It had an impact on democratically incompatible systems, and it forced people to address those problems.
What is democracy?Democracy is defined as a system of government where the populace has the power to decide legislation and debate issues. Supporting democracy helps build a more safe, stable, and affluent global environment where the United States may advance its national interests in addition to advancing such core American principles as worker rights and freedom of religion.
Democratically incompatible system is defined as a type of government where the people themselves hold the ultimate power and exercise it either directly or indirectly through a system of representation that often includes regular free elections. Democracy that is guided by a constitution. Deliberative democracy is a system in which the key to making moral decisions is real deliberation, not just voting.
Thus, it had an impact on democratically incompatible systems, and it forced people to address those problems.
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In a first-order decomposition reaction. 36.8% of a compound decomposes in 7.6 min. How long (in min) does it take for 88.8% of the compound to decompose?
Answer: 36.5 minutes
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant = 100
a - x = amount left after decay process
a) for completion of 36.8 % of reaction
[tex]7.6=\frac{2.303}{k}\log\frac{100}{100-36.8}[/tex]
[tex]k=\frac{2.303}{7.6}\times 0.19[/tex]
[tex]k=0.060min^{-1}[/tex]
b) for completion of 88.8 % of reaction
[tex]t=\frac{2.303}{k}\log\frac{100}{100-88.8}[/tex]
[tex]t=\frac{2.303}{0.060}\log\frac{100}{11.2}[/tex]
[tex]t=\frac{2.303}{0.060}\times 0.95[/tex]
[tex]t=36.5min[/tex]
It will take 36.5 minutes for 88.8% of the compound to decompose.
Final answer:
To accurately determine the time required for 88.8% of a compound to decompose in a first-order reaction, knowing the rate constant is essential. While the question provides a specific data point for decomposition over time, it lacks the rate constant needed for a direct calculation. Therefore, accurately answering this specific question based on the provided context is challenging without additional information.
Explanation:
The question asks for the time required for 88.8% of a compound to decompose in a first-order decomposition reaction given that 36.8% decomposes in 7.6 minutes. In first-order reactions, the time it takes for a certain percentage of the reactant to decompose does not depend on the initial concentration but on the rate constant (k) of the reaction. The integrated rate law for first-order reactions is given by the formula ln([A]0/[A]) = kt, where [A]0 is the initial concentration, [A] is the concentration at time t, and k is the rate constant.
To determine the time required for 88.8% decomposition, we would need to know the rate constant of the reaction. However, with the information provided, we can infer the rate constant by using the time and percent decomposition already given (36.8% in 7.6 minutes), but to find the exact time for 88.8% decomposition without additional specific information (e.g., actual rate constant value) is not directly feasible based on the provided context alone. Knowing the rate constant, we could then apply the formula to find the time for 88.8% of the compound to decompose.
Without the rate constant, an alternative but less precise method involves understanding that the time to reach a certain level of decomposition is related to the half-life of the reaction. Since the time for half of a reactant to decompose (its half-life) in a first-order reaction is constant, we can indirectly estimate times for specific percentages of decomposition if we know the half-life, which again depends on knowing the rate constant.
1-propanol (P1° = 20.9 Torr at 25 °C) and 2-propanol (P2° = 45.2 Torr at 25 °C) form ideal solutions in all proportions. Let x1 and x2 represent the mole fractions of 1-propanol and 2-propanol in a liquid mixture, respectively, and y1 and y2 represent the mole fractions of each in the vapor phase. For a solution of these liquids with x1 = 0.450, calculate the composition of the vapor phase at 25 °C.
y1= y2=
To find the vapor phase composition of a mixture of 1-propanol and 2-propanol at 25 °C, we apply Raoult's Law and Dalton's Law using the given vapor pressures and mole fraction. The calculated vapor phase composition is y1 = 0.2744 for 1-propanol and y2 = 0.7256 for 2-propanol.
The question asks for the composition of the vapor phase at 25 °C for a liquid mixture of 1-propanol and 2-propanol with a given mole fraction of 1-propanol (x1 = 0.450). According to Raoult's Law, the partial pressure of each component in an ideal solution is equal to the mole fraction of the component in the liquid phase times the vapor pressure of the pure component. The total vapor pressure (Ptot) of the solution is the sum of the partial pressures. We can then use Dalton's Law to find the mole fractions of the components in the vapor phase.
To calculate the composition of the vapor phase, we will use the following steps:
Calculate the partial pressure of 1-propanol (P1) using the formula P1 = x1 × P1°, where P1° is the vapor pressure of pure 1-propanol.
P1 = x1 × P1° = 0.450 × 20.9 Torr = 9.405 Torr
Calculate the partial pressure of 2-propanol (P2) using the formula P2 = x2 × P2°, where P2° is the vapor pressure of pure 2-propanol and x2 is the mole fraction of 2-propanol in the liquid phase.
Since the mole fractions must sum to 1,
x2 = 1 - x1 = 1 - 0.450 = 0.550.
Now we calculate P2 = x2 × P2° = 0.550 × 45.2 Torr = 24.86 Torr.
Ptot = P1 + P2 = 9.405 Torr + 24.86 Torr = 34.265 Torr
Calculate the mole fraction of 1-propanol in the vapor phase (y1) using y1 = P1 / Ptot.
y1 = P1 / Ptot = 9.405 Torr / 34.265 Torr = 0.2744
Calculate the mole fraction of 2-propanol in the vapor phase (y2) using y2 = P2 / Ptot.
y2 = P2 / Ptot = 24.86 Torr / 34.265 Torr = 0.7256
Thus, the composition of the vapor phase at 25 °C for the solution with x1 = 0.450 is y1 = 0.2744 and y2 = 0.7256.
The surface tension of water is 7.28 ✕ 10−2 J/m2 at 20°C. Predict whether the surface tension of heptane would be higher or lower than that of water at the same temperature. Explain your answer.
Answer:
Lower
Explanation:
Surface tension occurs because molecules at the surface do not have molecules above them, so they cohere more strongly to their neighbours.
The stronger cohesive forces make it more difficult to move an object through the surface than when it is beneath the surface.
The attractive forces in water are strong because of hydrogen bonding.
A hexane molecule is nonpolar, so the only attractions are the weak London dispersion forces.
The cohesive forces at the surface are much lower than those in water, so the surface tension of hexane is lower than that of water at the sane temperature.
A chemist prepares a solution of silver perchlorate by measuring out of silver perchlorate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's silver perchlorate solution. Round your answer to significant digits.
To calculate the concentration of a solution, moles of solute are divided by the final volume in liters. Without the specific mass of silver perchlorate, we cannot determine the concentration of the solution.
Explanation:To calculate the concentration of the chemist's silver perchlorate solution, we need to know the mass of silver perchlorate used and the final volume of the solution. The details of the mass have been omitted from the question, but typically, you would use the molar mass of silver perchlorate to determine the number of moles. To prepare a solution of desired concentration, the chemist would dissolve silver perchlorate in a volumetric flask and add water to the mark, ensuring the final volume accounts for the space occupied by the dissolved solute.
Once a known mass of silver perchlorate is dissolved to make a known final volume, the molarity (M) can be calculated by dividing the number of moles of silver perchlorate by the final volume of the solution in liters. For example, if we dissolved 50 grams of silver perchlorate and the molar mass is X g/mol, we would first calculate the moles as 50 g / X g/mol. Then if the final volume of the solution is 1 L, the concentration would be 50 g / X g/mol / 1 L = Y M.
If the question had provided specific values, we would use them in the calculation and round the answer to the appropriate number of significant digits. However, without the exact mass of silver perchlorate used, we cannot complete this calculation.
Identify the functional group(s) that appear in betaxolol. This compound is in a class of drugs called beta-blockers, which are used to lower blood pressure, lower heart rate, reduce angina (chest pain), and reduce the risk of recurrent heart attacks Alcohol Ether Arene Carboxylic Acid Aldehyde Ester Amine Alkene KetoneFigure:contains some chemical structures
Answer:
AlcoholEtherAreneAmineExplanation:
In the attached picture you may find the structure of betaxolol.
You can see the alcohol group C-O-H as well as the ether group C-O-C.
The arene -or aromatic ring- can also be seen.
There's also a secondary amine group, C-NH-C.
The dissolving of ammonium nitrate is often used in an instant cold pack used to relieve swelling.
If this process is an endothermic reaction, i.e. absorbs heat, then why does it feel cold to the touch instead of hot?
Answer:
Here's what I get
Explanation:
Does an ice cube feel cold to the hand?
The melting of ice is an endothermic process.
If you hold an ice cube, heat flows from your hand to the ice. When your hand loses heat, it feels cold.
In the same way, the endothermic dissolving of ammonium nitrate removes the heat from your hand, so the pack feels cold.
Given the reactions below, answer the following questions.
Cl_2(g) + F_2(g) rlhar 2ClF(g) delta G degree_rxn = 115.4 kJ/mol
Cl_2(g) + Br_2(g) rlhar 2ClBr(g) delta G degree_rxn = -2.0 kJ/mol
Calculate the delta G degree_rxn for 2ClF(g) + Br_2(g) rlhar 2ClBr(g) + F_2(g) __________ kJ/mol
Answer:
[tex]\Delta G_{rxn}^{0}=-117.4kJ/mol[/tex]
Explanation:
Gibbs free energy is an additive property.
[tex]2ClF(g)\rightarrow Cl_{2}(g)+F_{2}(g)[/tex] ; [tex]\Delta G_{1}^{0}=-115.4kJ/mol[/tex]
[tex]Cl_{2}(g)+Br_{2}(g)\rightarrow 2ClBr(g)[/tex] ; [tex]\Delta G_{2}^{0}=-2.0kJ/mol[/tex]
----------------------------------------------------------------------------
[tex]2ClF(g)+Br_{2}(g)\rightarrow 2ClBr(g)+F_{2}(g)[/tex] ; [tex]\Delta G_{rxn}^{0}=\Delta G_{1}^{0}+\Delta G_{2}^{0}=(-115.4-2.0)kJ/mol=-117.4kJ/mol[/tex]
So, standard gibbs free enrgy change for the given reaction is -117.4 kJ/mol
The change in free energy is called G (∆G). The [tex]\rm \Delta G^\circ rxn= -117.4\; kj/mol[/tex]
What is Gibbs free energy?Gibb free energy is the measure of the maximum amount of reversible work done in a thermodynamic system.
It means only when the temperature change, but the pressure is constant.
[tex]\rm Cl_2(g) + F_2(g) \longrightarrow 2ClF(g) \;delta\; G\;degree_rxn = 115.4 kJ/mol[/tex]
[tex]\rm Cl_2(g) + Br_2(g) \longrightarrow 2ClBr(g) \;delta\; G \;degree_r_x_n = -2.0 kJ/mol[/tex]
[tex]\rm \Delta G^\circ rxn= \Delta G^\circ1 +G^\circ2[/tex]
[tex]2ClF(g) + Br_2(g) \longrightarrow 2ClBr(g) + F_2(g)[/tex]
[tex]\rm \Delta G^\circ rxn= -115.4\;kj/mol - 2.0 \;kj/mol = -117.4\;kj/mol[/tex]
Thus, the [tex]\rm \Delta G^\circ rxn= -117.4\; kj/mol[/tex].
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Predict the major organic product for the reaction below. Draw the product in the planar (overhead) representation, clearly showing stereochemistry by drawing in a wedge or hashed bond per stereocenter.
Answer:
The main organic product for the reaction is shown in the following figure.
Explanation:
This is due to the presence of the substituent -CH3 in the molecule, which makes it impossible to leave the hydrogen to form the double bond in the elimination.
The question involves predicting the major product of a given reaction based on stereochemistry, utilizing concepts from organic chemistry such as skeletal structure and the reactivity of alkenes. Understanding these principles can help predict the product, though without the specified diagram, an exact prediction cannot be made.
Explanation:Your question relates to predicting the major organic product of a given reaction by examining the stereochemistry, which involves the three-dimensional arrangement of atoms in a molecule. Indeed, stereochemistry is a critical aspect of organic chemistry because it influences the properties and reactions of molecules.
Organic chemists often represent large molecules using a skeletal structure or line-angle structure. In this drawing style, carbon atoms are at the ends or bends of lines. Hydrogens attached to carbons are not drawn, and atoms other than carbon and hydrogen are represented by their symbols.
Alkenes are particularly reactive due to the presence of a C=C moiety, a reactive functional group known to undergo addition reactions. For instance, in the halogenation of alkenes, halogens add to the double bond of the alkene instead of replacing hydrogen as they would in an alkane. The stereochemistry of the reaction is significant because the spatial arrangement of atoms can influence the ability of halogens to add across the C=C bond.
Unfortunately, without the specific planar diagram or additional information about the reaction in question, I cannot predict the exact major product; however, this general information about the reactivity and stereochemistry of alkenes might help you predict it.
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Specify which atoms, if any, bear a formal charge in the Lewis structure given and the net charge for the species. Be sure to answer all parts.
a. Formal charge C:___________.
b. Formal charge N:________.
d. Net charge:________.
This is an incomplete question, here is a complete question.
Specify which atoms, if any, bear a formal charge in the Lewis structure given and the net charge for the species. Be sure to answer all parts.
[tex]:C\equiv N:[/tex]
a. Formal charge C:___________.
b. Formal charge N:________.
d. Net charge:________.
Answer :
a. Formal charge C : (-1) charge
b. Formal charge N : (0) charge
d. Net charge : (-1) charge
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, [tex]CN^-[/tex]
As we know that carbon has '4' valence electrons and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in [tex]CN^-[/tex] = 4 + 5 + 1 = 10
According to Lewis-dot structure, there are 6 number of bonding electrons and 4 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :
[tex]\text{Formal charge on C}=4-2-\frac{6}{2}=-1[/tex]
[tex]\text{Formal charge on N}=5-2-\frac{6}{2}=0[/tex]
Thus, the net charge will be = -1 + 0 = -1
In what way are planned economies and command economies similar?
O
A. They both are characteristic of the United States.
O
B. They both rely on private ownership.
C. They both involve strong governmental control.
O
D. They both require open exchange.
SUBMIT
Answer: D
Explanation:
Answer:
I believe the answer is C I took the test a little bit ago
Explanation:
Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product?A. 2C(s)+H2(g)→C2H2(g); ΔG∘f=209.2 kJ/molB. N2(g)+3H2(g)→2NH3(g); ΔG∘f=−33.30 kJ/molC. 2C(s)+2H2(g)→C2H4(g); ΔG∘f=68.20 kJ/molD. 2SO(g)+O2(g)→2SO2(g); ΔG∘f=−600.4 kJ/mol
Answer:
Option B and D
Explanation:
The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system.
A feasible way to synthesize the product entails the spontaneity of the reaction.
The spontaneity of a reaction is explained by the standard gibb's free energy.
If Delta-G = -ve ( the reaction is spontaneous)
if Delta -G = +ve ( the reaction is non-spontaneous)
if Delta-G = 0 ( the reaction is at equilibrium)
Hence the option (B and D) with negative Gibb's free energy are the reaction that are spontaneous.
In the following reaction, which component acts as an oxidizing agent? 10 I− (aq) + 2 MnO4− (aq) + 16 H+ (aq) → 5 I2 (s) + 2 Mn2+ (aq) + 8 H2O (l)
Answer:
The oxidizing agent is the MnO₄⁻
Explanation:
This is the redox reaction:
10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)
Let's determine the oxidation and the reduction.
I⁻ acts with -1 in oxidation state and changes to 0, at I₂.
All elements in ground state has 0 as oxidation state.
As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.
In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine
Sometimes, as we learned in this lesson, a single Lewis structure does not describe a molecule, and it instead resonates between two or more structures. What symbol do we place between these structures to show this?
A. ←
B. ≡
C. ↔
D. →
Answer:
C. ↔
Explanation:
Resonating structure -
These are the set of lewis structures of the same compound .
Where the structure helps to show the delocalization of the electrons over the structure .
The charge on all the resonating structure remains the same .
All the structures can be interconverted to each other , and are shown by the arrow ↔ .
Hence , from the given information of the question,
The correct option is C. ↔
Consider that you have a 100 mM stock solution and you need to prepare 10 mL of a 30 mM solution. How many milliliters of the stock solution do you need? mL How many milliliters of deionized water do you need? mL
Answer:
V₁ = 3.0 mL
V(H₂O) = 7 mL
Explanation:
Given data
Initial concentration (C₁): 100 mMInitial volume (V₁): ?Final concentration (C₂): 30 mMFinal volume (V₂): 10 mLIn order to determine the volume of the stock solution, we will use the dilution rule.
C₁ × V₁ = C₂ × V₂
100 mM × V₁ = 30 mM × 10 mL
V₁ = 3.0 mL
The volume of deionized water required is:
V(H₂O) = 10 mL - 3.0 mL = 7 mL
The volume of the stock solution required is 3 mL.
The stock solution is defined as the solution from which other solutions are prepared. We have the following information;
concentration of stock solution = 100 mMConcentration of diluted solution = 30 mM Volume of diluted solution = 10 mLUsing the formula;
M1V1 = M2V2
V1 = M2V2/M1
V1 = 30mM × 10 mL/100 mL
V1 = 3 mL
The volume of the stock solution required is 3 mL.
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By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monopro-tic
Answer: The molar mass of monoprotic acid is 135.9 g/mol
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]
Molarity of NaOH solution = 0.1008 M
Volume of solution = 15.0 mL
Putting values in above equation, we get:
[tex]0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol[/tex]
As, the acid is monoprotic, it contains 1 hydrogen ion
1 mole of [tex]OH^-[/tex] ion of NaOH neutralizes 1 mole of [tex]H^+[/tex] ion of monoprotic acid
So, 0.00151 moles of [tex]OH^-[/tex] ion of NaOH will neutralize [tex]\frac{1}{1}\times 0.00151=0.00151mol[/tex] of [tex]H^+[/tex] ion of monoprotic acid
Moles of monoprotic acid = 0.00151 moles
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of monoprotic acid = 0.00151 mole
Given mass of monoprotic acid = 0.2053 g
Putting values in above equation, we get:
[tex]0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.9g/mol[/tex]
Hence, the molar mass of monoprotic acid is 135.9 g/mol
The molar mass of the monoprotic acid, which can be calculated using the titration method with sodium hydroxide, was found to be approximately 135.7 g/mol.
Explanation:The subject of your query concerns on calculating the molar mass of an organic acid using the titration method with known information about sodium hydroxide. First, we need to look into the moles of sodium hydroxide (NaOH) used. Since we know the volume V=15.0 mL and concentration C=0.1008 M of NaOH used, moles (n) can be calculated using the equation n=CV, converting volume to liters, gives n = 0.1008 mol/L * 0.015 L = 0.001512 mol.
Because the sodium hydroxide and monoprotic acid react in a 1:1 ratio (as given by the neutralization equation NaOH + HA -> NaA + H2O), the moles of acid are also 0.001512.
The molar mass of the substance can then be determined by dividing the given mass of the sample by the number of moles. The mass of the organic acid is 0.2053g, so the molar mass (M) is M = mass / n = 0.2053g / 0.001512 mol = 135.7 g/mol.
This means that the molar mass of the monoprotic organic acid is approximately 135.7 g/mol.
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If you are asked to make 40 mL of a 1.0% (w/v %) agarose gel, how many grams of agarose will you add to the 40ml of buffer
Answer:
0.4g
Explanation:
1.0% (w/v%) = 1 g of agarose 100 ml of Tris-Acetate-EDTA, this is the buffer that agarose is run with
the amount of agarose for 40 ml = 1 /100 × 40 ml = 0.4 g
Final answer:
To prepare 40 mL of a 1.0% (w/v %) agarose gel, measure and mix 0.4 grams of agarose powder with 40 mL of buffer solution in an Erlenmeyer flask.
Explanation:
If you are asked to make 40 mL of a 1.0% (w/v %) agarose gel, you will add 0.4 grams of agarose to the 40 mL of buffer. To made this calculation, you need to understand the meaning of w/v %, which stands for weight/volume percentage. It is defined as the mass of a solute (in this case agarose powder) divided by the volume of the solution, and then multiplied by 100 to get the percentage.
For a 1.0% (w/v %) agarose gel, this means that you need 1 gram of agarose powder for every 100 mL of solution. Therefore, for 40 mL of solution, you simply use the proportion:
1 gram/100 mL = X grams/40 mL
Solving for X gives you:
X = (1 gram/100 mL) * 40 mL = 0.4 grams
Conclusion: To prepare a 40 mL agarose gel at 1.0% w/v concentration, weigh out 0.4 grams of agarose powder using an electronic scale, and then blend it with the appropriate buffer in an Erlenmeyer flask.
The highest concentration of aqueous HCL is ~ 37% by weight, and the density of this solution is 1.18 g/cm3. What is the molarity of HCl of this 37% aqueous HCl solution?
Answer: The molarity of HCl of this 37% aqueous HCl solution is 12.0 M
Explanation:
Given : 37 g of HCl is dissolve in 100 g of solution.
Mass of solute (HCl) = 37 g
Mass of solution = 100 g
Density of solution = 1.18g/ml
Volume of solution =[tex]\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.18g/ml}=84.7ml[/tex]
Molarity is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]V_s[/tex] = volume of solution in ml = 1000 ml
[tex]moles of solute =\frac{\text {given mass}}{\text {molar mass}}=\frac{37g}{36.5g/mol}=1.01moles[/tex]
Now put all the given values in the formula of molarity, we get
[tex]Molarity=\frac{1.01moles\times 1000}{84.7ml}=12.0mole/L[/tex]
The molarity of HCl solution is 12.0 M
A driver with a nearly empty fuel tank may say she is "running on fumes."
If a 15.0 gallon automobile gas tank had only gasoline vapor remaining in it, what is the farthest the vehicle could travel if it gets 20.0 miles per gallon on liquid gasoline?
Assume the average molar mass of molecules in gasoline is 101 g/mol , the density of liquid gasoline is 0.75 g/mL, the pressure is 747 mmHg , and the temperature is 25 ∘C.
Answer:
The farthest the vehicle could travel (if it gets 20.0 miles per gallon on liquid gasoline) is 1.62 miles.
Explanation:
The automobile gas tank has a volume capacity of 15 gallons which can be converted to liters: 15 × 3.7854 = 56.781 liters
We can find the moles of gasoline by using the ideal gas equation: PV = nRT.
Make n (number of moles) the subject of the formula: n = PV/RT, where:
P = 747 mmHg
V = 56.781 liters
R (universal gas constant) = 0.0821 liter·atm/mol·K
T = 25 ∘C = (273 + 25) K = 298 K
1 atm (in the unit of R) = 760 mmHg
Therefore n = 747 × 56.781/(0.0821 × 760 × 298) = 2.281 mol.
Given that the molar mass of the gasoline = 101 g/mol,
the mass of gasoline = n × molar mass of gasoline = 2.281 mol × 101 g/mol = 230.38 g
the density of the liquid gasoline = 0.75 g/mL
In order to calculate the distance the vehicle can travel, we have to calculate volume of gasoline available = mass of the liquid gasoline ÷ density of liquid gasoline
= 230.38 g ÷ 0.75 g/mL = 307.17 mL = 0.3071 liters = 0.3071 ÷ 3.7854 = 0.0811 gallons
since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by the car = gallons available × miles per gallon = 0.0811 × 20 = 1.62 miles.
A substance that is required to run the vehicle and the machine is known as fuel. These substance are as follows:-
PetrolDiesalsKerosineAccording to the data given in the question.
The farthest the vehicle could travel is 1.62 miles.
The volume capacity of 15 gallons which can be converted to liters: [tex]15 * 3.7854 = 56.781 liters[/tex]
The formula used is as follows:-[tex]PV =nRT[/tex]
After putting the value:-
P = 747 mmHg
V = 56.781 liters
R (universal gas constant) = 0.0821 liter·atm/mol·K
[tex]T = 25C = (273 + 25) K = 298 K[/tex]
Therefore, the value of [tex]n = \frac{747 * 56.781}{(0.0821 * 760 * 298} = 2.281 mol.[/tex]
Given that the molar mass of the gasoline = 101 g/mol,
Mass of gasoline = n × molar mass of gasoline
[tex]= 2.281 mol * 101 g/mol = 230.38 g[/tex]
the density of the liquid gasoline = 0.75 g/mL
[tex]= \frac{230.38}{0.75}= 307.17 mL \\\\= 0.3071 liters\\\\= \frac{0.3071}{3.7854} = 0.0811 gallons[/tex]
Since the vehicle gets 20.0 miles per gallon on liquid gasoline, the distance traveled by car = gallons available × miles per gallon =
[tex]0.0811 * 20 = 1.62 miles.[/tex]
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(b) Data has been collected to show that at a given wavelength in a 1 cm pathlength cell, Beer's Law for the absorbance of Co2+ is linear. If a 0.135 M solution of Co2+ has an absorbance of 0.350, what is the concentration of a solution with an absorbance of 0.460?
Answer : The concentration of a solution with an absorbance of 0.460 is, 0.177 M
Explanation :
Using Beer-Lambert's law :
[tex]A=\epsilon \times C\times l[/tex]
where,
A = absorbance of solution
C = concentration of solution
l = path length
[tex]\epsilon[/tex] = molar absorptivity coefficient
From this we conclude that absorbance of solution is directly proportional to the concentration of solution at constant path length.
Thus, the relation between absorbance and concentration of solution will be:
[tex]\frac{A_1}{A_2}=\frac{C_1}{C_2}[/tex]
Given:
[tex]A_1[/tex] = 0.350
[tex]A_2[/tex] = 0.460
[tex]C_1[/tex] = 0.135 M
[tex]C_2[/tex] = ?
Now put all the given values in the above formula, we get:
[tex]\frac{0.350}{0.460}=\frac{0.135}{C_2}[/tex]
[tex]C_1=0.177M[/tex]
Therefore, the concentration of a solution with an absorbance of 0.460 is, 0.177 M
Beer's Law states that there is a linear relationship between the concentration of a substance in a solution and its absorbance. We can use the equation c = A/(εb) to solve for the concentration of a solution with a given absorbance.
Explanation:Beer's Law states that there is a linear relationship between the concentration of a substance in a solution and its absorbance. The equation for Beer's Law is A = εbc, where A is the absorbance, ε is the molar absorptivity, b is the path length of the cell, and c is the concentration.
In this case, we have the absorbance (0.350) and concentration (0.135 M) for a solution of Co2+. We can rearrange the equation to solve for the concentration: c = A/(εb). Plug in the given values to find the molar absorptivity and path length, and then substitute those values into the equation to calculate the concentration of a solution with an absorbance of 0.460.
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Calculate the theoretical oxygen demand (mg/L). of a solution containing 450mg of glucose (C6H12O6). in 2 L of distilled water.
Answer:
ThOD =239.792 mg/L
Explanation:
Theorical Oxigen demand (ThOD):
is the theoretical amount of oxygen
required to oxidize the organic fraction of a
waste up to carbon dioxide and water.
⇒ C sln = 450 mg C6H12O6 / 2 L H2O = 225 mg/L sln
∴ mm C6H12O6 = 180.156 g/mol
balanced reaction:
C6H12O6 + 6O2 → 6CO2 + 6H2O∴ mol C6H12O6 = 1 mol
⇒ mass C6H12O6 = (180.156 g/mol)( 1 mol) = 180.156 g
∴ the value of ThOD is determined when 180.156 g C6H12O6 consume mass O2 = 6(32) = 192 g Oxygen; then in a solution of 225 mg/L, you have:
⇒ ThOD = (192/180.156)×225 mg/L
⇒ ThOD = 239.792 mg/L
The theoretical oxygen demand is 0.24 g/L or 240mg/L.
To calculate the theoretical oxygen demand of a solution containing 450mg of glucose (C6H12O6) in 2 L of distilled water, we need to start by determining the stoichiometry of the complete combustion of glucose, which can be represented by the balanced chemical equation:
C6H12O6 + 6O2 → 6CO2 + 6H2O
From the equation, we can see that 1 mole of glucose requires 6 moles of oxygen to completely react. Therefore, to calculate the oxygen demand, we first need to find the number of moles of glucose present in the solution:
The molar mass of glucose (C6H12O6) is approximately 180 g/mol. So, 450 mg of glucose is equal to 0.450 g or 0.450/180 = 0.0025 moles of glucose.
Using the stoichiometry from the equation, the moles of oxygen required is 0.0025 moles of glucose imes 6 moles of oxygen/mole of glucose = 0.015 moles of oxygen.
Now, the molar mass of oxygen (O2) is approximately 32 g/mol, hence 0.015 moles imes 32 g/mol = 0.48 grams of oxygen. Since the solution volume is 2 L, we need to find the amount per liter, which gives us 0.48 g / 2 L = 0.24 g/L or 240 mg/L of theoretical oxygen demand.
Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0 C, the pressure is 0.230 atm. What is the likely formula of the compound?
The question is incomplete, here is the complete question:
Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0°C, the pressure is 0.230 atm. What is the likely formula of the compound?
A. XeO
B. [tex]XeO_4[/tex]
C. [tex]Xe_2O_2[/tex]
D. [tex]Xe_2O_3[/tex]
E. [tex]Xe_3O_2[/tex]
Answer: The chemical formula of the compound is [tex]XeO_4[/tex]
Explanation:
To calculate the molecular weight of the compound, we use the equation given by ideal gas equation:
PV = nRT
Or,
[tex]PV=\frac{w}{M}RT[/tex]
where,
P = Pressure of the gas = 0.230 atm
V = Volume of the gas = 50.0 mL = 0.050 L (Conversion factor: 1 L = 1000 mL)
w = Weight of the gas = 0.100 g
M = Molar mass of gas = ?
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = [tex]0^oC=273K[/tex]
Putting value in above equation, we get:
[tex]0.230\times 0.050=\frac{0.100}{M}\times 0.0821\times 273\\\\M=\frac{0.100\times 0.0821\times 273}{0.230\times 0.050}=194.9g/mol\approx 195g/mol[/tex]
The compound having mass as calculated is [tex]XeO_4[/tex]
Hence, the chemical formula of the compound is [tex]XeO_4[/tex]
Calculate the solubility of carbon dioxide in water at an atmospheric pressure of 0.400 atm (a typical value at high altitude).
Answer:
1.40*10⁻² M
Explanation:
We have the solubility formula
Solubility,
S = KH*P
where
KH = measure of hardness of water / carbonate hardness = 3.50*10⁻² mol/L.atm
P = atmospheric pressure = 0.400 atm
Hence, we have
S = KH*P
= (3.50*10⁻² mol/L.atm)*(0.400 atm)
= 1.40*10⁻² mol/L
But 1 mol/L = 1 M,
Hence, the answer (1.40*10⁻² mol/L ) is equivalent to
= 1.40*10⁻² M
Classify each of the following particulate level illustrations as a representation of either a pure substance, a homogeneous mixture, or a heterogeneous mixture.Figure:three compounds were drawn
Answer:
The classification and illustrations are attached in the drawing.
Explanation:
It is possible to identify the pure substance observing the figure, since it is the only one that has 2 joined atoms (purple and blue) which forms a single compound.
On the other hand, the homogeneous mixture is identified by noting that its atoms are more united with respect to the heterogeneous mixture, highlighting that in homogenous mixtures the atoms, elements or substances are not visible to the naked eye and are in a single phase, instead in the heterogeneous mixture if they can be differentiated.
Heterogeneous mixture: Visible multiple chemicals with distinct parts.
Homogeneous mixture: Appears as one material, uniform with indistinguishable atoms and elements.
Pure material distinguished by figure with linked blue and purple atoms forming a single compound.
Heterogenous combination of visible compounds. Homogenous mixture comprised of various materials that seem like one substance Each sample portion has the same makeup and properties. The graphic shows that the pure material is the only one with two connected atoms (blue and purple) that form a compound.
The homogeneous mixture is distinguished by its more unified atoms, emphasizing that while in homogeneous mixtures the atoms, elements, and substances are not visible to the eye and are in a single phase, in the heterogeneous mixture they can be distinguished.
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Under what conditions does an increase in temperature turn a nonspontaneous process into a spontaneous process?
Choose one or more:
a. ΔH < 0, ΔS > 0
b. ΔH > 0, ΔS > 0
c. ΔH > 0, ΔS < 0
d. ΔH < 0, ΔS < 0
(What I'm trying to figure out is if you need to consider T being negative OR positive to start)
Answer:ΔH < 0, ΔS > 0
Explanation:
Consider ∆G= ∆H-T∆S
If ∆H is negative and and ∆S is positive, when T is increased, T∆S becomes more positive until the result of the difference between ∆H-T∆S becomes negative. Remember that a reaction is only spontaneous when ∆G is negative. Hence under these conditions, ∆G becomes negative and the reaction becomes spontaneous.
Write the following numbers in standard notation, maintaining the same number of significant figures.
6.104 x10^2
9.5 x 10^-3
Move the decimal by a certain number of spaces, according to the exponent of 10
Answer:
6.104x10^2=610.4
9.5x10^3=9500
Explanation:
To convert from scientific notation (6.104 x10^2, 9.5 x 10^-3) to standard notation, move the decimal point in accordance with the power of 10 (right for positive, left for negative). The converted numbers become 610.4 and 0.0095, respectively.
Explanation:In scientific notation, the base number is expressed as a decimal between 1 and 10, and tens are represented by powers of 10. In standard notation, we write the number in usual decimal representation.
To convert 6.104 x10^2 to standard notation, move the decimal two places to the right (because the exponent is positive), giving you 610.4.
For 9.5 x 10^-3, move the decimal three places to the left (because the exponent is negative). So, it becomes 0.0095.
Remember that the number of significant figures must remain the same when converting to standard notation.
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Arrange the following compounds in order of increasing solubility in water: O2, LiCl, Br2, methanol (CH3OH).
The order of increasing solubility in water for O2, Br2, LiCl, and methanol (CH3OH) is: O2 < Br2 < LiCl < Methanol (CH3OH). This is based on whether the molecules are nonpolar or polar and the intermolecular forces present.
Explanation:The solubility of compounds in water ultimate depends on the specific intermolecular interactions. The key principle here is 'like dissolves like', with polar substances dissolving in polar solvents (like water) and nonpolar substances dissolving in nonpolar solvents.
In your list: O2, LiCl, Br2, and methanol (CH3OH), the order of increasing solubility in water would in fact be: O2 < Br2 < LiCl < Methanol (CH3OH)
O2 and Br2 are both nonpolar molecules and hence have a very limited solubility in polar solvents such as water. LiCl is a polar ionic compound, it will easily dissolve in water through ion-dipole interactions. Methanol (CH3OH) is the most soluble because, in addition to being polar, it can form hydrogen bonds with water molecules, which is a stronger intermolecular force. Learn more about Solubility of Compounds here:
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The order of increasing solubility in water is O₂ < Br₂ < LiCl < CH₃OH.
To arrange these compounds in order of increasing solubility in water, we need to consider their polarity and ability to form hydrogen bonds with water. Water is a polar solvent and dissolves polar compounds and ionic compounds well, but non-polar compounds poorly.
O₂ (Oxygen): Non-polar molecule, very low solubility in water.Br₂ (Bromine): Non-polar molecule, low solubility in water, but slightly more soluble than oxygen due to its larger size and polarizability.LiCl (Lithium Chloride): Ionic compound, highly soluble in water due to dissolution into Li+ and Cl- ions.Methanol (CH₃OH): Polar molecule, high solubility in water due to hydrogen bonding with water molecules.Therefore, the order of increasing solubility in water is O₂ < Br₂ < LiCl < CH₃OH.
) A 0.907 M lead II nitrate solution has a density of 1.252 g/cm3. Find the molality of the solution.
Answer: The molality of the solution is 0.953 m
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]V_s[/tex] = volume of solution in ml
Given : 0.907 moles of lead(ii)nitrate is dissolved in 1000 ml of the solution.
density of solution= 1.252 g/ml
Thus mass of solution = [tex]Density\times volume=1.252\times 1000 ml=1252g[/tex]
mass of solute =[tex]moles\times {\text {Molar mass}}=0.907mol\times 331g/mol=300g[/tex]
mass of solvent =mass of solution - mass of solute = (1252-300)g= 952 g
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
[tex]Molality=\frac{n\times 1000}{W_s}[/tex]
where,
n = moles of solute = 0.907 moles
[tex]W_s[/tex]= weight of solvent in g = 952 g
[tex]Molality = \frac{0.907\times 1000}{952g}=0.953m[/tex]
Thus molality of the solution is 0.953 m
Final answer:
To find the molality of the 0.907 M lead(II) nitrate solution, we first calculate the mass of the solvent in kilograms and then divide the amount in moles of the solute by this mass. The final molality is approximately 0.953 mol/kg.
Explanation:
The question is asking to find the molality of a 0.907 M lead(II) nitrate solution with a density of 1.252 g/cm3. To calculate the molality, we need the number of moles of solute and the mass of the solvent in kilograms.
First, calculate the mass of 1 litre (1000 cm3) of solution using the density:
Mass of solution = Density imes Volume = 1.252 g/cm3 imes 1000 cm3 = 1252 g
Next, the mass of the solute in 1 litre of solution:
Mass of solute = Molarity imes Molar Mass imes Volume = 0.907 mol/L imes 331.2 g/mol imes 1 L = 300.2794 g
Now, find the mass of the solvent (water) by subtracting the mass of solute from the total mass of the solution:
Mass of solvent = Mass of solution - Mass of solute = 1252 g - 300.2794 g = 951.7206 g
Convert mass of solvent to kilograms:
Mass of solvent (kg) = 951.7206 g imes (1 kg / 1000 g) = 0.9517206 kg
Finally, calculate the molality (m):
Molality (m) = Moles of solute / Mass of solvent (kg) = 0.907 mol / 0.9517206 kg = 0.953 m
Therefore, the molality of the lead(II) nitrate solution is approximately 0.953 mol/kg.
In ionic bonds the atom that contributes an electron and has a positive charge as a result is called the . The atom in ionic bonding that accepts an electron and has a negative charge as a result is called the .
Answer:
a. Cation
b. Anion
Explanation:
I 1 mL o a 0.02% w/v isoproterenol hydrochloride solution is diluted to 10 mL with sodium chloride injection be ore intravenous administration, calculate the percent concentration o the diluted solution
Answer:
0.002% w/v
Explanation:
The unit w/v means that mass (in g) per volume (in mL). When the solution is diluted, the mass of the solvent will not change, and the mass can be calculated by the concentration (C) multiplied by the volume (V). So, if 1 is the initial solution, and 2 the diluted solution:
C1*V1 = C2*V2
C1 = 0.02%
V1 = 1 mL
V2 = 10 mL
0.02*1 = C2*10
10C2 = 0.02
C2 = 0.002% w/v
For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibrium?
a. double the concentrations of both products and then double the container volume
b. double the container volume
c. add more A
d. double the concentration of B and halve the concentration of C
e. double the concentrations of both products
f. double the concentrations of both products and then quadruple the container volume
Answer:
a. No change.
b. The equilibrium will shift to the right.
c. No change
d. No change
e. The equilibrium will shift to the left
f. The equilibrium will shift to the right
Explanation:
We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining Keq again.
The equilibrium constant for A(s)⇌B(g)+C(g)
Keq = Kp = pB x pC
where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )
We also use Q which has the same form as Kp but denotes the system is not at equilibrium:
Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.
a. double the concentrations of Q which has the same form as Kp but : products and then double the container volume
Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.
Initially the system will decrease the partial pressures of B and C by a half:
Q = pB´x pC´ ( where pB´and pC´are the changed pressures )
Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp ⇒ Kp = Q/4
But then when we double the volume ,the sistem will react to double the pressures of A and B. Therefore there is no change.
b. double the container volume
From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change reduced the pressures by a half :
Q = pB´x pC´ = ( 1/2 pB ) x ( 1/2 pC ) = 1/4 pB x pC = 1/4 Kp
c. add more A
There is no change in the partial pressures of B and C since the solid A does not influence the value of kp
d. doubling the concentration of B and halve the concentration of C
Doubling the concentrantion doubles the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:
Q = pB´x pC´ = ( 2 pB ) x ( 1/2 pC ) = K
e. double the concentrations of both products
We learned that doubling the concentration doubles the pressure so:
Q = pB´x pC´ = ( 2 pB ) x ( 2 pC ) = 4 Kp
Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.
f. double the concentrations of both products and then quadruple the container volume
We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:
Q = pB´x pC´ = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16 Kp = 1/4 Kp
So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.
If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.
Final answer:
The decomposition reaction of A into B and C responds to changes at equilibrium by shifting in a direction that opposes the imposed change, as explained by Le Chatelier's principle. The effects of changes in concentration, container volume, and addition of reactants are used to predict shifts in equilibrium.
Explanation:
Response to Changes in Chemical Equilibrium
For the decomposition of a solid A into gases B and C, represented by the reaction A(s) ⇌ B(g) + C(g), here is how the reaction will respond to changes at equilibrium:
a. Doubling the concentrations of both products B and C and then doubling the container volume will initially cause the reaction to shift to the left (toward A), as the increase in volume will decrease the pressure and favor the formation of fewer gas molecules. However, because the concentrations of the gases are also doubled, the net effect may be smaller, and the system will work to re-establish equilibrium according to Le Chatelier's principle.b. Doubling the container volume alone will shift the equilibrium to the left because the system will try to increase the pressure by forming more gas molecules (favoring the formation of B and C).c. Adding more of the solid reactant A will shift the equilibrium to the right, increasing the production of B and C.d. Doubling the concentration of B and halving the concentration of C will lead to a shift in the equilibrium toward the left, as the system tries to counteract the changes and restore equilibrium.e. Doubling the concentrations of both products, B and C, without changing the volume will shift the equilibrium to the left, as the system tries to reduce the concentrations of the products by forming more A.f. Doubling the concentrations of B and C and then quadrupling the container volume will have a more pronounced effect in shifting the equilibrium to the left than simply doubling the volume, as the change in pressure will be more significant.Each of these actions will induce a response from the reaction to maintain the established equilibrium according to Le Chatelier's principle. The reaction will typically shift in the direction that opposes the change imposed on the system.