Answer: The acetaminophen
Explanation: Liquid-Liquid is a very important and commercial separation method used for the chemical separation and analyst of chemical mixtures. It is also known as PARTITIONING. In this technique the solute is transferred from one solvent to another of which both solvents are IMMISCIBLE OR PARTLY MISCIBLE. example in the mixture involving sodium bicarbonate,acetylsalicylic acid and acetaminophen and a binder after the binder is removed and you mix the two other components with the sodium bicarbonate the solution the acetylsalicylic acid dissolved but the acetaminophen did not meaning you could separate it from the other.
Measurements show that the ph of a particular lake is 4.0. What is the hydroxide ion concentration of the lake?
Answer:
[OH⁻] → 1×10⁻¹⁰
Explanation:
pH = - log [H⁺]
pOH = - log [OH⁻]
pH + pOH = 14
4 + pOH = 14
14 - 4 = 10 → pOH
10^-pOH = [OH⁻]
10⁻¹⁰ = [OH⁻] → 1×10⁻¹⁰
The pOH of the lake is 10 (calculated by subtracting the pH from 14), and the hydroxide ion concentration of the lake is 10^(-10) moles per liter, obtained by raising 10 to the power of the negative pOH.
Explanation:The pH of a solution is a measure of the hydrogen ion concentration, and the pOH is a measure of the hydroxide ion concentration. In water, pH and pOH are related to each other, and have a sum of 14 at room temperature. When the pH of the solution is given, you can find the pOH by subtracting the pH from 14. In this case, the pH of the lake is 4.0, so the pOH would be 14 - 4 = 10. The concentration of hydroxide ions in a solution can be calculated using the formula 10^(-pOH). So, the hydroxide ion concentration of the lake is 10^(-10) moles per liter.
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If an enclosure of 0.432 L has a partial pressure of O2 of 3.4×10−6 torr at 28 ∘C, what mass of magnesium will react according to the following equation? 2Mg(s)+O2(g)→2MgO(s)
The mass of the magnesium that is produced is [tex]2.1 * 10^{-8[/tex] g.
What is ideal gas equation?The ideal gas equation can also be expressed in different forms to solve for different parameters. For example, it can be rearranged to find the molar volume (V/n) or the number of moles (n) when other variables are known.
We have that;
PV = nRT
n = PV/RT
n = [tex]3.4 * 10^{-6[/tex] * 0.432/62.4 * 55
n = [tex]4.3 * 10^{-10[/tex] moles
If 2 moles of Mg produces 1 mole of oxygen
x moles of oxygen produces [tex]4.3 * 10^{-10[/tex] moles of oxygen
x =[tex]8.6 * 10^{-10[/tex] moles
Mass of magnesium = [tex]8.6 * 10^{-10[/tex] moles * 24 g/mol
= [tex]2.1 * 10^{-8[/tex] g
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Which type of interaction does NOT contribute to the stabilization of the tertiary structure of a protein? a.Disulfide bond b.Phosphodiester bond c.Hydrogen bond d.Salt bridge
Answer:
Option b, Phosphodiester bond
Explanation:
Proteins are sequence of amino acids. Two amino acids are joined by peptide bond. Therefore, proteins are also known as sequence of polypeptide chains. These polypeptides chains have four level of structures:
Primary structure
secondary structure
Tertiary structure
Quaternary structure
Primary structure is simply a sequence of amino acids. In secondary, tertiary and Quaternary structure, various interactions are present.
Disulfide bond, hydrogen bond and salt bridge stabilizes tertiary structure of the protein.
Phosphodiester bond is present as link between two nucleotides and thus, present in the backbone of nucleic acid (RNA and DNA)
Therefore, the correct option is option b
Among disulfide bonds, hydrogen bonds, salt bridges, and phosphodiester bonds, the latter does not contribute to the stabilization of the tertiary structure of a protein. Instead, phosphodiester bonds are crucial in the formation of nucleic acids.
Explanation:In regards to protein structure, the interaction types that contribute to the stabilization of the tertiary structure of a protein include disulfide bonds, hydrogen bonds, and salt bridges.
Disulfide bonds are covalent bonds between two sulfur atoms that stabilize the protein structure. Hydrogen bonds are weak forces of attraction between the hydrogen atom in one molecule and an electronegative atom in another. Salt bridges are ionic bonds between amino acid side chains that also stabilize the protein structure.
However, the phosphodiester bond does not play a role in tertiary protein structure. This type of bond is important in the formation of nucleic acids, such as DNA and RNA, where it links the 3' carbon of one nucleotide to the 5' carbon of another. It serves a different purpose in molecular biology and does not contribute to the stabilization of the tertiary structure of a protein.
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You have 10.0 g each of Na, C, Pb, Cu and Ne. Which contains the smallest number of moles? You have 10.0 g each of Na, C, Pb, Cu and Ne. Which contains the smallest number of moles?
1. Pb
2. Cu
3. Ne
4. Na
Answer:
The answer to your question is number 1. Pb
Explanation:
Data
10 grams of Na, C, Pb, and Ne
Process
Calculate the moles of each element
23 g of Na ----------------- 1 mol
10 g of Na ------------------- x
x = 0.43 moles of Na
20 g of Ne ---------------- 1 mol
10 g of Ne ----------------- x
x = 0.5 moles of Ne
64 g of Cu -------------- 1 mol
10 g of Cu -------------- x
x = 0.16 moles of Cu
207 g of Pb ------------- 1 mol
10 g of Pb -------------- x
x = 0.048 moles of Pb
Pb has the smallest number of moles
Answer:
Pb contains the smallest no of moles
Mole = mass/atomic mass
For lead no of mole = 10g/207.2g/mol
= 0.04826mol
For Cu no of mole = 10g/63.546g/mol = 0.1574mol
For Ne no of mole = 10g/20.1797g/mol = 0.4955mol
For Na no of mole = 10g/22.9898g/mol = 0.5350mol
If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is A. 705°C. B. 327°C. C. 273°C. D. 873°C.
Answer:
Option B. 327°C
Explanation:
Absolute T° = T° in K
T° in K = T° in C + 273
T° in K - 273 = T° in C
600 K - 273 = 327°C
You have a total body concentration of 300 mOsM and total body volume of 3 liters. If you add 0.5 L of a solution containing 150 mosmol of NaCl to the body, what would the new total body concentration be?
Answer:
The new total body concentration would be 300 mOsM
Explanation:
In order to do this, we need to convert all concentrations to moles.
First, with the total body concentration, we have the initial volume of 3 liters and the concentration of 300 M (I will omit til the end the part of mOs)
The moles of the body concentration in this volume is:
moles = M * V
moles = 300 * 3 = 900 moles
To this moles, we add 150 moles of NaCL so, the total moles now is:
moles = 900 + 150 = 1050 moles
Finally, we can calculate the concentration with the new volume of 3.5 L (the sum of 3 and 0.5 liters added):
M = 1050 / 3.5
M = 300 mOsM
So the concentration remains the same as initial
Which of the following is a basic solution? Select one: a. HCl dissolved in water b. household ammonia c. vinegar d. pure water Incorrect
Answer:
b. household ammonia
Explanation:
Basic solution -
A solution is considered to be basic in nature , if it is capable to release [tex]OH^-[/tex]ions .
The pH of a basic solution is always greater than 7 .
The taste of a basic solution is bitter .
From the given options of the question ,
The basic solution is the household ammonia.
Rest HCl , Vinegar are acidic in nature ,
And ,
Pure water is neutral. ( where, pH = 7 ) .
Balance the equation. Mg3(PO4)2+? C → ? Mg3P2+? CO , using the smallest possible integers. What is the sum of the coefficients in the balanced equation?1. 14.
2. 16.
3. 12.
4. 18.
5. 22.
Answer:
Sum of coefficients in balanced equation is 18
Explanation:
Unbalanced equation: [tex]Mg_{3}(PO_{4})_{2}+C\rightarrow Mg_{3}P_{2}+CO[/tex]
Balance O: [tex]Mg_{3}(PO_{4})_{2}+C\rightarrow Mg_{3}P_{2}+8CO[/tex]
Balance C: [tex]Mg_{3}(PO_{4})_{2}+8C\rightarrow Mg_{3}P_{2}+8CO[/tex]
Balanced equation:[tex]Mg_{3}(PO_{4})_{2}+8C\rightarrow Mg_{3}P_{2}+8CO[/tex]
Sum of coefficients in balanced equation = (1+8+1+8) = 18
So, option (4) is correct.
In today’s experiment, some lab students let their chromatograms elute slightly longer than others. If one student’s solvent front traveled to 9.80 cm, and another student’s solvent front traveled 8.00 cm, should these students obtain different Rfs for the known cations? Why or why not?
Answer:
They should obtain the same Rf for the same compounds.
Explanation:
The Rf is defined as A/B. Where A is the displacement of the substance of interest, and B is the solvent front.
By dividing the substance's displacement by B, we make it so that the Rf factor is equal for identical compounds in the same mobile phase, no matter what the solvent front is.
For the endothermic formation of sulfuric acid H2O(g) + SO3(g) ↔ H2SO4(g) _______ the pressure of the system and ________ the temperature of the system would shift the reaction to the left.
1. decreasing, lowering
2. decreasing, raising
3. increasing, raising
4. increasing, lowering
Answer: Option (1) is the correct answer.
Explanation:
As per Le Chatelier's principle, any disturbance caused in an equilibrium reaction will tend to shift the equilibrium in a direction away from the disturbance.
For example, [tex]H_{2}O(g) + SO_{3}(g) \rightleftarrow H_{2}SO_{4}(g)[/tex]
As this given reaction is endothermic in the forward direction and exothermic in the backward direction. Thus, in order to shift the reaction on left side we need to decrease the temperature.
Also, the number of moles are more on reactant side as compared to product side. So, when we decrease the number of moles on reactant side then the equilibrium will shift on left side.
Therefore, we can conclude that for the given reaction decreasing the pressure of the system and lowering the temperature of the system would shift the reaction to the left.
Final answer:
Decreasing the pressure of the system and lowering the temperature for the reaction H₂O(g) + SO₃(g) ⇔ H₂SO₄(g) both shift the equilibrium to the left because Le Chatelier's Principle dictates that the system will counteract changes by shifting towards more gas molecules (when pressure decreases) and by absorbing heat (when temperature is lowered for an endothermic reaction).
Explanation:
To determine how changes in pressure and temperature affect the equilibrium of the reaction H₂O(g) + SO₃(g) ⇒H₂SO₄(g), Le Chatelier's Principle can be applied. This principle states that if a system at equilibrium is subjected to a change in conditions, the system will adjust to partially counteract the effect of the change.
For the reaction given, if we decrease the pressure, the equilibrium will shift towards the side with more gas molecules to increase the pressure again. Since the left side has two moles of gas and the right side has only one, decreasing the pressure will shift the equilibrium to the left.
Regarding temperature, since it's an endothermic reaction, heat can be considered a reactant. So, if we lower the temperature, the system will shift towards the side that absorbs heat to counteract the decrease in temperature, which would be shifting the equilibrium to the left. Therefore, the correct answer is 1. decreasing, lowering.
For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed to come to equilibrium, and the equilibrium concentration of PCl3(g) is 0.27 M, what is the equilibrium concentration of PCl5(g)?
The equilibrium concentration of PCl5 is 0.0036 M.
The equation of the reaction is; PCl5(g) ⇄ PCl3(g) + Cl2(g)
The number of moles of PCl3 at equilibrium is = 0.27 M × 1.00-L = 0.27 moles
Now;
Kc = [PCl3] [Cl2]/[PCl5]
Kc = 2.0 × 10^1 or 20
We can see that;
Number of moles of PCl3 = Number of moles of Cl2 = 0.27 moles
Let the equilibrium concentration of PCl5 be x
20 = (0.27)^2/x
x = (0.27)^2/20
x = 0.0036 moles
Since the volume does not change;
equilibrium concentration of PCl5 = 0.0036 moles/1.00-L = 0.0036 M
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For the given equilibrium, the reaction PCl5(g) PCl3(g) + Cl2(g) is staged. Given Kc = 2.0 x 101 and [PCl3] = 0.27M, we find that [Cl2] is also 0.27M due to a 1:1 ratio in the reaction. We can then solve for [PCl5], finally getting [PCl5] = 0.364 × 10-2 M.
Explanation:The equilibrium for the reaction PCl5(g) PCl3(g) + Cl2(g) is defined by the ratio of concentrations of the products to the reactants, expressed as Kc = [PCl3][Cl2] / [PCl5]. Given that Kc = 2.0 × 101 and [PCl3] = 0.27 M, we can solve for the equilibrium concentration of [Cl2] using the same Kc equation, resulting in [Cl2] = 0.27 M (since it's a 1:1 ratio in this particular reaction). Finally, substituting these concentrations back into the equilibrium constant expression, we can solve for [PCl5] by rearranging the equation to [PCl5] = [PCl3][Cl2] / Kc = 0.27 x 0.27 / 2.0 x 101 = 0.364 × 10-2 M.
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If He gas has an average kinetic energy of 4310 J/mol under certain conditions, what is the root mean square speed of O2 gas molecules under the same conditions? (given that He and O_2 gas are at the same temperature what can you conclude about 0_2's average kinetic energy?) variable equations used to find the solution are greatly appreciated!!!
Answer:
The root mean square speed of O2 gas molecules is
519.01 m/s
Explanation:
The root mean square velocity :
[tex]v_{rms}=\sqrt{\frac{3RT}{M}}[/tex]
[tex]K.E_{avg}=\frac{3}{2}RT[/tex]
[tex]K.E =\frac{1}{2}mv_{rms}^{2}[/tex]
Molar mass , M
For He = 4 g/mol
For O2 = 2 x 16 = 32 g/mol
O2 = 32/1000 = 0.032 Kg/mol
First calculate the temperature at which the K.E of He is 4310J/mol
K.E of He =
[tex]K.E_{avg}=\frac{3}{2}RT[/tex]
[tex]T=\frac{2(K.E)}{3(R)}[/tex]
K.E of He = 4310 J/mol
[tex]T=\frac{2(4310J/mol)}{3(8.314J/Kmol)}[/tex]
[tex]T=345.60K[/tex]
Now , Use Vrms to calculate the velocity of O2
[tex]v_{rms}=\sqrt{\frac{3(8.314J/Kmol)(345.60K)}{0.032Kg/mol}}[/tex]
[tex]v_{rms}=\sqrt{\frac{8619.9552}{0.032}}[/tex]
[tex]v_{rms}=\sqrt{26935.001}[/tex]
[tex]v_{rms}=519.01m/s[/tex]
The root mean square speed of O2 gas molecules can be calculated using the average kinetic energy of He gas and the formula Urms = sqrt(3 * kB * T / m).
Explanation:The root mean square speed (Urms) of gas molecules can be calculated using the equation:
Urms = sqrt(3 * kB * T / m)
Where kB is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, and m is the molar mass of the gas.
Since the average kinetic energy (KEavg) of helium gas (He) is given as 4310 J/mol, we can assume that the temperature is the same for He and oxygen gas (O2). We know that the root mean square speed of He gas is close to 500 m/s. By using the equation and the given data, we can calculate the root mean square speed of O2 gas molecules.
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In the presence of oxygen, the cells convert glucose into energy through a process called: Select one: a. anaerobic metabolism. b. respiration. c. perfusion. d. aerobic metabolism.
Answer:
b. respiration
Explanation:
Food is converted into energy that can be used by the cells of the body by the process of cellular respiration. In cellular respiration, glucose and oxygen are combined and converted into water and carbon dioxide and energy. The produced energy is transferred to ATP.
In the presence of oxygen, cells convert glucose into energy through a process called aerobic metabolism, which is also referred to as cellular respiration. This process involves several stages and results in the formation of ATP, carbon dioxide, and water.
Explanation:Cellular Respiration and Energy Conversion
In the presence of oxygen, cells convert glucose into energy via a process known as aerobic metabolism (option d), which is a type of cellular respiration. The term 'respiration' is often used to describe this process as well (option b). Aerobic metabolism involves the conversion of glucose and oxygen into carbon dioxide and water, releasing energy that is stored in molecules of ATP. This energy conversion process comprises several stages, including Glycolysis, Transformation of Pyruvate, the Krebs Cycle, and Oxidative phosphorylation. In contrast, anaerobic metabolism (a) occurs in the absence of oxygen and involves different pathways such as anaerobic glycolysis or fermentation.
Cellular respiration is a crucial part of energy metabolism for all aerobic organisms, allowing them to harness energy from food. It is a complex yet beautifully orchestrated series of chemical reactions that take place within the cells, more specifically within the mitochondria. The process ultimately results in the production of ATP, which is the primary energy carrier within the cell.
A gas sample in a closed, expandable container of initial volume 5.00 L was allowed to warm from 25°C to 35°C. What was its new volume?
Answer : The new volume of was, 5.17 L
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles of gas.
Mathematically,
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where,
[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.
[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.
We are given:
[tex]V_1=5.00L\\T_1=25^oC=(25+273)K=298K\\V_2=?\\T_2=35^oC=(35+273)K=308K[/tex]
Putting values in above equation, we get:
[tex]\frac{5.00L}{298K}=\frac{V_2}{308K}\\\\V_2=5.17L[/tex]
Therefore, the new volume of was, 5.17 L
. A solution is prepared by dissolving 23.7 g of CaCl2 in 375 g of water. The density of the resulting solution is 1.05 g/mL. The concentration of CaCl2 in this solution is __________ molar.
a. 0.564
b. 0.571
c. 0.569
d. 0.537
e. 0.214
Answer:
The concentration of CaCl2 in this solution is 0.564 molar (option A)
Explanation:
Step 1: Data given
Mass of CaCl2 = 23.7 grams
Mass of water = 375 grams
Density of solution is 1.05 g/mL
Step 2: Calculate total mass
Total mass = mass of CaCl2 + mass of water
Total mass = 23.7 grams + 375 grams = 398.7 grams
Step 3: Calculate volume of the solution
Density = mass / volume
Volume = mass / density
Volume = 398.7 grams / 1.05 g/mL
Volume = 379.7 mL = 0.3797 L
Step 4: Calculate moles CaCl2
Moles CaCl2 = mass CaCl2 / molar mass CaCl2
Moles CaCl2 = 23.7 grams / 110.98 g/mol
Moles CaCl2 = 0.214 moles
Step 5: Calculate concentration
Concentration of CaCl2 = moles / volume
Concentration of CaCl2 = 0.214 moles / 0.3797 L
Concentration of CaCl2 = 0.564 mol / L = 0.564 molar
The concentration of CaCl2 in this solution is 0.564 molar (option A)
Why are the Nobel Gases so unreactive?
A.) Because they are used for cancer treatment.
B.) They have a full valence shell (outer electrons) of 8.
C.) Because they have a high boiling point.
Answer:B
Explanation:
They have a full valence shell of 8
Solid lithium hydride reacts with water to form aqueous lithium hydroxide and hydrogen gas. When this equation is written and balanced, the coefficient of lithium hydride is?
Answer:
The coefficient of LiH is 1 (non written)
Explanation:
The equation is this:
LiH (s) + H₂O (l) → LiOH (aq) + H₂ (g)
Ratio is 1:1 between reactants and products.
The coefficient of LiH is 1
Li2O(s)+H2O(l)→2LiOH(aq)
What mass of sodium bicarbonate would be needed to neutralize the spill if a bottle containing 1.75 L of concentrated HCl dropped on a lab floor and broke open?
Approximately 1764.21 grams of sodium bicarbonate are needed to neutralize a spill of 1.75 L of concentrated HCl, assuming a concentration of 12 M for the HCl.
Explanation:To determine the mass of sodium bicarbonate (NaHCO₃) needed to neutralize a spill of 1.75 L of concentrated HCl, one must first know the concentration of the hydrochloric acid (HCl). However, if we assume that 'concentrated' HCl is approximately 12 M (molar), we could proceed with the calculation. The balanced chemical equation for the reaction between sodium bicarbonate and hydrochloric acid is:
NaHCO₃ (s) + HCl (aq) → NaCl (aq) + H₂O (l) + CO₂ (g)From this equation, there is a one-to-one mole ratio between HCl and NaHCO₃. Thus, the moles of HCl that spilled are:
1.75 L HCl × 12 mol/L = 21 moles HClSince the mole ratio is 1:1, 21 moles of sodium bicarbonate are also required. The molar mass of NaHCO₃ is 84.01 g/mol, so you would need:
21 moles × 84.01 g/mol = 1764.21 grams of NaHCO₃Therefore, to neutralize the acid spill, you would need approximately 1764.21 grams of sodium bicarbonate.
Which of the following substances would you predict to have the highest ΔHvap?
1.H2
2.F2
3.SiF4
4.H2O
5.Ar
Answer: Option (4) is the correct answer.
Explanation:
Heat of vaporization is defined as the heat energy which is necessarily added to a liquid substance in order to transform the quantity of the substance into a gas.
For example, in [tex]H_{2}O[/tex] there will be presence of strong hydrogen bonding and in order to break this bond high amount of heat energy is required.
Whereas [tex]H_{2}[/tex], [tex]F_{2}[/tex] and [tex]SiF_{4}[/tex] are all covalent compounds which are bonded together by Vander waal forces. As these forces are weak in nature hence, they require less amount of heat energy to convert into vapor state.
Hence, they have low value of [tex]\Delta H_{vap}[/tex]. Also, Ar is a noble gas and it has only Vander waal forces. So, it will also have low value of [tex]\Delta H_{vap}[/tex].
Therefore, we can conclude that out of the given options [tex]H_{2}O[/tex] have the highest [tex]\Delta H_{vap}[/tex].
4. H₂O
Enthalpy of vaporization
The enthalpy of vaporization (symbol ∆Hvap), also known as the (latent) heat of vaporization or heat of evaporation, is the amount of energy (enthalpy) that must be added to a liquid substance to transform a quantity of that substance into a gas.
In terms of formula it can be written as:
[tex]\Delta H_\mathrm{vap}=\Delta U_\mathrm{vap}+p \Delta V[/tex]
Lets look at all the options one by one:
1. In case of H₂O molecule, there is a strong hydrogen bonding thus greatest energy is required to break this bonding.
2. While in case of H₂, F₂ and SiF₄ molecules are all covalent compounds that are bonded via weak vanderwaal forces thus it needs lesser heat energy to convert into vapor state.
3. Noble gases usually have weak vanderwaal forces thus Ar has lower ∆Hvap.
So, we can conclude that H₂O has the the highest ΔHvap.
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The results of an experiment in which teaching method is the experimental treatment, used with a class of low achievers, do not generalize to heterogeneous ability students. What type of threat to validity would this be______?
Answer:
Reduces the accuracy of external validity
Explanation:
This is because the research is a small sample of low achiever which cannot be used to extrapolate with whole population because of different level achiever groups. To eradicate this problem, there is Need to increase the size of sample group.
What mass of CO2 is produced from the combustion of 2.00 gallons of gas? Assume the gas is all octane (C8H18) with a density of 0.703 g/mL. 1 gallon = 3785 mL
Answer : The mass of [tex]CO_2[/tex] produced form the combustion is, 16.43 kg
Explanation :
Density : It is defined as the mass contained per unit volume.
Formula used for density :
[tex]Density=\frac{Mass}{Volume}[/tex]
First we have to calculate the mass of octane.
Given :
Density of octane = 0.703 g/mL
Volume = 2.00 gallons = 7570 mL
conversion used : 1 gallon = 3785 mL
Now put all the given values in the above formula, we get:
[tex]0.703g/mL=\frac{Mass}{7570mL}[/tex]
[tex]Mass=5321.71g[/tex]
Now we have to calculate the moles of octane.
[tex]\text{Moles of octane}=\frac{\text{Mass of octane}}{\text{Molar mass of octane}}[/tex]
Molar mass of octane = 114 g/mole
[tex]\text{Moles of octane}=\frac{5321.71g}{114g/mole}=46.68mole[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The balanced chemical combustion reaction of octane will be:
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
From the balanced chemical reaction we conclude that:
As, 2 moles of [tex]C_8H_{18}[/tex] react to give 16 moles of [tex]CO_2[/tex]
So, 46.68 moles of [tex]C_8H_{18}[/tex] react to give [tex]\frac{46.68}{2}\times 16=373.44[/tex] moles of [tex]CO_2[/tex]
Now we have to calculate the mass of [tex]CO_2[/tex].
[tex]\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2[/tex]
Molar mass of [tex]CO_2[/tex] = 44 g/mol
[tex]\text{ Mass of }CO_2=(373.44moles)\times (44g/mole)=16431.36g=16.43kg[/tex]
Thus, the mass of [tex]CO_2[/tex] produced form the combustion is, 16.43 kg
Atom economy is Atom economy is counting the atoms in the starting material and product. the calculation of the number of atoms conserved in the desired product rather than in waste. having green by-products. having green starting materials.
Answer: the calculation of the number of atoms conserved in the desired product rather than in waste.
Explanation:
Atom economy gives how much desired product is obtained compared to amount of starting materials.
[tex]\text {Atom economy}=\frac{\text {molecular weight of desired product}}{\text {molecular weight of all products}}\times 100%[/tex]
For example:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
[tex]\text {atom economy of water}=\frac{\text {molecular weight of desired product}}{\text {molecular weight of all products}}\times 100%[/tex]
[tex]\text {atom economy of water}=\frac{36}{44+36}\times 100%=45\%[/tex]
Thus atom economy for water is 45%
How many molecules of glucose (C6H12O6) are present in 23.3 g of the substance?
Answer:
The answer to your question is 7.77 x 10²² molecules of Glucose
Explanation:
Data
Number of molecules = ?
Mass = 23.3 g
Molecular mass of Glucose = 180 g
Avogadro's number = 6.023 x 10²³
Process
1.- Calculate the moles of Glucose in 23.3
180 g ------------------ 1 mol
23.3 g ---------------- x
x = (23.3 x 1) / 180
x = 0.129 moles
2.- Use Avogadro's number to calculate the number of molecules
1 mol of Glucose ------------- 6.023 x 10²³ molecules
0.129 moles -------------- x
x = (0.129 x 6.023 x 10²³) / 1
x = 7.77 x 10²² molecules
The number of molecules of glucose in 23.3 g of the substance is 7.77 x 10^23
Explanation:To determine the number of molecules of glucose (C6H12O6) in 23.3 g of the substance, you need to use the concept of moles. First, calculate the molar mass of glucose by adding up the atomic masses of its constituent elements. This gives you a molar mass of 180.16 g/mol. Next, use the formula:
moles = mass (g) / molar mass (g/mol)
Substituting the values, you get:
moles = 23.3 g / 180.16 g/mol = 0.129 moles of glucose
Since one mole of any substance contains Avogadro's number of molecules (6.022 x 10^23), multiply the number of moles by Avogadro's number to get the number of molecules:
0.129 moles x 6.022 x 10^23 molecules/mol = 7.77 x 10^23 molecules of glucose
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To achieve an octet, the phosphorus atom forms an ion. The name of this ion is: phosphoride ion phosphic ion phosphorous ion phosphide ion________-
Answer:
Phosphide ion
Explanation:
To achieve an octet, the phosphorus atom forms an ion. The name of this ion is Phosphide ion
Phosphorous have 15 electrons in shell. This means it has 5 electrons in its outer most shell well this means it requires 3 more electrons to complete its octate. Thus Phosphorus changes to phosphide ion with negative 3 sign on it.
Which amino acid would, when incorporated into a polypeptide chain (not at the N or C terminus), make the charge of the polypeptide more positive?
Answer: Arginine.
Explanation: Arginine is an amino acid that is used for the biosynthesis of protein. It is made up of an α-amino group, an α-carboxylic acid group, & a side chain consisting of a 3-carbon aliphatic straight chain ending in a guanidino group. At physiological pH, the carboxylic acid is deprotonated (−COO−), the amino group is protonated (−NH3+), and the guanidino group is also protonated to give the guanidinium form (-C-(NH2)2+), making arginine a charged, aliphatic amino acid.
The amino acid when incorporated into a polypeptide chain (not at the N or C terminus), make the charge of the polypeptide more positive is Arginine.
Arginine Like all of the amino acids, arginine has two functional groups, a carboxyl group (COOH) and an amine group (NH₂).Arginine is an amino acid that is used for the biosynthesis of protein. It is made up of an α-amino group, an α-carboxylic acid group, & a side chain consisting of a 3-carbon aliphatic straight chain ending in a guanidino group. At physiological pH, the carboxylic acid is deprotonated [tex](COO)[/tex], the amino group is protonated [tex](NH_3^+)[/tex], and the guanidino group is also protonated to give the guanidinium form [tex](-C-(NH_2)^{2+})[/tex], making arginine a charged, aliphatic amino acid.Thus, the amino acid is Arginine.
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You are studying the equilibirum between S8 and S2 gases at 0 celcius, you place a smaple of s2 in an otherwise empty, rigid container at an initial pressure of 3.-0 atm. wher eit reactsto form S8. at equilimrium, the partial pressure of s2 is 2.07 atm. calculate kp and kc.
Answer:
Kp and Kc are 0.01266 and 145.17, respectively.
Explanation:
Please check document attached.
An experiment calls for 7.57g of sugar. You have a sugar son that is 5% by weight (for every 100g there is 5g of sugar). How many mL of the sugar sold will be needed to deliver 7.57 g of sugar?The density of sugar is 1.157 g/mL. The molecular weight of sugar is 342 g/mol.
Answer:
mL sugar sold needed = 2.2215 mL sugar
Explanation:
mass sugar = 7.57 g
∴ wt% = 5% = (g sugar/g sln)×100
⇒ 0.05 = g sugar/g sln
∴ g sln = 100 g
⇒ g sugar = 5 g
∴ δ sugar = 1.157 g/mL
⇒ mL sugar = (5 g)×(mL/1.157 g) = 4.322 mL
⇒ mL sugar needed = (7.57 g)/(1.157 g/mL) = 6.543 mL
mL of the sugar sold needed = 6.543 mL - 4.322 mL = 2.2215 mL sugar
To find the mL of the sugar sold needed to deliver 7.57 g of sugar, calculate the grams of sugar in the solution and convert it to volume using the density of sugar.
Explanation:To find the number of mL of the sugar solution needed to deliver 7.57 g of sugar, we first need to calculate the number of grams of sugar in the solution. Since the sugar solution is 5% by weight, for every 100g of solution, there is 5g of sugar. Therefore, the number of grams of sugar in the solution is 5% of the solution's weight.
Next, we need to find the volume of the solution in mL. The density of sugar is given as 1.157 g/mL. So, to convert the weight of the sugar solution to volume, we divide the weight by the density.
Using the given information, we can calculate the number of mL of the sugar solution needed to deliver 7.57 g of sugar.
A coffee cup calorimeter initially contains 125g of water at 24.2oC. 10.5g of potassium bromide also at 24.2oC is added to the water. After the KBr dissolves the final temperature is 21.1oC. Calculate the enthalpy change for dissolving the salt in J/g and kJ/mol. Assume specific heat of solution is 4.18J/goC.
To calculate the enthalpy change for dissolving potassium bromide, use the equation q = mcΔT. Then divide the heat gained or lost by the mass of the substance to get the enthalpy change in J/g. To convert to kJ/mol, divide by the molar mass of potassium bromide.
Explanation:To calculate the enthalpy change for dissolving the salt, we need to use the equation q = mcΔT, where q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. First, we calculate the heat gained by the water using m = 125g, c = 4.18J/goC, and ΔT = 24.2oC - 21.1oC. Next, we calculate the heat lost by the potassium bromide using m = 10.5g, c = 4.18J/goC, and ΔT = 24.2oC - 21.1oC. Finally, we can calculate the enthalpy change in J/g by dividing the heat gained or lost by the mass of the substance. To convert to kJ/mol, we need to use the molar mass of potassium bromide and divide the enthalpy change in J/g by the molar mass.
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Part IV. For each trial, calculate the number moles of 6.0 M HCl used in the reaction. Report your answer using 4 digits. Note that it is 1 or 2 digits beyond the number of significant figures, i.e. 0.2345 moles.
This is an incomplete question, here is a complete question.
For each trial, calculate the number moles of 6.0 M HCl used in the reaction?
Trial 1 : Volume of HCl = 15.0ml
Trial 2 : Volume of HCl = 14.9ml
Trial 3 : Volume of HCl = 15.2ml
Answer :
The number moles of HCl for trial 1, 2 and 3 is, 0.090 mol, 0.089 mol and 0.091 mol
Explanation :
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :
[tex]\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
In this question, the solute is HCl.
Now we have to calculate the number of moles of HCl for trial 1.
Volume of HCl = 15.0 mL = 0.015 L
[tex]6.0M=\frac{\text{Moles of HCl}}{0.015L}[/tex]
[tex]\text{Moles of HCl}=0.090mol[/tex]
Now we have to calculate the number of moles of HCl for trial 2.
Volume of HCl = 14.9 mL = 0.0149 L
[tex]6.0M=\frac{\text{Moles of HCl}}{0.0149L}[/tex]
[tex]\text{Moles of HCl}=0.089mol[/tex]
Now we have to calculate the number of moles of HCl for trial 3.
Volume of HCl = 15.2 mL = 0.0152 L
[tex]6.0M=\frac{\text{Moles of HCl}}{0.0152L}[/tex]
[tex]\text{Moles of HCl}=0.091mol[/tex]
Consider the following molecules and the description of the bonding present in each: CH3CH2CH2CH3CH3CH2CH2CH3 (C−CC−C and C−HC−H bonds only) CH3CH2CH2OH CH3CH2CH2OH (C−CC−C and C−HC−H bonds, and one C−OHC−OH bond) CH3CH2CHO CH3CH2CHO (C−CC−C and C−HC−H bonds, and one C=OC=O bond) Based on the general trends of intermolecular forces, rank these molecules in order of increasing boiling point.
CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3
Explanation:
Boiling point trend of Butane, Propan-1-ol and Propanal.
Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.
In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.
Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.
In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol
The order of intermolecular interaction is; CH3CH2CH2OH > CH3CH2CHO > CH3CH2CH2CH3CH3CH2CH2CH3
Intermolecular forces are forces of attraction that hold a molecule together in a particular state of matter. The nature of intermolecular forces in a molecule depends on the kind of bonds between the atoms in the compounds. Compounds that contain nonpolar bonds often have weaker intermolecular interaction between molecules while molecules that have polar bonds experience a greater magnitude of intermolecular forces.
The order of intermolecular interaction is; CH3CH2CH2OH > CH3CH2CHO > CH3CH2CH2CH3CH3CH2CH2CH3
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