2}}{\text{[CO]$^{2}$} \times 0.02857} & = & 1.4 \times10^{2}\\\\0.01306 & = & 4.000\text{[CO]$^{2}$}\\\\\text{[CO]$^{2}$} & = &\dfrac{0.01306}{4.000}\\\\\text{[CO]$^{2}$} & = & 0.003265\\\text{[CO]} & = & \mathbf{0.05714}\\\end{array}[/tex]
3. Calculate the moles of CO
n = 3.50 L × 0.057 14 mol·L⁻¹ = 0.200 mol
[tex]\text{At equilibrium, there are }\boxed{\textbf{0.200 mol}} \text{ of CO in the flask}[/tex]
Kc is the equilibrium constant and depicts the reactants or the product concentration. At equilibrium, the moles of the carbon monoxide is 0.200 moles.
What are moles?Moles is the ratio of the mass of the substance to that of the molar mass of the substance.
The balanced chemical reaction is given as,
[tex]\rm 2CO + O_{2} \rightleftharpoons 2CO_{2}[/tex]
The known concentration of oxygen is calculated as:
[tex]\begin{aligned}\rm [O_{2}] &= \dfrac{0.100\;\rm mol}{3.50\;\rm L}\\\\&= 0.0285\;\rm mol/L\end{aligned}[/tex]
The known concentration of carbon dioxide is calculated as:
[tex]\begin{aligned}\rm [CO_{2}] &= \dfrac{0.400\;\rm mol}{3.50\;\rm L}\\\\&= 0.1143\;\rm mol/L\end{aligned}[/tex]
The concentration of the carbon monoxide is calculated as:
[tex]\begin{aligned}\rm k_{eq} = \rm \dfrac{[CO_{2}]^{2}}{[CO]^{2}[O_{2}]} &= 1.4 \times 10^{2}\\\\\rm \dfrac{(0.1143)^{2}}{[CO]^{2}\times 0.0285} &= 1.4 \times 10^{2}\\\\\rm [CO]^{2}& = 0.00326\\\\&= 0.0571\end{aligned}[/tex]
Moles of the carbon monoxide is calculated as:
[tex]\begin{aligned}\rm n &= \rm mass \times concentration\\\\&= 3.50 \;\rm L \times 0.057 14 \;\rm mol\; L^{-1}\\\\&= 0.200\;\rm mol\end{aligned}[/tex]
Therefore, at equilibrium, the moles of the carbon monoxide is 0.200 mol.
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Iron oxide (FeO) has the rock salt crystal structure and a density of 5.70 g/cm3. The atomic weights of iron and oxygen are 55.85 g/mol and 16.00 g/mol, respectively. (a) Determine the unit cell edge length
The edge length of a unit cell of a face-centered cubic structure is equal to the cube root of the volume of the unit cell. Hence, the edge length of the unit cell of Iron oxide (FeO) is [tex]\rm 4.3745 \ nm[/tex].
The rock salt crystal structure is a face-centered cubic structure. This means that there are atoms at each corner of the cube, and there is also an atom in the center of each face of the cube.
To find the unit cell edge length:
n = number of atoms
= 4
Atomic weight of [tex]\rm Fe = A_f[/tex]
= 55.85 g/mol
Atomic weight of [tex]\rm O = A_0[/tex]
= 16 g/mol
Density [tex]\rm = 5.70 g/cm^3[/tex]
Avogadro's number (NA) [tex]= 6.023\times10^2^3[/tex]
Volume of crystal,
[tex]\rm V = n \left(\frac{A_f + A_0}{{density} \times NA \right)}[/tex]
[tex]\frac{4\left((55.85 + 16)}{5.70 \times 6.023 \times 10^{23}}\\\right)}=8.3714 \times 10^{-23} \text{ cm}^3[/tex]
Unit cell edge length,
[tex]\rm a = \left( V \right)^{1/3}\\\left\\=( 8.3714 \times 10^{-23} \text{ cm}^3 \right)^{1/3}\\\\=4.3745 \times 10^{-8} \text{ cm} \times 10^{7} \frac{\text{nm}}{\text{cm}}\\= 0.43745\ nm[/tex]
Thus, the unit cell edge length is [tex]\rm 0.43745\ nm[/tex].
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The unit cell edge length of iron oxide (FeO) can be calculated using its density and the formula for mass and volume of a unit cell in the rock salt crystal structure.
Explanation:The rock salt crystal structure is a type of cubic lattice structure. In this type of structure, the atoms are arranged in a face-centered cubic (FCC) unit cell. The unit cell edge length can be calculated using the formula:
Edge length = (4 * volume)^(1/3)
Given that the density of FeO is 5.70 g/cm3, the mass of the unit cell can be determined using the formula:
Mass = density * volume
By substituting the values of density and volume, the mass of the unit cell can be calculated. Finally, by rearranging the equation for mass to solve for volume, the unit cell edge length can be determined.
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Identify the base in this acid-base reaction:
NaOH + HCI - NaCl + H2O
o NaOH
O HCI
o Naci
• H20
Answer:
[tex]\boxed{\text{NaOH}}[/tex]
Explanation:
[tex]\rm \underbrace{\hbox{NaOH }}_{\hbox{base}} + \underbrace{\hbox{HCl}}_{\hbox{acid}} \longrightarrow \, NaCl + H$_{2}$O[/tex]
A hydroxide of a metal is a base.
[tex]\text{The base is }\boxed{\textbf{NaOH}}[/tex]
B is wrong. HCl is an acid.
C is wrong. NaCl is a salt.
D is wrong. Water is neutral.
Answer:
A. NaOH
A. Is The Correct Answer
B. HCl
C. NaCl
D. H2O
complete and balance the molecular equation for the reaction of aqueous copper(II) chloride, CuCl2,CuCl2, and aqueous potassium phosphate, K3PO4.K3PO4. Include physical states. molecular equation: CuCl2(aq)+K3PO4(aq)⟶CuCl2(aq)+K3PO4(aq)⟶ Enter the balanced net ionic equation for this reaction. Include physical states.
Answer : The net ionic equation will be,
[tex]3Cu^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)[/tex]
Explanation :
First we have to balance the chemical reaction.
The given balanced ionic equation will be,
[tex]3CuCl_2(aq)+2K_3PO_4(aq)\rightarrow Cu_3(PO_4)_2(s)+6KCl(aq)[/tex]
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The ionic equation in separated aqueous solution will be,
[tex]3Cu^{2+}(aq)+6Cl^-(aq)+6K^+(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)+6K^+(aq)+6Cl^-(aq)[/tex]
In this equation, [tex]K^+\text{ and }Cl^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]3Cu^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Cu_3(PO_4)_2(s)[/tex]
How do you calculate the boiling point of pure water at 653.7 torr?
Explanation:
It is known that atmospheric pressure is equal to 760 torr.
And, at atmospheric pressure that is, 760 torr the boiling point of pure water is 100 degree celsius.
So, calculate the boiling point of pure water at 653.7 torr as follows.
[tex]\frac{100^{o}C}{760 torr} \times 653.7 torr[/tex]
= [tex]0.131 \times 653.7 torr[/tex]
= [tex]86.01^{o}C[/tex]
Therefore, we can conclude that the boiling point of pure water at 653.7 torr is [tex]86.01^{o}C[/tex].
One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.00 liters against a constant external pressure of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore significant figures for this problem. (T = 300 K; 1 L·atm = 101.3 J)
Answer : The work done on the surroundings is, 709.1 Joules.
Explanation :
The formula used for isothermally irreversible expansion is :
[tex]w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)[/tex]
where,
w = work done
[tex]p_{ext}[/tex] = external pressure = 1.00 atm
[tex]V_1[/tex] = initial volume of gas = 1.00 L
[tex]V_2[/tex] = final volume of gas = 8.00 L
Now put all the given values in the above formula, we get :
[tex]w=-p_{ext}(V_2-V_1)[/tex]
[tex]w=-(1.00atm)\times (8.00-1.00)L[/tex]
[tex]w=-7L.atm=-7\times 101.3J=-709.1J[/tex]
The work done by the system on the surroundings are, 709.1 Joules. In this, the negative sign indicates the work is done by the system on the surroundings.
Therefore, the work done on the surroundings is, 709.1 Joules.
The work done by the gas on the surroundings when it expands is -709.1 J.
Explanation:The subject of this question is the work done by a gas during isothermal expansion. In physics, the work done by a gas is given by the formula W = -PΔV, where P is the pressure and ΔV is the change in volume. For this question, the external pressure P is 1.00 atm (which is equivalent to 101.3 J/L), the initial volume V₁ is 1.00 L, and the final volume V₂ is 8.00 L.
Thus, the change in volume ΔV is V₂ - V₁ = 8.00 L - 1.00 L = 7.00 L. Substituting into the formula for work, we find W = -(1.00 atm)(7.00 L) = -7.00 L⋅atm.
One last step needed is to convert from liters-atmospheres to joules. As given in the question, 1 L⋅atm is approximately 101.3 J, so W = -7.00 L⋅atm * 101.3 J/L⋅atm = -709.1 J. The negative sign indicates work was done by the gas on the surroundings.
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Which one of the following statements BEST explains what is meant by the “dual nature” of the electron? A : An electron may be either in a quasi-free state as in a metal or in a tightly-bound state deep within an atom. B : An electron plays a role in the production of both magnetic and electric fields. C : An electron can be transmitted in a beam as in a television or through a wire. D : An electron may act with either particle-like or wave-like characteristics. E : An electron can travel at very small velocities or be at rest. It’s motion adequately described by Newton’s laws of motion. The electron can also travel at speeds close to the speed of light; and its motion is described by relativistic laws of motion.
Answer:
D : An electron may act with either particle-like or wave-like characteristics.
Explanation:
This is the whole basis of the Schrödinger equation.
The other options are correct, but they do not state the dual nature of the electron.
What is the product of the reaction of 3-isopropylcyclodecanol with hydrobromic acid? a. 1-bromo-3-isopropylcyclodecane b. 1-bromo-2-isopropylcyclodecane c. 1-bromo-3-isopropylcyclodecene d. none of these choices
Answer: Correct answer is a - bromo-3-isopropylcyclodecane
6 C(s) + 3 H2(g) → 2 C6H6(l) Δ H = 49 kJ
TRUE or FALSE
Answer:
True => ΔH°f for C₆H₆ = 49 Kj/mole
Explanation:
See Thermodynamic Properties Table in appendix of most college level general chemistry texts. The values shown are for the standard heat of formation of substances at 25°C. The Standard Heat of Formation of a substance - by definition - is the amount of heat energy gained or lost on formation of the substance from its basic elements in their standard state. C₆H₆(l) is formed from Carbon and Hydrogen in their basic standard states. All elements in their basic standard states have ΔH°f values equal to zero Kj/mole.
Why are H2S, mercaptans, sulphur compounds and salt removed from the crude oils?
Answer & Explanation:
1- The sulfur present in the crude oil can poison the catalysts that are used in the refining of crude oil into the many useful products obtained from crude oil.
2- The crude oil, containing sulfur compounds, introduced into a car or truck engine, it will again interfere with the catalytic converter installed in the exhaust system of the vehicle.
3- if sulfur is present in fuel, when the fuel is burned it will be converted to SO₂ which is a known precursor to acid rain.
4- Sulfur compounds can lead to increased corrosion in distillation equipment at the high temperatures used.
The correct answer is that H2S, mercaptans, sulphur compounds, and salt are removed from crude oils to prevent corrosion, environmental pollution, and to meet product specifications.
H2S (hydrogen sulphide), mercaptans (thiols), and other sulphur compounds are removed from crude oil for several reasons:
1. Corrosion Prevention: H2S is a highly corrosive gas that can cause significant damage to pipelines, storage tanks, and refinery equipment. It can lead to the formation of sulphuric acid when it reacts with moisture, which accelerates corrosion. Removing H2S and other sulphur compounds helps to mitigate this risk.
2. Environmental Concerns: Sulphur compounds in fuels contribute to the formation of acid rain when burned, as they can react with moisture and oxygen in the atmosphere to form sulphuric acid. Additionally, H2S is toxic and can be harmful to human health and the environment if released into the atmosphere.
3. Product Specifications: Many refined products, such as gasoline, diesel, and jet fuel, have strict sulphur content specifications. Low sulphur fuels burn cleaner and are less harmful to the environment. The removal of sulphur compounds is necessary to meet these specifications.
4. Safety: H2S is also a toxic gas that can be lethal at high concentrations. Its removal is essential for the safety of workers and for the safe operation of refineries and transportation systems.
5. Catalyst Poisoning: Sulphur compounds can act as poisons for catalysts used in various refining processes, reducing their effectiveness and lifetime. Removing sulphur before processing increases the efficiency and longevity of these catalysts.
Salt is removed from crude oil primarily because it can cause corrosion and scaling in refinery equipment, which can lead to reduced efficiency and increased maintenance costs. Additionally, salt can contaminate the final products, affecting their quality and usability.
The process of removing these impurities typically involves a combination of physical and chemical treatments, such as distillation, adsorption, hydrodesulphurization, and caustic washing, depending on the nature and concentration of the contaminants.