Partial molar properties of components in a mixture can vary and be different from their pure-component properties. In the case of an ethanol-water mixture, the partial molar volumes and enthalpies can be calculated using data from the International Critical Tables.
Explanation:Partial molar properties of components in a mixture may differ from their corresponding pure-component properties and can vary with composition. In the case of ethanol-water mixture, the partial molar volumes and enthalpies of ethanol and water can be calculated using data from the International Critical Tables. The partial molar volumes can show how the volume changes when the mole fraction of a component is changed, while the partial molar enthalpies can show the change in enthalpy with respect to mole fraction.
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Final answer:
Partial molar volume and partial molar enthalpy are calculated for components in a mixture and represent the change in volume and enthalpy when one mole of a substance is added to a mixture. They differ from pure substances' molar volumes and enthalpies and may vary with composition.
Explanation:
Partial molar properties help us understand how the addition of a component affects a mixture's overall properties, such as volume and enthalpy. The partial molar volume (VM) is the volume change upon adding one mole of a substance to a mixture, at constant temperature and pressure. It differs from molar volume which is the volume occupied by one mole of a pure substance. The notation ∆V/(∆nB)T,p,nA, represents the partial molar volume of methanol (substance B) with respect to water (substance A). An example is given with values of VA = 17.74 cm³ mol⁻¹ for water and VB = 38.76 cm³ mol⁻¹ for methanol at a certain composition.
Similarly, we can define partial molar enthalpy (HM), which is the enthalpy change upon adding one mole of a substance. This property can be complex, as interactions between molecules can cause it to differ significantly from the pure component's molar enthalpy. Without specific data on ethanol-water mixtures, we can only conclude that like partial molar volume, the partial molar enthalpy would have to be measured or calculated for different compositions and cannot be directly determined from this example.
pOh = 5.3 is it a acid, basic or neutral susbatance
Answer:
The solution is basic
Explanation:
Low pH means that a solution is acidi while low pOH means that a solution is basic.
A rigid tank contains an ideal gas at 40°C that is being stirred by a paddle wheel. The paddle wheel does 150 kJ of work on the ideal gas. It is observed that the temperature of the ideal gas remains constant during this process as a result of heat transfer between the system and the surroundings at 30°C. Determine the entropy change of the ideal gas.
Answer:
Entropy change is zero
Explanation:
∆S=S2-S1=0
We're S is entropy
Metal can not conduct electricity True or False?
Answer:
False
Explanation:
Metals are made up of atoms bonded to each other by metallic bonds. Such bonds have a sea of delocalized electrons, which are free to flow. Electricity requires an unobstructed flow of electrons in order to be able to be conducted. Because all metals provide this inherently, then all metals can conduct electricity, making the statement false.
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Calculate the amount of heat needed to melt of solid octane () and bring it to a temperature of . Round your answer to significant digits. Also, be sure your answer contains a unit symbol.
Answer:
85.0 kJ
Explanation:
Calculate the amount of heat needed to melt 160. g of solid octane (C8H18 ) and bring it to a temperature of 99.2 degrees c. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
To melt 160 g of octane and bring it's temperature to 99.2°C
(from literature)
Heat of fusion of Octane = 20.740 kJ/mol
Melting point of octane = -57°C
Boiling point of Octane = 125.6 °C
Molar mass of octane = 114.23 g/mol
Heat capacity of octane = 255.68 J/K.mol
So, it is evident that Octane is still a liquid at 99.2°C.
So, the required heat is the heat required to melt octane and raise the temperature of Octane liquid to 99.2°C
First, we convert the mass of octane given to number of moles as the heat parameters provided by literature are given in molar units.
Number of moles = (mass)/(Molar mass)
Number of moles of octane = (160/114.23) = 1.401 moles
Heat required to melt the octane = nL = (1.401×20.740) = 29.05674 kJ
Heat required to raise the temperature of already melted octane from its melting temperature of -57°C to 99.2°C
= nCΔT
n = 1.401 moles
C = 255.68 J/K.mol
ΔT = (99.2 - (-57)) = 156.2°C (same as a temperature difference of 156.2 K)
Heat required to raise the temperature of already melted octane from -57°C to 99.2°C
= (1.401×255.68×156.2)
= 55,952.04 J = 55.952 kJ
Total heat required to melt the 160 g of Octane and raise its temperature to 99.2°C
= 29.05674 + 55.952 = 85.01 kJ = 85.0 kJ
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The heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process - melting and then heating. You use the latent heat of fusion and specific heat capacity of octane to calculate the total heat required.
Explanation:
The amount of heat needed to melt the solid octane and bring it to the desired temperature involves a two-step process – melting, and heating. The process should be looked at in separate steps, similar to how the enthalpy of combustion is used to calculate the heat produced in the combustion of 1.00 L of isooctane in the reference provided. Note that different substances have different heat capacities and heat of fusion, so the associated values (likely given in the problem or textbook) for octane specifically would need to be used to get the final answer.
First, you would need to know the latent heat of fusion, which is the amount of energy required to change one kilogram of the substance from solid to liquid without changing its temperature. Then, you would multiply the mass of the octane by the latent heat of fusion to find out how much energy is required for it to go from solid to liquid.
Once the octane is in liquid form, we then need to heat it to the desired temperature. To do this, you use the specific heat capacity formula – the amount of energy required to raise one kilogram of the substance by one degree Celsius. Multiply the mass of the material by the specific heat capacity of the octane and the change in temperature (final temperature – initial temperature).
The added heat required for both of these processes would give you your final answer.
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The surface of a solid sphere is covered by a monolayer of receptors for a ligand. When the ligands diffuse to the surface of the sphere, they are captured instantaneously by the receptors. The ligand diffusion coefficient is 5.2*10-9m2/s. Assume the bulk concentration far away from the sphere is constant of 6.6 nM and the sphere radius is 9.8 μm. (a) Find the ligand concentration C at the location of r=4.9 μm at steady state, where r is the distance to the surface of the sphere. Please enter the numerical value with a unit of nM.
Answer:
The ligand concentration C is 0.274 nM
Explanation:
According to the image, the bulk concentration is inversely proportional to the radius, thus:
[tex]B=\frac{B_{\alpha }r^{2} }{\alpha }[/tex]
Where
Bα = 6.6 nM
r = 14.7 um = 1.47x10⁻⁵m
[tex]B=\frac{6.6*(1.47x10^{-5})^{2} }{5.2x10^{-9} } =0.274nM[/tex]
What is the main greenhouse gas responsible for influencing the world climate
Answer:
carbon dioxide
Explanation:
In all neutral atoms, there are equal numbers of ___________________________. Group of answer choices electrons and protons neutrons and electrons protons and neutrons allotropes and electrons
In a neutral atom, there are equal numbers of electrons and protons, which gives the atom its neutral charge. Neutrons do not factor into the atom's charge, and allotropes are different forms of an element, not part of its atomic structure.
Explanation:In a neutral atom, there are equal numbers of electrons and protons. The number of protons in an atom's nucleus determines the atomic number, and the atom's charge is neutral because the positive charge of the protons is balanced by the negative charge of the electrons. Neutrons, on the other hand, do not carry an electronic charge and are not necessarily equal in number to the protons or electrons. Concerning allotropes, they are different forms of the same element and it's not directly related to this question about atomic structure.
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In neutral atoms, there are equal numbers of electrons and protons. Each element, when neutral, has a characteristic number of electrons that matches its atomic number (the number of protons). Neutrons are electrically neutral and their number does not affect the atom's charge.
Explanation:In all neutral atoms, there are equal numbers of electrons and protons. This is because the number of protons, which carry a positive charge, and the number of electrons, which carry a negative charge, balance each other in a neutral atom, resulting in an overall charge of zero.
Each element, when electrically neutral, has a characteristic number of electrons equal to its atomic number, which is the number of protons it contains. For example, in a carbon-12 atom (which is neutral), there are six protons and six electrons, yielding a net charge of zero.
It's important to note that while neutrons also exist within an atom, they are electrically neutral. Therefore, the number of neutrons does not affect the atom's charge and doesn't necessarily match the number of protons or electrons.
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AB2 is a molecule that reacts readily with water. Calculate the bond energy of the A–B bond using the standard enthalpy of reaction and the bond energy data provided. Enter a number in kJ to 0 decimal places. 2AB2(g) + 2H2O(g) ⟶ O=O(g) + 4HB(g) + A2 ΔH° = –142 kJ
Given question is incomplete. The complete question is as follows.
[tex]AB_{2}[/tex] is a molecule that reacts readily with water. Calculate the bond energy of the A–B bond using the standard enthalpy of reaction and the bond energy data provided. Enter a number in kJ to 0 decimal places.
[tex]2AB_{2}(g) + 2H_{2}O(g) \rightarrow O=O(g) + 4HB(g) + A_{2}[/tex] [tex]\Delta H^{o}[/tex] = –142 kJ
Bond: O–H O=O H–B [tex]A \rightarrow A^{+}[/tex]
Bond energy (kJ/mol): 467 498 450 321
Explanation:
The given reaction is as follows.
[tex]2AB_{2} + 2H_{2}O \rightarrow O_{2} + 4HB + A_{2}[/tex]
Now, we will calculate the enthalpy of reaction as follows.
[tex]\Delta H^{o}_{R}[/tex] = -142 kJ
Also, we know that
[tex]\Delta H^{o}_{R} = \Delta H_{reactants} - \Delta H_{products}[/tex]
= [tex][(2 \times 2 (A-B) + 2 \times 2 (O-H)] - [(O=O) + 4(H-B) + (A-A)][/tex]
-142 = [tex]4(A-B) + 4 \times 467 - 498 - 4(450) - 321[/tex]
[tex]4(A-B)[/tex] = -142 - 1868 + 498 + 1800 + 321
= 609
(A-B) = 152.25 kJ/mol
Thus, we can conclude that the bond energy of the A–B bond is 152.25 kJ/mol.
Be sure to answer all parts. The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2AlCl3(s) Assume that 0.80 g Al is mixed with 0.23 g Cl2. (a) What is the limiting reactant? Cl2 Al (b) What is the maximum amount of AlCl3, in grams, that can be produced? g AlCl3
Answer: a) [tex]Cl_2[/tex] is the limiting reagent
b) 0.27 g of [tex]AlCl_3[/tex] will be produced.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles[/tex]
[tex]\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles[/tex]
[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]
According to stoichiometry :
3 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]Al[/tex]
Thus 0.003 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{3}\times 0.003=0.002moles[/tex] of [tex]Al[/tex]
Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.
b) As 3 moles of [tex]Cl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]
Thus 0.003 moles of [tex]Cl_2[/tex] give =[tex]\frac{2}{3}\times 0.003=0.002moles[/tex] of [tex]AlCl_3[/tex]
Mass of [tex]AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g[/tex]
Thus 0.27 g of [tex]AlCl_3[/tex] will be produced.
Balance the following redox reaction occurring in an acidic solution. The coefficient of Mn2+(aq) is given. Enter the coefficients (integers) into the cells before each substance below the equation. ___ AsO2−(aq) + 3 Mn2+(aq) + ___ H2O(l) \rightarrow→ ___ As(s) + ___ MnO4−(aq) + ___ H+(aq)
Answer:
_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)
Explanation:
Step 1:
The unbalanced equation:
AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
Step 2:
Balancing the equation.
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
The above equation can be balanced as follow:
There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)
There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)
Now the equation is balanced
The balanced redox reaction is
5AsO⁻₂(aq) + 3Mn²⁺(aq) + 2H₂O(l) → 5As(s) + 3MnO₄−(aq) + 4H+(aq)
The unbalanced redox reaction is :
AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
The above equation can be balanced by ensuring the atom of the elements
on the left hand side is equal to those on the right hand side.
We have 3 atoms of Mn on the left side of the equation and 1 atom on the
right hand side.This will be balanced by putting 3 in front of MnO₄− as shown
below:
AsO₂−(aq) + 3 Mn²⁺(aq) + H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
We have 3 atoms of O on the left hand side and 12 atoms of O on the right
hand side. This is balanced by putting 5 in front of AsO₂− and 2 in front of
H₂O as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + H⁺(aq)
We have 4 atoms of H on the left hand side and 1 atom of H on the right
hand side.We can balance by putting 4 in front of H⁺ as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → As(s) + MnO₄−(aq) + 4H⁺(aq)
We have 5 atoms of As on the left hand side and 1 atom of As on the right
hand side. We can balance by putting 5 in front of As as shown below:
5AsO₂−(aq) + 3 Mn²⁺(aq) + 2H₂O(l) → 5As(s) + MnO₄−(aq) + 4H⁺(aq)
The equation is therefore now balanced as the number of atoms of the
element on the left hand side are equal with those on the right hand side.
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Combustion of hydrocarbons such as decane () produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Earth's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide. 1. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid decane into gaseous carbon dioxide and gaseous water. 2. Suppose of decane are burned in air at a pressure of exactly and a temperature of . Calculate the volume of carbon dioxide gas that is produced. Round your answer to significant digits.
The missing part of the question is shown in the image attached
Answer:
C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)
V= 70.4L of CO2
Explanation:
Equation of the reaction is:
C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)
Number of moles of decane = mass/ molar mass
Molar mass of decane= 122gmol-1
n= 0.370×10^3g/122gmol-1= 3.0 moles
T= 13°C +273=286K
P= 1atm
R= 0.082 atmLK-1mol-1
From :
PV= nRT
V= nRT/P
V= 3.0×0.082×286/1
V= 70.4L of CO2
The balanced chemical equation for the combustion of liquid decane [tex](C\_{10}H\_{22}\))[/tex] into gaseous carbon dioxide [tex](CO\_{2}\)[/tex]and gaseous water [tex](H\_{2}O\) is: \[ 2C_{10}H_{22(l)} + 31O_{2(g)} \rightarrow 20CO_{2(g)} + 22H_2O_{(g)} \][/tex]
To calculate the volume of carbon dioxide gas produced when 0.500 moles of decane are burned, we use the stoichiometry of the balanced equation and Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules (or moles). From the balanced equation, 1 mole of decane produces 20 moles of CO\_{2}\. Therefore, 0.500 moles of decane will produce:
[tex]\[ 0.500 \text{ moles } C_{10}H_{22} \times \frac{20 \text{ moles } CO_2}{2 \text{ moles } C_{10}H_{22}} = 5.00 \text{ moles } CO_2 \][/tex]
According to Avogadro's law, at the same temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas. Since the pressure and temperature are not specified in the question, we cannot calculate the exact volume of CO\_{2}\ produced. However, if we assume standard temperature and pressure (STP) conditions, where 1 mole of an ideal gas occupies 22.414 liters, the volume of CO\_{2}\ produced would be:
[tex]\[ 5.00 \text{ moles } CO_2 \times \frac{22.414 \text{ L}}{1 \text{ mole } CO_2} = 112.07 \text{ L} \][/tex]
Since the question specifies that the pressure is exactly 1 atmosphere but does not specify the temperature, we cannot use the value of 22.414 L/mol for the molar volume of CO\_{2}\. We need the temperature to determine the molar volume of CO\_{2}\ at that specific condition. If the temperature is also at STP (0°C or 273.15 K), then the volume calculated above is correct. If the temperature is different, we would need to use the ideal gas law, PV = nRT, to find the volume V, where P is the pressure, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the temperature is not provided, we can only provide the volume of [tex]CO\_{2}\[/tex] at STP, which is 112.07 L, rounded to four significant digits as 112.1 L. If the temperature were provided, we would use the ideal gas law to calculate the volume at that specific temperature and pressure.
An aqueous solution has [C6H5COOH] = 0.110 M and [Ca(C6H5COO)2] = 0.200 M. Ka = 6.3 × 10-5 for C6H5COOH. The solution volume is 5.00 L. What is the pH of the solution after 10.00 mL of 5.00 M NaOH is added? Group of answer choices 4.81 4.86 4.75 4.70 4.65
Answer:
The answer is 4.8659
Explanation:
Given:
[C₆H₅COOH]=0.11M
[C₆H₅COO]=2*0.2=0.4M
Ka=6.3x10⁻⁵
First, calculate the pKa:
[tex]pKa=-logKa=-log(6.3x10^{-5} )=4.2007[/tex]
The pH is:
[tex]pH=pKa+log\frac{C6H5COO]}{[C6H5COOH]} =4.2007+log\frac{0.4}{0.11} =4.7614[/tex]
Like the volume is 5L, the volume of C₆H₅COO is x, then, the volume of C₆H₅COOH is 5-x
[tex]4.7614=4.2007+log\frac{0.4x}{0.11*(5-x)}[/tex]
[tex]0.5607=log\frac{0.4x}{0.11*(5-x)}[/tex]
Solving for x:
x=2.49L=2490mL of C₆H₅COO
2510mL of C₆H₅COOH
The milimoles of C₆H₅COOH and C₆H₅COO is:
nC₆H₅COOH=(0.11*2510)-50=226.1mmol
nC₆H₅COO=(0.4*2490)+50=1046mmol
The pH is:
[tex]pH=4.2007+log\frac{1046}{226.1} =4.8659[/tex]
Answer:
4.86
Explanation:
The pH of a solution is a measure of the molar concentration of hydrogen ions in the solution and as such is a measure of the acidity or basicity of the solution.
Please kindly check attachment for the step by step solution of the given problem.
Because of the radioactive decay of uranium and thorium in rocks and soil, radium-228, a decay product of Thorium-232, can be found in drinking water. This isotope has a half-life of 5.75 years and an atomic number of 88. If Ra-228 undergoes beta decay, what would the atomic number of the new element be? What would the mass number of this isotope be? Explain your reasoning (e.g. Explain what happens during beta decay).
Answer: The atomic number and mass number of the new element formed is 89 and 228 respectively.
Explanation:
Beta decay is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron. The released beta particle is also known as electron.
In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]
The chemical equation for the beta decay of Ra-228 follows:
[tex]_{88}^{228}\tyextrm{Ra}\rightarrow _{89}^{228}\textrm{Ac}+_{-1}^0\beta[/tex]
Hence, the atomic number and mass number of the new element formed is 89 and 228 respectively.
At a given temperature, the elementary reaction A − ⇀ ↽ − B , A↽−−⇀B, in the forward direction, is first order in A A with a rate constant of 0.0100 s − 1 . 0.0100 s−1. The reverse reaction is first order in B B and the rate constant is 0.0610 s − 1 . 0.0610 s−1. What is the value of the equilibrium constant for the reaction A − ⇀ ↽ − B A↽−−⇀B at this temperature
Answer:
The equilibrium constant for the reversible reaction = 0.0164
Explanation:
At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.
The reaction is given as
A ⇌ B
Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]
The rate of forward reaction = |r₁| = k₁ [A]
The rate of backward reaction = |r₂| = k₂ [B]
(Taking only the magnitudes)
where k₁ and k₂ are the forward and backward rate constants respectively.
k₁ = 0.010 s⁻¹
k₂ = 0.0610 s⁻¹
|r₁| = 0.010 [A]
|r₂| = 0.016 [B]
At equilibrium, the rate of forward and backward reactions are equal
|r₁| = |r₂|
k₁ [A] = k₂ [B] (eqn 1)
Note that equilibrium constant, K, is given as
K = [B]/[A]
So, from eqn 1
k₁ [A] = k₂ [B]
[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164
K = [B]/[A] = (k₁/k₂) = 0.0164
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Which best describes the bacteria discovered by the scientist? They are probably archaebacteria because they are autotrophs. They are probably eubacteria because they are autotrophs. They are probably eubacteria because the colony lives in an extreme environment. They are probably archaebacteria because the colony lives in an extreme environment.
Answer:
They are probably archeabacteria because the colony lives in an extreme environment.
Explanation:
Archeabacteria are singled cell microorganisms that live in extreme environment. They are found in hot springs, salt lakes, oceans, soils and marshlands. They posses different shapes like rods, spheres, spiral and plates. Thermophiles, halophiles, and methanogens are the three types of archeabacteria.
Eubacteria are microorganisms that are found in most of the earth's habitats like soil, water, etc. They are have different shapes like cocci, bacilli, filaments, vibro,etc. They do not live in extreme environment unlike the archeabacteria. This is the major difference between the archeabacteria and eubacteria.
Both archeabacteria and eubacteria
are prokaryotes. Archeabacteria can both be autotrophic or heterotrophic and can live in places without oxygen. Some eubacteria are autotrophs and some are heterotrophs.
Answer:
They are probably archaebacteria because the colony lives in an extreme environment.
Explanation:
In the PhET simulation window, click the Macro menu in the top left corner of the screen. This view gives a view of the beaker at a macroscopic level (as your naked eye would see it). The Micro menu shows what happens to sugars and salts at the molecular level when they dissolve in water (note that you can use the arrows to switch to other type of solutes). Use both the Macro and Micro menus in the PhET simulation to help complete the following statements regarding solutions.
Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
Hints
Help
Reset
ions
solute
negatively
opposite
Partical Charges
positively
NaCl
sugar
CaCl2
1. Pure water contains only water molecules that interact strongly with each other due to their ___________, which are graphically depicted as δ+ and δ−.
2. Solutions are formed when a _______ like a salt or sugar becomes homogeneously distributed in a solvent like water, and this distribution can be viewed in the Micro view.
3. When salts dissolve, they separate into individua________that strongly interact with the water molecules.
4. Binary salts are made up of two elements at varying ratios, where one element is a charged cation, and the other is a_________ charged anion.
5. In the Micro view, each shake of the container releases 6 molecules of its respective solute, but 6 molecules of the salt________actually produce more ions in solution than 6 molecules of the salt__________.
6. Not all soluble molecules are salts, e.g., a covalent species like__________readily dissolves in water without forming ions.
7. The reason they dissolve is because their partially charged atoms are able to associate with the partially charged atoms of water molecules, and these attractive forces occur as long as they are between atoms with_________charges.
This question is about the Macro and Micro views of the PhET simulation for solutions and the concepts related to dissolving salts and sugars in water.
Explanation:1. Pure water contains only water molecules that interact strongly with each other due to their partial charges, which are graphically depicted as δ+ and δ−.
2. Solutions are formed when a solute like a salt or sugar becomes homogeneously distributed in a solvent like water, and this distribution can be viewed in the Micro view.
3. When salts dissolve, they separate into individual ions that strongly interact with the water molecules.
4. Binary salts are made up of two elements at varying ratios, where one element is a charged cation, and the other is a negatively charged anion.
5. In the Micro view, each shake of the container releases 6 molecules of its respective solute, but 6 molecules of the salt NaCl actually produce more ions in solution than 6 molecules of the salt CaCl2.
6. Not all soluble molecules are salts, e.g., a covalent species like sugar readily dissolves in water without forming ions.
7. The reason they dissolve is because their partially charged atoms are able to associate with the partially charged atoms of water molecules, and these attractive forces occur as long as they are between atoms with opposite charges.
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This answer explains the properties of solutions, including partial charges and the formation of ions. It also provides examples of different solutes and their behavior in water.
Explanation:1. Pure water contains only water molecules that interact strongly with each other due to their partial charges, which are graphically depicted as δ+ and δ−.
2. Solutions are formed when a solute like a salt or sugar becomes homogeneously distributed in a solvent like water, and this distribution can be viewed in the Micro view.
3. When salts dissolve, they separate into individual ions that strongly interact with the water molecules.
4. Binary salts are made up of two elements at varying ratios, where one element is a charged cation, and the other is a negatively charged anion.
5. In the Micro view, each shake of the container releases 6 molecules of its respective solute, but 6 molecules of the salt NaCl actually produce more ions in solution than 6 molecules of the salt sugar.
6. Not all soluble molecules are salts, e.g., a covalent species like sugar readily dissolves in water without forming ions.
7. The reason they dissolve is because their partially charged atoms are able to associate with the partially charged atoms of water molecules, and these attractive forces occur as long as they are between atoms with opposite charges.
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Consider the following data on some weak acids and weak bases:
acid base
Ka name formula Kb name formula
6.8x10-4 hydrofluoric acid HF 1.8x10-5 ammonia NH3
4.9x10-10 hydrocyanic acid HCN 1.7x10-9 pyridine C5H5N
Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on.
solution
a.0.1 M NaCN
b.0.1M NH4Br
c.0.1M KF
d.0.1M KBr
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Final answer:
The question involves ranking solutions of weak acids and bases by their pH. After assessing the ionization and hydrolysis of the compounds, the order from lowest to highest pH is 0.1 M NH4Br, 0.1 M NaCN, 0.1 M KF, and 0.1 M KBr.
Explanation:
To rank the solutions in order of increasing pH, we must consider the ionization of each compound in water and the corresponding Ka or Kb values. A lower Ka or a higher Kb value usually results in a higher pH.
0.1 M KF: KF will form from the dissolution of a weak acid (HF) and a strong base. The resulting F- ions will hydrolyze to form OH- ions, creating a basic solution with a pH higher than 7 but not as high as that of a strong base.
Therefore, the order from lowest to highest pH should be b, a, c, d.
What is the hydrogen-ion concentration of the ph is 3.7
Answer:
the answer 37
Explanation:
Identify the option below that exemplifies how the intrinsic chemical properties of a reactant affect the rate of reaction. Select the correct answer below: When exposed to air, sodium reacts completely overnight, whereas iron barely reacts. Five 4-inch diameter saplings burn more rapidly than a 20-inch tree of the same height. Food spoils more quickly on a kitchen counter than in a refrigerator. A Bunsen burner is used in the laboratory to increase the speed of reactions.
The intrinsic chemical properties of a reactant can affect the rate of reaction, as seen in the example of food spoiling more quickly on a kitchen counter than in a refrigerator due to temperature differences.
Explanation:The option that exemplifies how the intrinsic chemical properties of a reactant affect the rate of reaction is Food spoils more quickly on a kitchen counter than in a refrigerator. This is because the temperature difference between the kitchen counter and the refrigerator affects the rate of chemical reactions. Chemical reactions typically occur faster at higher temperatures, so food spoils more quickly when exposed to the higher temperature of the kitchen counter compared to the lower temperature inside the refrigerator.
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What is the pH of a solution with hydroxide ion concentration of 0.005
Answer:
pH = 11.7
Explanation:
pOH= -log [OH]=-log[0.05]
=2.3
pOH+pH= 14
pH= 14-2.3= 11.7
A gaseous hydrocarbon (a compound that contains only hydrogen and carbon) is found to be 11 % hydrogen by mass. a. Find the empirical formula for the compound.A gaseous hydrocarbon (a compound that contains only hydrogen and carbon) is found to be 11 % hydrogen by mass. a. Find the empirical formula for the compound.
Final answer:
To find the empirical formula of the gaseous hydrocarbon, assume 100 grams of the compound. Since it is 11% hydrogen by mass, it contains 11 grams of hydrogen. The remaining mass is carbon. Divide the moles of each element by the smallest number of moles to get the empirical formula CH₂. The molecular formula cannot be determined without the molar mass.
Explanation:
Empirical Formula Calculation
To find the empirical formula, assume 100 grams of the gaseous hydrocarbon. Since it is 11% hydrogen by mass, this means it contains 11 grams of hydrogen. The remaining mass, 89 grams, must be carbon. The molar mass of hydrogen is 1 g/mol, and the molar mass of carbon is 12 g/mol. Divide the moles of each element by the smallest number of moles to get the empirical formula. In this case, 11 g H / 1 g/mol = 11 mol H and 89 g C / 12 g/mol = 7.42 mol C. Dividing both by the smallest number of moles, we get the empirical formula CH₂.
To determine the molecular formula, you'll need the molar mass of the compound. However, this information is not provided in the question. Without the molar mass, it is not possible to determine the molecular formula.
Which of the following series of scientific steps is most likely in the correct order?
Question 6 options:
observation, theory, experiment
scientific law, experiment, theory
observation, hypothesis, experiment
hypothesis, theory, experiment
Answer:
C
Explanation: you have to observe to have an educated guess about something you test it
Which statements describe the death of stars? Check all that apply.
A. Astar's final stages of life depend on its mass.
Stars die when they run out of fuel.
B. When a star starts to die, its core expands.
C. The outer portion of the star contracts.
D. All stars become black holes when they die.
Answer:A and B
Explanation:a stats final stages of life depend on its mass and stars die when they run out of fuel
write a balanced equation for the reaction between dichromate Ion and iron 2 to yield iron 3 and chromium 3 in acidic solution Cr2O72- + Fe2+ -> Cr3+ + Fe3+
Answer:
Cr2O72- + 5Fe2+ +14H+ -> 2Cr3+ + 5Fe3+ +7H2O
Explanation:
Fe2+ is oxidized to Fe3+ and Cr6+ is reduced to Cr3+
Final answer:
Explains the balanced redox reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution.
Explanation:
Balanced Equation: 7H₂O(l) + 6Fe²+(aq) + Cr₂O72-(aq) → 6Fe³+(aq) + 2Cr³+(aq) + 14H+(aq)
Explanation: The reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution is represented by this balanced redox equation. By balancing the atoms and charges on both sides, the equation shows the transformation of Fe²+ to Fe³+ and Cr2O7²- to Cr³+.
Example: In a similar process, Cr atoms are oxidized to form Cr³+ while Fe atoms are reduced to Fe³+, showcasing the transfer of electrons in the redox reaction.
If a solution containing 118.08 g of mercury(II) chlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will form
Answer:
49.544 g.
Explanation:
The balanced equation of reaction is given below;
Hg(ClO3)2 (aq) + Na2S (aq) --------> 2 NaClO3 (aq) + HgS (s).
So, the parameters given in the question are; Mass of Hg(ClO3) = 118.08 and the mass of Na2S = 16.642 g.
Therefore, the first thing we are going to be looking at is the reactant which is the limiting reagent from the number of moles
(1). For Hg(ClO3), the molar mass = 367.5 g/mol. Therefore, the number of moles, n = mass/ molar mass.
Number of moles = 118.08/ 367.5.
Number of moles = 0.321 moles.
(2). For Na2S, the molar mass = 78.05 g /mol.
The number of moles = 16.643 / 78.05.
The number of moles= 0.213.
Therefore, the limiting reagent = Na2S.
This means that the excess reagent is Hg(ClO3).
==> In the balanced equation of reaction above, the solid precipitate = HgS.
Hence, the mass of HgS formed = 0.213 × 232.6 g/mol. = 49.544 g.
This problem involves calculating the mass of a precipitate from a doubly replacement reaction. Sodium sulfide is the limiting reactant and by using stoichiometry, the moles of sodium sulfide will amount to the same moles of HgS, the precipitate. Multiplying the moles of HgS by its molar mass gives the mass of the precipitate.
Explanation:This question requires a two-step process that starts with writing balanced chemical equations for the reaction and then using stoichiometry to calculate the mass of the precipitate. The balanced reaction is 2Na2S + Hg2(ClO3)2 -> 2NaClO3 + Hg2S2. Using stoichiometry and details given in the question, you can calculate the molar mass of the substances and find the limiting reactant.
Mercury(II) chlorate (Hg2(ClO3)2) has a molar mass of about 511.63 g/mol. Sodium sulfide (Na2S) has a molar mass of about 78.04 g/mol. Using stoichiometry, it's clear that sodium sulfide is the limiting reactant.
Using the moles of sodium sulfide calculated, you find that the reaction will produce the same number of moles of HgS as a precipitate. By multiplying this number by the molar mass of HgS (232.66 g/mol), you obtain the mass of the precipitate that will be formed. Deduce the exact values yourself for a better understanding of this chemical reaction involving stoichiometry and solubility product.
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Liquefied natural gas (LNG) is transported in very large tankers, stored as liquid in equilibrium with its vapor at approximately atmospheric pressure. If LNG is essentially pure methane, the storage temperature then is about 111.4 K, the normal boiling point of methane. The enormous amount of cold liquid can in principle serve as a heat sink for an onboard heat engine. Energy discarded to the LNG serves for its vaporization. If the heat source is ambient air at 300 K, and if the efficiency of a heat engine is 61% of its Carnot value, estimate the vaporization rate in moles vaporized per kJ of power output. For methane,
Answer:
0.2 mol/kJ
Explanation:
Methane is stored at a temperature of 111.4K, [tex]T_c[/tex]. The heat source to vapourization of methane is ambient air which is at 300 K. [tex]T_H[/tex]
Estimate the vaporization rate at the efficiency of heat engine 60% of its carnot value.
Calculate the vaporization rate from the given data by relation shown below:
[tex]Vaporization rate = \frac{Q_c}{[\frac{\delta H_n^{lv}}{W}]} ..........(1)[/tex]
here,
[tex]Q_c[/tex] is the heat at temperature [tex]T_c, \delta H_n^{lv}[/tex] is the phase transition enthalpy of methane and W is the work
Calculate [tex]Q_c[/tex] from the equation shown below:
[tex]Q_c = Q_n (1 - \eta_{HE}) .............(2)[/tex]
where Q_n is the heat at temperature of [tex]T_n[/tex] and [tex]\eta_{HE}[/tex] is the efficiency of heat engine
calculate [tex]Q_H[/tex] from the relation shown below:
[tex]Q_H = \frac{W}{\eta_{HE}} ..........(3)[/tex]
calculate the heat engine efficiency from the given carnot engine efficiency as shown below
[tex]\eta_{HE} = 0.6 \times \eta_{carnot} ............(4)[/tex]
here, [tex]\eta_{carnot}[/tex] is the carnot engine efficiency
[tex]\eta{carnot} = 1 - \frac{T_c}{T_H}[/tex]
substituting the values of temperature, we have
[tex]= 1 - \frac{111.4K}{300K}\\= 0.629\\[/tex]
substitute values of [tex]\eta_{carnot}[/tex] in equation 4, we get
[tex]\eta_{HE} =0.6 \times \eta_{carnot}\\ = 0.6 \times 0.629\\ = 0.3774\\\\[/tex]
check the attached file for additional solution
If the pH is 10 what is the concentration of hydroxide ion
Answer:
[OH-] = 10^-4 M
Explanation:
pOH= 14- pH= 14-10=4
[OH-]= antilog (-4)= 10^-4M
A rectangular corral of widths Lx = L and Ly = 2L contains seven electrons. What is the energy of (a) the first excited state, (b) the second excited state, and (c) the third excited state of the system? Assume that the electrons do not interact with one another, and do not neglect spin. State your answers in terms of the given variables, using h and me (electron mass) when needed.
Answer:
See explaination
Explanation:
The invariant mass of an electron is approximately9. 109×10−31 kilograms, or5. 489×10−4 atomic mass units. On the basis of Einstein's principle of mass–energy equivalence, this mass corresponds to a rest energy of 0.511 MeV.
Check attachment for further solution to the exercise.
The solubility of aspirin in water is 1 g per 300 mL at 25 degrees celsius. Assuming that your crystallization and washing with water were done at this temperature, what weight of aspirin did you lose in the filtrate and washings? How much was your percent yield lowered by this loss?
Answer:
The weight of aspirin lost is [tex]W =0.03267(V_A +V_B) N[/tex]
The percent yield is lowered by [tex]A = \frac{Amount \ of \ aspirin \ obtained - Lost Amount }{Expected\ Amount\ of \ Aspirin } *\frac{100}{1}[/tex]
Explanation:
From the question we are told that
Since the solubility of aspirin in water is 1 g per 300 mL it implies that after crystallization the solution would contain 1 g for every 300mL of water at 25°
Let assume that the volume of the solution is [tex]V_A[/tex]
The the aspirin lost after filtration would be
[tex]= \frac{1}{300} * V_A[/tex]
[tex]=\frac{V_A}{300}g[/tex]
Let assume that you used water of volume [tex]V_B[/tex] to wash the crystallized aspirin then the lost during washing would be
[tex]= \frac{1}{300} * V_B[/tex]
[tex]= \frac{V_B}{300}[/tex]
So the total loss is
[tex]= \frac{V_A}{300} + \frac{V_B}{300}[/tex]
[tex]\frac{V_A +V_B}{300}g[/tex]
So the weight of aspirin lost denoted by W is
[tex]W = \frac{V_A +V_B}{300} *9.8[/tex]
[tex]W =0.03267(V_A +V_B) N[/tex]
Let denote How much was your percent yield lowered by this loss by A
So
[tex]A = \frac{Amount \ of \ aspirin \ obtained - Lost Amount }{Expected\ Amount\ of \ Aspirin } *\frac{100}{1}[/tex]
Nylon-6 is a polymer constructed by a ring-opening polymerization, unlike most nylon polymers, which are synthesized using condensation polymerization. Use the structure of Nylon-6 to deduce and draw its cyclic monomer.
Answer: The monomer of nylon-6 is caprolactam.
Explanation:
From the question, we are asked to draw the cyclic monomer from the structure of Nylon-6.
Two repeating monomer units can be seen in the polymeric structure given in the problem.
From this we can see and conclude that The monomer of nylon-6 is caprolactam.
attached is a diagrammatic representation of the Structure (Nylon-6 monomer).
cheers i hope this helps!!!!
Final answer:
Nylon-6 is synthesized from the cyclic monomer epsilon-caprolactam through a process called ring-opening polymerization, unlike other nylons that are produced by condensation polymerization. Epsilon-caprolactam features a 6-carbon ring with an amide linkage, making it unique.
Explanation:
Nylon-6 is a unique type of polymer that is derived from a cyclic monomer known as epsilon-caprolactam. Unlike the Nylon 6,6 polymer, which is made from condensation polymerization of hexanedioic acid and 1,6-diaminohexane, Nylon-6 utilizes a process called ring-opening polymerization. This process involves breaking open the ring structure of the epsilon-caprolactam to form linear chains, which then link together head-to-tail to form the polymer. This is a key difference as it does not release a small molecule (like water or hydrogen chloride) during the polymerization process, making it a growth polymerization.
Epsilon-caprolactam, the monomer for Nylon-6, is a 6-carbon cyclic amide. To deduce and draw its cyclic monomer, one should recognize that the repeating unit in Nylon-6 consists of 6 carbons in a chain with an amide linkage connecting the ends. Thus, depicting epsilon-caprolactam involves drawing a ring structure containing 5 methylene (-CH2-) groups and one carbonyl group (=O) adjacent to an NH group, forming a lactam.