Answer:
Part A
The volume of the gaseous product is [tex]V = 787L[/tex]
Part B
The volume of the the engine’s gaseous exhaust is [tex]V_e = 2178 \ L[/tex]
Explanation:
Part A
From the question we are told that
The temperature is [tex]T = 350^oC = 350 +273 =623K[/tex]
The pressure is [tex]P = 735 \ torr = \frac{735}{760} = 0.967\ atm[/tex]
The of [tex]C_8 H_{18} = 100.0g[/tex]
The chemical equation for this combustion is
[tex]2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}[/tex]
The number of moles of [tex]C_8 H_{18}[/tex] that reacted is mathematically represented as
[tex]n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }[/tex]
The molar mass of [tex]C_8 H_{18}[/tex] is constant value which is
[tex]M = 114.23 \ g/mole[/tex]
So [tex]n = \frac{100 }{114.23} }[/tex]
[tex]n = 0.8754 \ moles[/tex]
The gaseous product in the reaction is [tex]CO_2_{(g)}[/tex] and water vapour
Now from the reaction
2 moles of [tex]C_8 H_{18}[/tex] will react with 25 moles of [tex]O_2[/tex] to give (16 + 18) moles of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
So
1 mole of [tex]C_8 H_{18}[/tex] will react with 12.5 moles of [tex]O_2[/tex] to give 17 moles of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
This implies that
0.8754 moles of [tex]C_8 H_{18}[/tex] will react with (12.5 * 0.8754 ) moles of [tex]O_2[/tex] to give (17 * 0.8754) of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
So the no of moles of gaseous product is
[tex]N_g = 17 * 0.8754[/tex]
[tex]N_g = 14.88 \ moles[/tex]
From the ideal gas law
[tex]PV = N_gRT[/tex]
making V the subject
[tex]V = \frac{N_gRT}{P}[/tex]
Where R is the gas constant with a value [tex]R = 0.08206 \ L\cdot atm /K \cdot mole[/tex]
Substituting values
[tex]V = \frac{14.88* 0.08206 *623}{0.967}[/tex]
[tex]V = 787L[/tex]
Part B
From the reaction the number of moles of oxygen that reacted is
[tex]N_o = 0.8754 * 12.5[/tex]
[tex]N_o = 10.94 \ moles[/tex]
The volume is
[tex]V_o = \frac{10.94 * 0.08206 *623}{0.967}[/tex]
[tex]V_o = 579 \ L[/tex]
No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as
[tex]V_e = V_o * \frac{0.79}{0.21}[/tex]
Substituting values
[tex]V_e = 579 * \frac{0.79}{0.21}[/tex]
[tex]V_e = 2178 \ L[/tex]
The total volume of gaseous products produced from the combustion of 100g C8H18 is approximately 382 L. By considering the composition of air and that N2 and Ar do not react, you can calculate the total volume of the gaseous exhaust.
Upon seeing this question, the first step is to write the balanced chemical equation for the combustion of C8H18, which is ubiquitous in gasoline:
C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O
This looks complex, but the Ideal Gas Law simplifies everything. According to the stoichiometric ratio, 1 mole of C8H18 will produce 8 moles of CO2 and 9 moles of H2O when burnt, totalling 17 moles of gas. The molar mass of C8H18 is around 114 g/mol, so 100g is approximately 0.877 moles.
Therefore, upon combustion, 0.877 * 17 = 14.9 moles of gas are produced. Applying the Ideal Gas Law (PV=nRT), keeping in mind that 350°C is 623K and 1 atm is 760 torr, we have: (nR(623K)/735torr) = V. Plugging the numbers and making sure to keep the units consistent, we get V = 14.9 x 0.0821 x 623 / 0.967 = approximately 382 L.
Coming to part (b), considering the composition of air and assuming no N2 or Ar reacts, you find that for every L of O2 burned, 1L of air is consumed. Adding the total volume of gas from part (a) and the extra nitrogen and argon, you find the total volume of the engine's gaseous exhaust.
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what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C
Answer:
The volume of the air is 0.662 L
Explanation:
Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:
[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]
In this case:
V1= 0.730 LT1= 28 °C= 301 °K (0°C= 273°K)V2= ?T2= 0°C= 273 °KReplacing:
[tex]\frac{0.730 L}{301K}=\frac{V2}{273K}[/tex]
Solving:
[tex]V2=273K*\frac{0.730L}{301K}[/tex]
V2=0.662 L
The volume of the air is 0.662 L
Footballs used by the NFL are inflated to 12.5 psi. 11 of 12 footballs measured 10.5 psi after the playoff game when the temperature was 5°C. At what temperature were the balls inflated to explain this pressure drop?
Final answer:
After calculating T1, we will have our answer in Kelvin, which can be converted back to degrees Celsius by subtracting 273.15.
Explanation:
To determine at what temperature the NFL footballs were inflated before the pressure dropped to 10.5 psi from 12.5 psi, we can use the ideal gas law and assume that the volume of the footballs remains constant (since they're made of materials that don't expand or contract much with temperature). The ideal gas law implies a direct relationship between pressure and temperature, meaning that if the temperature drops, the pressure will also drop, provided the number of moles of gas and the volume remain constant.
Starting with the initial condition, where P1 = 12.5 psi and the final condition where P2 = 10.5 psi at T2 = 5°C, we can find the initial temperature (T1) using the following relation from Charles's Law:
P1/T1 = P2/T2
However, to use this formula, we need to convert the temperatures into Kelvin:
T1 = (12.5 psi × 278.15 K) / 10.5 psi
After calculating T1, we will have our answer in Kelvin, which can be converted back to degrees Celsius by subtracting 273.15.
Question 7 When of benzamide are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezing point of pure . On the other hand, when of iron(III) chloride are dissolved in the same mass of , the freezing point of the solution is lower than the freezing point of pure . Calculate the van't Hoff factor for iron(III) chloride in . Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.
Answer:
See explaination
Explanation:
moles of benzamide = mass / Molar mass of it = 70.4g / ( 121.14g/mol) = 0.58 mol
Molality = moles of solute ( benzamide) / ( solvent mass in kg) = 0.58 mol / ( 0.85kg) = 0.6837
we have formula dT = i x Kf x m , where dT = change in freezing point = 2.7C , i = vantoff factor = 1 for non dissociable solutes , Kf = freezing oint constant of solvent , m = 0.6837
hence 2.7C = 1 x Kf x 0.6837m
Kf = 3.949 C/m
we use this Kf value for calculating i for NH4Cl , where moles of NH4Cl = ( 70.4g/53.491g/mol) =1.316 mol
molality = ( 1.316mol) / ( 0.85kg) = 1.5484 , dT = 9.9
hence 9.9 = i x 3.949C/m x 1.5484 m
i = 1.62
In the reaction Pb + Cu2+ → Pb2+ + Cu, the Cu2+
A. gains protons
B. loses protons
C. gains electrons
D. loses electrons
Answer:
The answer is C.
Explanation:
Cu 2+ is reduced to Cu by gaining 2 electrons, so reduction occurs .
Proton cannot be gain or lose .
How is energy or particles released from a nucleus?
Answer: The decay energy is the energy released by a radioactive decay. Radioactive decay is the process in which an unstable atomic nucleus loses energy by emitting ionizing particles and radiation.
Question 63 of 70 1.5 Points When 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1-butene is produced. When potassium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base
Answer: between 45% and 70%
Explanation:
taking this step by step, let us analyze this question carefully.
From the original image in the question, as can be seen in the second uploaded image, we observe that;
potassium hydroxide having being less bulky produces about 45% of 2-methyl-butane whereas potassium tert-butoxide having being bulky produces 70% of this product.
Given the potassium propoxide as the base and intermediate to that of potassium hydroxide and potassium tert-butoxide, it produces 2-methyl-1-butane in between 45% to 70%.
Cheers i hope this helped!!!!
Final answer:
Without additional information, it is not possible to determine the exact percentage of 2-methyl-1-butene produced using potassium propoxide as the base for the reaction of 2-bromo-2-methylbutane.
Explanation:
When 2-bromo-2-methylbutane is treated with a base to form a mixture of 2-methyl-2-butene and 2-methyl-1-butene, the specific base used influences the proportion of each product. The use of potassium hydroxide results in 45% yield of 2-methyl-1-butene, while potassium tert-butoxide increases this yield to 70%. If potassium propoxide were used as the base, without further experimental data or specific trends indicated in the question, it is not possible to accurately predict the exact percentage of 2-methyl-1-butene in the mixture. Generally, the bulkiness of the base can influence the product distribution in elimination reactions due to steric hindrance, but without additional information about the reaction conditions or the specific mechanism with potassium propoxide, we can only hypothesize about the potential outcome based on trends we see with other bases.
Be sure to answer all parts. Which of the following ions possess a dipole moment? (a) ClF2+ (b) ClF2− has a dipole moment has no dipole moment cannot be determined has a dipole moment has no dipole moment cannot be determined (c) IF4+ has a dipole moment has no dipole moment cannot be determined (d) IF4− has a dipole moment has no dipole moment cannot be determined
Answer:
(A). Has Dipole moment, (B). No dipole moment, (C). Has dipole moment, (D). It has no Dipole moment.
Explanation:
In order to determine if a specie has dipole moment or not there is the need for us to draw the Lewis structure, please check attached file for the Lewis structures of each species.
(A). ClF2+: it HAS dipole moment because of Asymmetry. Note that the Fluorine atoms are more Electronegative that the chlorine atom.
(B). ClF2− : Normally, it should have a dipole moment because Fluorine atoms are more Electronegative that the chlorine atom on each side BUT due to its geometry (according to VSEPR theory) which makes it to have a linear geometry, the dipole moment will cancel out ,hence, NO DIPOLE MOMENT.
(C). IF4+ : it HAS dipole moment. Although, the dipole moment of two out of the four Fluorine will cancel out the other two will not cancel out making it to have Dipole moment.
(D). IF4-: it has no dipole moment because the four bonds are in opposite directions
ClF2+ and IF4+ has dipole moment whereas ClF2− and IF4− has no dipole moment.
ClF2+ has dipole moment because of Asymmetry i.e. the Fluorine atoms has higher electronegativity value that the chlorine atom.
ClF2− has no dipole moment due to its geometry i.e. it has a linear geometry, the dipole moment will cancel out and no dipole moment occur.
IF4+ has dipole moment because the dipole moment of two out of the four Fluorine will cancel out but the other two will not cancel out making it to have Dipole moment.
IF4- has no dipole moment because the four bonds are in opposite directions.
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Consider an electrochemical cell based on the spontaneous reaction 2AgCl(s) + Zn(s) → 2Ag(s) + 2Cl– + Zn2+. If the zinc ion concentration is kept constant at 1 M, and the chlorine ion concentration is decreased from 1 M to 0.001 M, the cell voltage should A) increase by 0.06 V. D) decrease by 0.18 V. B) increase by 0.18 V. E) increase by 0.35 V. C) decrease by 0.06 V.
Answer:
B) increase by 0.18 V
Explanation:
The given chemical spontaneous reaction is :
[tex]2 AgCl_{(s)} + Zn _{(s)} \to 2Ag (s) + 2Cl^- _{(aq)} + Zn ^{2+} _{(aq)}[/tex]
By applying Nernst Equation:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
here ;
n = number of electrons transferred in the reaction
n =2
[tex]E^0 = E^0_{cathode } - E^0_{anode}[/tex]
[tex]E^0 = E^0_{Ag^+/Ag } - E^0_{Zn^{2+}/Zn}[/tex]
[tex]E^0 =+0.80 \ V -(-0.76 \ V)[/tex]
[tex]E^0 =1.56 \ V[/tex]
When it happens to occur that the concentration of chlorine (aq) and Zn²⁺ (aq) is 1 M ;
[tex]E^0_{cell} = E^0[/tex] is as follows:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1) \ \ \ \ \ \ \ ( where \ log (1) = 0)[/tex]
[tex]E_{cell} = 1.56 \ V[/tex]
Now; the [tex]E_{cell}[/tex] value in the decreased concentration of chlorine (aq) ion is calculated as:
[tex]E_{cell} = E^0 - \frac{0.059}{n} log [\frac{product}{reactant}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log [\frac{[Zn^{2+}]}{[Cl^-]^2}][/tex]
[tex]E_{cell} = 1.56 - \frac{0.059}{n} log (1*0.001^2)[/tex]
[tex]E_{cell} = +1.737 \ V[/tex]
Hence; the change in voltage = [tex]E_{cell} - E^0[/tex]
= 1.737 - 1.56
= 0.177 V
≅ 0.18 V
We therefore conclude that: since the [tex]E^0_{cell}[/tex] value after the decreased concentration of Chlorine is greater than the [tex]E^0[/tex] before the change; then there is increase in the value by 0.18 V
50 g of NaCl is mixed with water to make .5 L of salt water. Which is the solute?
Answer:
The correct answer to the following question will be "NaCl".
Explanation:
A solute is being known as the soluble material within a liquid or a solution. The solvent is available in larger quantities than those of the solute among fluid. Concentration seems to be a calculation including its quantity of solvent contained in such a liquid solution, regard to something like the solvent volume.So that NaCl is the right answer.
A calorimeter weighing 123.7g has a quantity of 20C water added to it. The combined mass of the calorimeter + cold water is 198.3g. 61g of water is heated to 60C and is poured into the calorimeter. The temperature of the mixed cold and hot water and calorimeter is 38.5C. What is the calorimeter constant?
Answer:
The calorimeter constant would be 567.62 J/C
Explanation:
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What is the molarity of a hydrochloric acid solution, HCl (aq), if 30.00 mL of the solution is required to completely react with 25.00 mL of a 0.200 M magnesium hydroxide solution, Mg(OH)2 (aq)? Be sure to write out the balanced chemical reaction between these two compounds.
Answer:
0.333 M
Explanation:
Step 1: Write the balanced equation
2 HCl(aq) + Mg(OH)₂(aq) = MgCl₂(aq) + 2 H₂O(l)
Step 2: Calculate the reacting moles of Mg(OH)₂
25.00 mL of a 0.200 M magnesium hydroxide react. The reacting moles of Mg(OH)₂ are:
[tex]25.00 \times 10^{-3} L \times \frac{0.200mol}{L} = 5.00 \times 10^{-3} mol[/tex]
Step 3: Calculate the reacting moles of HCl
The molar ratio of HCl(aq) to Mg(OH)₂ is 2:1. The reacting moles of HCl are:
[tex]5.00 \times 10^{-3} molMg(OH)_2 \times \frac{2mol}{1molMg(OH)_2} =1.00 \times 10^{-2}molHCl[/tex]
Step 4: Calculate the molarity of HCl
[tex]\frac{1.00 \times 10^{-2}mol}{30.00 \times 10^{-3}L} = 0.333 M[/tex]
In an isothermal gas chromatography experiment using an ECD detector, 1.69 nmols of nchlorohexane, C6H13Cl, was added as an internal standard to an unknown amount of nchlorodecane, C10H21Cl. The area of the first peak to elute was 32434 units and the area of the second peak to elute was 2022 units. Calculate the amount of chlorodecane in the unknown.
Answer:
The amount of Chlorodecane in the unknown is 0.105nmols
Explanation:
a) Since the GC is in an isothermal state, Chlorohexane C6H13Cl (1.69 nmols) because of its lower boiling point will elute first and Chlorodecane C12H21Cl will elute second.
The area of the first peak corresponding to Chlorohexane is 32434 units.
The area of the second peak corresponding to chlorodecane is 2022 units.
Since the response factor of the compound is not given in question and considering the response factor is same for both the compounds, the answer will be as follow:
1.69 nmols of Chlorohexane gives 32434 units
How much of chlorodecane gives 2022 units
By cross multiplication;
Moles of Chlorodecane = 2022*1.69/32434
=0.105nmols
The P:OP:O ratio is the amount of inorganic phosphate incorporated into ATP per atom of oxygen consumed. It represents the coupling of phosphorylation and oxidation. It takes into account the number of protons translocated per electron pair and the ratio of the number of protons needed to synthesize and transport ATP. Determine the relationship of the P:OP:O ratio to the ratio of the number of protons translocated per electron pair and the ratio of the number of protons needed to synthesize ATP and transport it to the cytoplasm.
Answer:
For NADH; P:O = 2.5
For FADH ₂; P : O = 1.5
Explanation:
The P:O (phosphate:oxygen) ratio represents the amount of inorganic phosphate, Pi used per atom of oxygen consume to synthesize ATP.
The Chemiosmotic theory predicts H⁺:O and H⁺:ATP ratios. Experimentally these appear to be 10 and 4 respectively when NADH is the substrate, equivalent to a P:O ratio of 2.5, and 6 and 4 respectively for FAD-linked substrates (e.g. succinate), equivalent to a P:O ratio of 1.5.
1. Electron flow from NADH to O₂ pumps protons at three sites to yield 3 ATP (P:O = 2.5)
For NADH: 10 H ⁺ translocated/O (2e -)
ATP/2e - = (10 H⁺/ 4 H +) = 2.5
2. Succinate (via FADH2) bypasses site 1 giving 2 ATP (P : O = 1.5)
For FADH ₂= 6 H ⁺/O(2e - )
ATP/2e - = (6 H +/ 4 H +) = 1.5
20.0 g of Nitrogen is produced when oxygen gas reacts with NO gas. 29.8 L of oxygen is required at STP to produce this 20.0 g of NO. A.yes B.false
Answer:
There will be produced 30.64 grams of nitrogen dioxide
The statement is false
Explanation:
Step 1: Data given
MAss of nitrogen produced = 20.0 grams
Molar mass of N2 = 28.0 g/mol
Mass of NO = 20.0 grams
Molar mass of NO = 30.01 g/mol
Volume of O2 at STP = 29.8 L
Step 2: the balanced equation
2NO + O2 → 2N02
Step 3: Calculate moles NO
Moles NO = mass NO / molar mas NO
Moles NO = 20.0 grams / 30.01 g/mol
Moles NO = 0.666 moles
Step 4: Calculate moles O2
1 mol O2 at STP = 22.4 L
29.8 L = 29.8/22.4 = 1.33 moles
Step 5: Calculate the limiting reactant
For 2 moles NO we need 1 mol O2 to produce 2 moles NO2
NO is the limiting reactant. IT will completely be consumed (0.666 moles).
O2 is in excess. There will react 0.666/2 = 0.333 moles O2
There will remain 1.333 - 0.333 = 0.997 moles O2
Step 6: Calculate moles NO2
For 2 moles NO we need 1 mol O2 to produce 2 moles NO2
For 0.666 moles NO we'll have 0.666 moles NO2
Step 7: Calculate mass NO2
Mass NO2 = moles NO2 * molar mass NO2
Mass NO2 = 0.666 moles * 46.0 g/mol
Mass NO2 = 30.64 grams
There will be produced 30.64 grams of nitrogen dioxide
The statement is false
In addition to displacing halide ions, the acetylide ion also adds to carbonyl groups. 2-Methyl-3-butyn-2-ol (MBI) is an acetylenic alcohol used in the manufacture of products for the agrochemical and specialty chemical industry. It can be synthesized by the addition of acetylene to acetone to form the alkoxide ion and, as a second step, protonation of the alkoxide ion to produce the alcohol. Complete the mechanism for 2-methyl-3-butyn-2-ol production by drawing in the products of each step and the missing curved arrows. Sodium amide deprotonates the terminal alkyne to form sodium ethynide. Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges where needed. Electron flow arrows should start on an atom or a bond and should end on an atom, bond, or location where a new bond should be create
Answer:
Explanation:
check below for explanation.
A 5.00L tank at −2.4°C is filled with 16.7g of sulfur hexafluoride gas and 8.88g of dinitrogen monoxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction of each gas. Be sure each of your answer entries has the correct number of significant digits. gas mole fraction sulfur hexafluoride dinitrogen monoxide
Answer:
X(SF₆) = 0.361
X(N₂O) = 0.639
Explanation:
Step 1: Calculate the moles of each gas
We use the following expression.
n = m / M
where,
n: molesm: massM: molar massSF₆: 16.7 g / 146.06 g/mol = 0.114 mol
N₂O: 8.88 g / 44.01 g/mol = 0.202 mol
The total number of moles is 0.114 mol + 0.202 mol = 0.316 mol
Step 2: Calculate the mole fraction of each gas
We use the following expression.
X = moles of gas / total number of moles
X(SF₆) = 0.114 mol / 0.316 mol = 0.361
X(N₂O) = 0.202 mol / 0.316 mol = 0.639
A container of gas is heated from 250 K to 303 K. What is the new pressure if the initial pressure is 880 kPa? (Assume constant volume).
Answer:
The new pressure at constant volume is 1066.56 kPa
Explanation:
Assuming constant volume, the pressure is diectly proportional to the temperature of a gas.
Mathematically, P1/T1 = P2 /T2
P1 = 880 kPA= 880 *10^3 Pa
T1 = 250 K
T2 = 303 K
P2 =?
Substituting for P2
P2 = P1 T2/ T1
P2 = 880 kPa * 303 / 250
P2 = 266,640 kPa/ 250
P2 = 1066.56 kPa.
The new pressure of the gas is 1066.56 kPa
At constant volume, if the temperature of the container is heated to the given value, the pressure of the gas increases to 1066.5kPa.
What is Gay-Lussac's law?Gay-Lussac's law states that the pressure exerted by a given quantity of gas varies directly with the absolute temperature of the gas.
It is expressed as;
P₁/T₁ = P₂/T₂
Given the data in the question;
Initial pressure P₁ = 880kPa = ( 880 / 101.325)atm = 8.68492atmInitial temperature T₁ = 250KInitial temperature T₂ = 303KFinal pressure P₂ = ?P₁/T₁ = P₂/T₂
P₁T₂ = P₂T₁
P₂ = P₁T₂ / T₁
P₂ = (8.68492atm × 303K) / 250K
P₂ = (2631.53atmK) / 250K
P₂ = 10.526atm
P₂ = (10.526 × 101.325)kPa
P₂ = 1066.5kPa
Therefore, at constant volume, if the temperature of the container is heated to the given value, the pressure of the gas increases to 1066.5kPa.
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The addition of 9.54 kJ of heat is required to raise the temperature of 225.0 g of a liquid hydrocarbon from 20.5 degree C to 45.0 degree C. What is the heat capacity of this hydrocarbon?
The heat capacity of the hydrocarbon is 0.389 kJ/°C.
Explanation:To find the heat capacity of the hydrocarbon, we can use the formula:
Heat Capacity = q / ∆T
where q is the heat added and ∆T is the change in temperature.
Given that q = 9.54 kJ and ∆T = 45.0 °C - 20.5 °C = 24.5 °C, we can plug these values into the formula:
Heat Capacity = 9.54 kJ / 24.5 °C = 0.389 kJ/°C
Therefore, the heat capacity of the hydrocarbon is 0.389 kJ/°C.
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A coordination complex is made up of _________________________________________. (Choose one) Selected Answer: IncorrectC. a transition metal center and alkaline earth metal ligands surrounding it. Answers: CorrectA. a Lewis acidic metal center with Lewis basic ligands surrounding it. B. an Arrhenius basic metal center with Lewis acidic ligands surrounding it. C. a transition metal center and alkaline earth metal ligands surrounding it. D. a Lewis basic metal center with Lewis acidic ligands surrounding it.
Answer:
a Lewis acidic metal center with Lewis basic ligands surrounding it.
Explanation:
A lewis acid is a chemical specie capable of accepting a lone pair of electrons while a lewis base is a chemical specie capable of donating a lone pair of electrons.
In a coordination compound, a lewis acid(central metal atom/ion) accepts electron pairs from lewis bases (ligands). Any chemical specie called a ligand must possess at least one electron pair available for coordinate covalent bonding.
Coordination compounds are usually formed between transition metal atoms/ions and neutral molecules having lone pairs of electrons or anionc species .
The correct option is: A. A coordination complex consists of a Lewis acidic metal center with Lewis basic ligands surrounding it.
A coordination complex features a central metal atom or ion, often a transition metal, functioning as a Lewis acid due to its ability to accept electron pairs.
Surrounding this central metal are ligands, which are molecules or ions that act as Lewis bases by donating electron pairs to the metal. This donor-acceptor interaction leads to the formation of coordinate covalent bonds, creating a stable structure.
The unique properties of the metal and ligands, as well as their interactions, shape the geometry and characteristics of the complex. These properties are crucial for various chemical reactions and biological processes, highlighting the importance of coordination complexes in science.
Complete Question: -
"A coordination complex is made up of ____________."
Then the options should be listed below:
A. a Lewis acidic metal center with Lewis basic ligands surrounding it.
B. an Arrhenius basic metal center with Lewis acidic ligands surrounding it.
C. a transition metal center and alkaline earth metal ligands surrounding it.
D. a Lewis basic metal center with Lewis acidic ligands surrounding it.
a student prepares a dilute solution of sodium hydroxidem, NaOH (aq), starting with 6 M sodium hydroxide. She then titrates a 1.372 g samle of KHP with the dilute sodium hydroxide solution, NaOH (aq), calculate the molar concentration of the sodium hydroxide solution, NAoH (aq)
Answer:
M = 0.3077 M
Explanation:
As I said in the comments, you are missing the required volume of the base to react with the KHP. I found this on another site, and the volume it used was 21.84 mL.
Now, KHP is a compound often used to standarize NaOH or KOH solutions. This is because it contains a mole ratio of 1:1 with the base, so it's pretty easy to use and standarize any base.
Now, as we are using an acid base titration, the general expression to use when a acid base titration reach the equivalence point would be:
n₁ = n₂ (1)
This, of course, if the mole ratio is 1:1. In the case of KHP and NaOH it is.
Now, we also know that moles can be expressed like this:
n = M * V (2)
And according to this, we are given the volume of base and the required mass of KHP. So, if we want to know the concentration of the base, we need to get the moles of the KHP, because in the equivalence point, these moles are the same moles of base.
The reported molar mass of KHP is 204.22 g/mol, so the moles are:
n = 1.372 / 204.22 = 6.72x10⁻³ moles
Now, we will use expression (2) to get the concentration of the diluted base:
n = M * V
M = n / V
M = 6.72x10⁻³ / 0.02184
M NaOH = 0.3077 M
This is the concentration of the dilute solution of NaOH
The molar concentration of sodium hydroxide ( NaOH ) is ; 0.3077 M
Given data :
Volume of base = 21.84 mL = 0.02184 L ( missing data )
Molar mass of KHP = 204.22 g/mol
mass of KHP = 1.372 g
At equivalence point
n1 = n2 ( ratio of KHP to NaOH )
note : n = M * V ---- ( 1 )
First step : calculate the number of moles of KHP
n = mass / molar mass
n = 1.372 / 204.22 = 6.72 * 10⁻³ moles
Determine the molar concentration of NaOH
From equation ( 2 )
M ( molar concentration ) = n / V
= 6.72 * 10⁻³ moles / 0.02184 L
= 0.3077 M
Hence we can conclude that The molar concentration of sodium hydroxide ( NaOH ) is 0.3077 M
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Make the appropriate conclusion. Choose the correct answer below. A. Reject Upper H 0. There is insufficient evidence to indicate blood lactate level is linearly related to perceived recovery at alphaequals0.10. B. Reject Upper H 0. There is sufficient evidence to indicate blood lactate level is linearly related to perceived recovery at alphaequals0.10. C. Do not reject Upper H 0. There is sufficient evidence to indicate blood lactate level is linearly related to perceived recovery at alphaequals0.10. D. Do not reject Upper H 0. There is insufficient evidence to indicate blood lactate level is linearly related to perceived recovery at alphaequals0.10.
Answer:
A. Reject Upper H 0. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the alphaequals0.10 level of significance.
Explanation:
H0: p1 = p2
Ha: p1 > p2
pcap = (101 + 72)/(489 + 625) = 0.1553
SE = sqrt(0.1553*(1-0.1553)*(1/489 + 1/625)) = 0.0219
Test statistic,
z = (101/489 - 72/625)/0.0219 = 4.1710
p-value = 0.0000
Reject H0
A. Reject Upper H 0. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced drowsiness as a side effect than subjects in group 2 at the alphaequals0.10 level of significance.
Acid chlorides can be converted into primary amines with the loss of a single carbon atom via the Curtius rearrangement. The reaction involves treating an acid chloride with sodium azide to form an acyl azide. The acyl azide then loses N2 to give an isocyanate, which is hydrolyzed to release CO2 and the primary amine. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Answer:
Find the given attachment.
What is my percent yield if I calculated that a reaction would yield 20.0 g of iron (II) chloride, but I only managed to get 4.0 g of iron (II) chloride in the laboratory?
Answer:
Theoretical yield is 20%.
Explanation:
Percent yield = [(actual yield)/(theoretical yield)][tex]\times 100[/tex] %
Theoretical yield is calculated amount of product by using stoichiometric ratio between reactant and product.
Actual yield is the amount of product which is experimentally obtained.
Here theoretical yield is 20.0 g and actual yield is 4.0 g.
So, percent yield = [tex]\frac{4.0g}{20.0}\times 100[/tex] % = 20%
Collagen is made of fibers and is found in tendons and cartilage. Collagen
contributes to meat toughness and is, therefore, an example of which type
of protein?
Answer:
Collagen is a "fibrous proteins" present in the extracellular matrix.
Explanation:
Protein are made up of hundreds or thousands of smaller units called amino acids and is made of carbon, hydrogen, oxygen, nitrogen and sometimes sulfur and is found in many foods. Proteins are organic molecules found in living organisms.
There are three types of proteins; fibrous, globular, and membrane.
Collagen which is a fibrous proteins, form muscle fiber, tendons, connective tissue and bone.
Collagen are naturally occurring proteins that consist of single molecules made up of amino-acids, which are in turn built of carbon, oxygen and hydrogen. It is mostly found in fibrous tissues such as tendons, ligaments, and skin.
The synthesis of collagen occurs in two stages: intracellular and extracellular.
Collagen is an abundant connective tissues such as cartilage, tendons, bones, and ligaments. Collagen is a contributing factor to variation in meat tenderness and texture.
a. What is the Gibbs free energy for this reaction at 298 K?
2NH3(g) N2(g) + 3H2(g)
ΔH = 91.8 kJ
ΔS = 0.1987 kJ/K
Is the reaction spontaneous at 298 K (room temperature)? (3 points)
Answer:
32.6 Jmol-1
Explanation:
Now we have to use the formula
∆G= ∆H-T∆S where
∆G= change in free energy (the unknown)
∆H= enthalpy change = 91.8KJ
∆S= entropy change= 0.1987 kJ/K
T= temperature= 298K
∆G= 91.8 - (298× 0.1987)
∆G= 91.8 - 59.2126
∆G= 32.6 Jmol-1
Answer:
the reaction is spontaneous
Explanation:
Describe how the metal probably increases the reaction rate, identify whether this an example of homogeneous or heterogeneous catalysts, and explain how you know.
Explanation:
The metal probably helps to speed up the reaction rate a/c to collision theory, reactant molecules must collide with a reasonable direction by either weakening bonds in reactant molecules to make them extra reactive or by attaching reactant molecules in the exact direction to react.
This is known as an example of heterogeneous catalysis because the catalyst is solid and the reactants are liquid or gases mixture. In this catalysis, the catalyst is present in a different phase compare to the reactants.
Answer:
The metal probably increases reaction rate by either holding reactant molecules in the correct orientation to react or by weakening or breaking bonds in reactant molecules to make them more reactive.This is an example of heterogeneous catalysis. It is heterogeneous catalysis because the catalyst is a solid and the reactants are gases. In heterogeneous catalysis, the catalyst is in a different phase than the reactants.
Explanation:
Determine the number of moles of H in each sample.
1). 8.48 mole NH3
2).0.111 mole N2H4
3). 36.0 mole C10 H22
Answer:
1) 25.44 moles H
2) 0.444 moles H
3) 792 moles H
Explanation:
Step 1: Number of moles H in 8.48 moles NH3
In 1 mol NH3 we have 3 moles H
For 8.48 moles NH3 we have 3*8.48 = 25.44 moles H
Step 2: Number of moles H in 0.111 moles N2H4
For 1 mol N2H4 we have 4 moles H
For 0.111 moles N2H4 we have 4*0.111 = 0.444 moles H
Step 3: Number of moles H in 36.0 moles C1OH22
For 1 mol C10H22 we have 22 moles of H
For 36.0 moles C1OH22 we have 22*36.0 = 792 moles H
2. What is the difference between a formula unit for an ionic compound and a molecular formula
for a molecule?
Final answer:
The formula unit for an ionic compound represents the simplest ratio of positive and negative ions that result in a neutral compound, typically involving metals and nonmetals. A molecular formula for a molecule indicates the actual number of atoms of each element in a molecule, formed by covalent bonds between nonmetals. The periodic table helps distinguish between the two.
Explanation:
The difference between a formula unit for an ionic compound and a molecular formula for a molecule lies in the types of elements involved and the nature of their bonding. A formula unit refers to the simplest, neutral collection of ions in an ionic compound, which is comprised of metals and nonmetals. The periodic table can be used to determine which elements fall into these categories. On the other hand, a molecular formula represents the actual number and type of atoms in a molecule, which are bonded covalently and generally involve nonmetal elements.
For ionic compounds, the formula mass is used, which is the sum of atomic masses of all the elements in the empirical formula (simplest ratio of ions), with each multiplied by its corresponding subscript. This contrasts with a molecular compound's molecular mass, which is computed using the molecular formula, representing the total mass of all atoms in the actual molecule.
To summarize, the key distinction is that molecular formulas represent covalently-bonded nonmetals and give the specific numbers and types of atoms, whereas formula units represent ionic compounds composed of metals and nonmetals and indicate the simplest ratio of ions that results in neutrality.
Since the hybridization of the central atom is usually the only factor needed to determine molecular shape, you can simplify by considering only the hybridization of the central atom. Using this simplified method, the O atom of CO2 would bond with one of its p or s orbitals. Use valence bond theory to devise a hybridization and bonding scheme for CO2.
Answer:
CO2 contains sp hybridized carbon and sp2 oxygen atoms (linear shape)
Explanation:
There are two sigma bonds and two pi binds in the CO2 molecule. Carbon in its ground state contains one outer 2s orbital filled with two electrons and two outer 2p orbitals which are singly filled. Oxygen contains in its ground state contains an outer 2s orbital and three 2p orbitals filled with four electrons.
When these orbitals on oxygen are sp2 hybridized, one orbital is left unhybridized in each oxygen atom. Recall that two hybrized sp2 orbitals on oxygen atom accommodate the two lone pairs and one sp2 hybridized orbital in each oxygen atom forms a sigma bond to carbon. The remaining unhybridized orbital on oxygen is used by each oxygen atom to overlap with each unhybridized p orbital on the sp hybridized carbon.
Carbon forms two sigma bonds to oxygen via the two hybridized sp orbitals on carbon. The two unhybridized orbitals overlap sideways and firm pi bonds with oxygen p orbitals.
Triatomic molecules must be either linear or bent. In CO2, the bond angle must be 180° giving a linear molecule.
The carbon atom in CO₂ has sp hybridization due to its two regions of high electron density. Each of these forms a sigma bond with an oxygen atom, resulting in a linear geometry. Each C-O bond consists of one sigma bond and one pi bond, forming double bonds.
Explanation:The question asks for a hybridization and bonding scheme for the molecule CO₂ based on the valence bond theory. In CO₂, the central atom is Carbon (C). This atom is surrounded by two regions of high electron density, each represented by the double bonds with oxygen atoms.
According to the valence bond theory, a central atom with two regions of electron density (Lone pairs or bonds) is associated with sp hybridization. Therefore, in the case of CO₂, the Carbon atom will have sp hybridization. Since sp hybridization generates two hybrid orbitals, each of these will form a σ (sigma) bond with one of the Oxygen atoms maintaining a linear geometry (180° angle).
Moving onto the molecule's bond character, each C-O bond in CO₂ is a double bond, consisting of a σ bond and a π bond. The σ bonds are formed by the overlapping of sp hybrid orbitals of carbon with p orbitals of oxygen, while the π bonds are formed by the side-by-side overlapping of unhybridized p orbitals from both Carbon and Oxygen atoms.
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Be sure to answer all parts. Calculate ΔH o for the following reaction in two ways, using the data given below. H2(g) + I2(g) → 2HI(g) (a) Using the equation ΔH o rxn = ∑(BE(reactants)) − ∑(BE(products)) kJ/mol (b) Using the equation ΔH o rxn = ∑(nΔH o f (products)) − ∑(mΔH o f (reactants)) kJ/mol Bond BE kJ/mol Substance ΔH o f kJ/mol H―H 436.4 H2(g) 0 I―I 151 I2(g) 61.0 H―I 298.3 HI(g) 25.9
Answer:
a) ΔH°rxn = -9.2kJ/mol
b) ΔH°rxn = -9.2kJ/mol
Explanation:
Using Hess's law, you can find ΔH of a reaction from ΔH of formation of the substances involved in the reaction, thus:
ΔH°rxn = ∑(BE(reactants)) − ∑(BE(products))
Or:
ΔH°rxn = ∑(nΔH°f (products)) − ∑(mΔH°f (reactants))
For the reaction:
H₂(g) + I₂(g) → 2HI(g)
a) Using the first equation:
ΔH°rxn = ΔH (H-H) + ΔH (I-I) - 2ΔHBE (H-I)
ΔH°rxn = 436.4kJ + 151kJ - 2×298.3kJ
ΔH°rxn = -9.2kJ/mol
b) Using the second equation:
ΔH°rxn = 2Δ°f (HI) − ΔH°f (H₂) - ΔH°f (I₂)
ΔH°rxn = 2×25.9kJ - 0kJ - 61.0kJ
ΔH°rxn = -9.2kJ/mol
The reaction H₂(g) + I₂(g) → 2HI(g) has a ΔHo value of -9.2 kJ/mol, which is calculated using both bond dissociation energies and enthalpies of formation. This indicates the reaction is exothermic.
Explanation:To calculate ΔHo for the reaction H₂(g) + I₂(g) → 2HI(g),first we will use the bond dissociation energies (BE), and second the enthalpies of formation (ΔHof).
(a) Using the equation ΔHorxn = ∑BE(reactants) − ∑BE(products), we get ΔHorxn = [1(436.4 kJ/mol) + 1(151 kJ/mol)] - 2(298.3 kJ/mol) = -9.2 kJ/mol. This asserts that the reaction is exothermic and energy is released in the process. The sum of bond energies of reactants is smaller than that of the products.
(b) Using the equation ΔHorxn = ∑nΔHof(products) − ∑mΔHof(reactants), we get ΔHorxn = 2(25.9 kJ/mol) - [1(0 kJ/mol) + 1(61.0 kJ/mol)] = -9.2 kJ/mol. This value is same as obtained by the first method and thus validates it.
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