Dmitri Mendeleev was the Russian scientist who first arranged the elements in the periodic table. He arranged elements in the periodic table by their atomic mass, and he also made sure that elements with similar properties were placed in the same group.
The periodic table is a tabular representation of the chemical elements, which are arranged by atomic number, electron configuration, and chemical properties. The rows of the periodic table are known as periods, and the columns are known as groups or families. Elements in the same group have similar chemical and physical properties.
Mendeleev's contributions to the periodic table
Mendeleev was a Russian chemist who published the first widely recognized periodic table in 1869. In the periodic table, Mendeleev arranged the elements according to their atomic mass. He also left gaps in the periodic table for unknown elements, and he predicted their properties based on the properties of the known elements.
For example, he predicted the properties of germanium, which was discovered later, and he even named it. He was also able to predict the existence and properties of some of the noble gases.
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A hard-working human brain, perhaps one that is grappling with physical chemistry, operates at about 25 W (1 W = 1J s-'). What mass of glucose must be consumed to sustain that power output for an hour?
Approximately 5.78 grams of glucose must be consumed to sustain a power output of 25 W for one hour.
Power = Energy/Time
25 W = Energy/3600 s
Energy = 25 W x 3600 s = 90000 J
C6H12O6 + 6O2 → 6CO2 + 6H2O + energy
The energy produced by the complete oxidation of glucose is approximately 2.8 x 10^6 J/mol. Therefore, to produce 90,000 J of energy, we need to divide 90,000 J by the energy produced per mole of glucose:
90,000 J / (2.8 x 10^6 J/mol) = 0.0321 mol
The molar mass of glucose is approximately 180 g/mol. Therefore, the mass of glucose required to sustain a power output of 25 W for one hour is:
0.0321 mol x 180 g/mol = 5.78 g
Power in physics is defined as the rate at which work is done or energy is transferred. It is a scalar quantity that measures how quickly a certain amount of energy is being transferred or converted from one form to another. The standard unit for power is the watt (W), which is equivalent to one joule per second (J/s).
In more mathematical terms, power is given by the formula P = W/t, where P represents power, W represents work, and t represents time. Power is also related to force and velocity through the equation P = Fv, where F represents force and v represents the velocity.
Power is an important concept in physics and engineering, as it is used to describe the performance of machines, engines, and other energy conversion systems. The greater the power of a system, the more work it can do in a given amount of time, and the faster it can accomplish a task.
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The codons in mRNA specify the amino acids that are used to make a protein. Mark the following statement concerning translation TRUE or FALSE.
TRUE. The codons in mRNA, which are collections of three nucleotides, stand for certain amino acids that are combined to produce proteins during translation.
In order to create a protein, the information contained in mRNA must be deciphered during the process of translation. The genetic code that regulates the order in which amino acids are put together to make proteins is found in the sequence of nucleotides in mRNA known as codons. A codon is made up of three nucleotides, each of which stands for an amino acid or a stop signal that denotes the completion of protein synthesis. The ribosome scans the mRNA's codon sequence during translation and matches each codon with the appropriate amino acid. A functional protein is produced when a chain of amino acids that have been joined together by peptide bonds folds into a three-dimensional structure. Hence, the codons in mRNA play a critical role in determining the amino acid sequence of a protein.
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citation chaining is a process for finding more articles that may be relevant for your research topic. which of these would be a good starting point for this process?
A good starting point for citation chaining would be a relevant and well-cited article or book that directly relates to your research the topic.
This article or book should have a comprehensive bibliography or the reference list that you can use to find additional sources. By examining the references cited in the original article, you can identify the other articles and books that are likely to be relevant to your research. Then, you can examine the references in those articles to find even more sources, continuing the process until you have a comprehensive set of relevant sources for your research.
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1. calculate the final concentration of sodium azide and dcmu in the locomotion chambers. show your work. The DCMU is a 10mM concentrationThe Sodium Azide is a 1M concentration.If you add 10mL of the Chlamydomonas, 100 microliters of sterile water, and 100 microliters of 10mM DCMU what is the final concentration of DCMU?If you add 10mL of the Chlamydomonas, 100 microliters of sterile water, and 100 microliters of 1M Sodium Azide what is the final concentration of DCMU?
The final concentration of DCMU in the locomotion chambers will be 0.1 mM. If 10mL of the Chlamydomonas, 100 microliters of sterile water, and 100 microliters of 10mM DCMU is added.
To Calculate the final concentration of Sodium Azide and DCMU in the locomotion chambers. The final concentration of Sodium Azide in the locomotion chambers will be 10mM (millimolar) if 10mL (milliliters) of the Chlamydomonas, 100 μL (microliters) of sterile water, and 100 μL of 1M (molar) Sodium Azide is added.
The final concentration of DCMU (3-(3,4-dichlorophenyl)-1,1-dimethylurea) in the locomotion chambers will be 0.1 mM (millimolar) if 10 mL (milliliters) of the Chlamydomonas, 100 μL (microliters) of sterile water, and 100 μL of 10 mM (millimolar) DCMU are added.
Calculating the final concentration of DCMU:
Formula: C1V1 = C2V2C1 = initial concentration of DCMU = 10 mMV1 = volume of DCMU added = 100 μL (microliters)C2 = final concentration of DCMU = ?V2 = final volume = 10 mL + 100 μL + 100 μL = 10.2 mL
(convert 100 μL to mL by dividing it by 1000)
Substituting the values in the formula:
C1V1 = C2V210 mM x 100 μL = C2 x 10.2 mL1000 (since 1 mL = 1000 μL)C2 = 0.098 mM (millimolar) = 0.1 mM (approx.)
Thus, the final concentration of DCMU in the locomotion chambers will be 0.1 mM if 10mL of the Chlamydomonas, 100 microliters of sterile water, and 100 microliters of 10mM DCMU is added.
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if the density of a gas is 1.87 grams/liter at 34.0 c and 745 mm hg, what will be its density at 84.0 c and 721 mm hg?
The density of the gas at 84° C and 721 mm Hg will be 2.50 g/L.
The density of a gas can be calculated using the following formula:
Density = (Pressure x Molar Mass) / (Gas Constant x Temperature)
Where, Density is the density of the gas in grams per liter. Pressure is the pressure of the gas in millimeters of mercury (mm Hg). Molar mass is the molar mass of the gas in grams per mole. Gas constant is the universal gas constant (0.08206 L atm / mole K). Temperature is the temperature of the gas in kelvin (K).
Now, let's find the density of the gas at 34° C and 745 mm Hg. The temperature should be converted from Celsius to Kelvin. Temperature (K) = 34 + 273 = 307 K
Density = (Pressure x Molar Mass) / (Gas Constant x Temperature)
Density = (745 x Molar Mass) / (0.08206 x 307)
Density = 28.91 x Molar Mass g/L
Also, we need to find the molar mass of the gas. Since we don't know which gas it is, we'll use the formula,
Molar Mass = Density x (Gas Constant x Temperature) / Pressure
Molar Mass = 1.87 x (0.08206 x 307) / 745
Molar Mass = 0.103 g/mol
Now, we can find the density of the gas at 84° C and 721 mm Hg.
Temperature (K) = 84 + 273 = 357 K
Density = (Pressure x Molar Mass) / (Gas Constant x Temperature)
Density = (721 x 0.103) / (0.08206 x 357)
Density = 2.50 g/L
Therefore, the density of the gas at 84° C and 721 mm Hg will be 2.50 g/L.
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What is [Al(H2O)5(OH) 2+] in a 0. 15 M solution of Al(NO3)3 that contains enough of the strong acid HNO3 to bring [H3O +] to 0. 10 M?
Al(NO3)3 solution concentration and the concentration of H3O+ ions in the solution following the addition of HNO3 are given in the problem. We can determine the presence of [Al(H2O)5(OH)2+] in the solution using this knowledge along with the known equilibria for the hydrolysis of Al3+.
For Al3+, the hydrolysis process may be expressed as follows:
Al(H2O)63+ + water becomes Al(H2O)5(OH)2+ + H3O+.
The reaction's equilibrium constant expression is as follows:
Al(H2O)5(OH)2+) = K
Al(H2O)63+ / [H3O+]
We must take into account the dissociation of Al(NO3)3 in water in order to determine [Al(H2O)5(OH)2+] in a 0.15 M solution of Al(NO3)3:
Al3+ (aq) + 3NO3- Al(NO3)3 (s) (aq)
Al3+ has a concentration of 0.45 M (3 times that of the Al(NO3)3 solution) in an Al(NO3)3 solution with a concentration of 0.15 M. H3O+ is present in the solution at a concentration of 0.10 M.
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A student makes three plots of their data and finds that a plot of [A] vs t is linear, a plot of ln[A] vs t is non-linear, and a plot of 1/[A] vs t is non-linear. What is the rate law of the reaction? Rate = k Rate = k[A] Rate = k[A]2 Rate = k[A]3
A student makes three plots of their data and finds that a plot of [A] vs t is linear, a plot of ln[A] vs t is non-linear, and a plot of 1/[A] vs t is non-linear. The rate law of the reaction is b. Rate = k[A]
The given question is related to the rate law of the reaction. The student makes three plots of their data and finds that a plot of [A] vs t is linear, a plot of ln[A] vs t is non-linear, and a plot of 1/[A] vs t is non-linear. The rate law of a reaction is a mathematical equation that relates the rate of the reaction to the concentrations of reactants and the reaction's constant of proportionality. The rate law is also called the rate equation or rate expression.
As per the given information, the plot of [A] vs t is linear, which means that the reaction is a first-order reaction. The plot of ln[A] vs t is non-linear, which means that the reaction is not zero-order or first-order. It could be a second-order or third-order reaction. The plot of 1/[A] vs t is non-linear, which means that the reaction is not a first-order reaction. It could be a second-order or third-order reaction. Therefore, the rate law of the reaction can be given as Rate = k[A]. This represents a first-order reaction. Hence, the correct option is Rate = k[A].
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suppose you experimentally calculate the value of the density of co2 as 2.03 g/l. the known value is 1.98 g/l. what is the percent error of your experimentally determined density?
The percent error of your experimentally determined density is that is an error of 2.53%.
It can be calculated using the following equation: Error % = (Experimentally Determined Value - Known Value)/Known Value x 100. So in your case, the equation would look like: Error % = (2.03 g/l - 1.98 g/l)/1.98 g/l x 100
This gives us an error of 2.53%.
The given value of density of CO2 is 2.03 g/L and the actual value of density of CO2 is 1.98 g/L. The percent error can be calculated using the below formula: Percent error = (|experimental value - actual value|/actual value) × 100Therefore, the percent error of experimentally determined density is Percent error = (|2.03 g/L - 1.98 g/L|/1.98 g/L) × 100= (0.05 g/L/1.98 g/L) × 100= 2.53%Thus, the percent error of the experimentally determined density is 2.53%.
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How many milliliters of 1.58 M HCl are needed to react completely with 23.2 g of NaHCO3 (= 84.02 g/mol)?
HCl(aq) + NaHCO3(s) ? NaCl(s) + H2O(l) + CO2(g)
a. 175 mL
b. 536 mL
c. 276 mL
d. 572 mL
e. 638 mL
c. 276 mL of 1.58 M HCl.
To answer this question, we need to use the mole ratio between the two reactants: 1 mole of HCl for every 1 mole of NaHCO3.
In this case, we need 23.2 g of NaHCO3, which is equal to 0.273 moles (23.2 g / 84.02 g/mol).
Since we need 1 mole of HCl for every 1 mole of NaHCO3, we can calculate the number of moles of HCl needed with the following equation: 0.273 moles of NaHCO3 x 1 mole HCl/1 mole NaHCO3 = 0.273 moles of HCl.
Now we can use the molarity of HCl (1.58 M) to calculate the volume of HCl needed. 1.58 M HCl x 0.273 moles HCl/1 L HCl = 0.433 L HCl, or 433 mL of HCl. Therefore, the correct answer is c. 276 mL of 1.58 M HCl.
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Diborane, B2H6, is a useful reagent in organic chemistry. One of the several ways it can be prepared is by the following reaction.
2 NaBH4(aq) + H2SO4(aq) 2 H2(g) + Na2SO4(aq) + B2H6(g)
What volume of 0.0865 M H2SO4, in milliliters, should be used to consume completely 1.05 g of NaBH4?
What mass of B2H6 can be obtained?
Answer:
Diborane, B2H6, is a useful reagent in organic chemistry. One of the several ways it can be prepared is by the following reaction 2 NaBH4(aq) H2SO4(aq) 2 H2 (g) + Na2SO4(aq) + B2H6(g) What volume of 0.0915 M H2SO4, in milliliters, should be used to consume completely 1.35 g of NaBH4? mL 200 What mass of B2H6 can be obtained? 0.51
Explanation:
hope its help
Describe a hybridization scheme for the central atom and molecular geometry of the triiodide ion,
Answer:
Explanation:
I_3^−
The triiodide ion, I3−, is a polyatomic anion composed of three iodine atoms. It has a central iodine atom, which is surrounded by two other iodine atoms in a trigonal planar geometry. The hybridization of the central atom is sp2. This is because the central atom has 3 electron pairs in its valence shell, which means it needs to form three bonds with the other atoms. This requires the central atom to use one s-orbital and two p-orbitals to form three sp2 hybrid orbitals. These three sp2 orbitals are then used to form the three bonds with the other two iodine atoms, resulting in a trigonal planar geometry.
When a utensil is stored in water between uses, what are the requirements?A. Running water at any temperature, or a container of water at 70 F (21 C) or lower.B. Running water at any temperature, or a container of water at 135 F (57 C) or lower.C. Running water at 70 F (21 C) or lower, or a container of water at 70 F (21 C) or lower.D. Running water at 135 F (57 C) or lower, or a container of water at 135 F (57 C) or lower.
D. Running water at 135 F (57 C) or lower, or a container of water at 135 F (57 C) or lower.
At the resting membrane potential, the membrane is most permeable to ________, which moves ________ the cell due to its A) chloride : into B) potassium : into C) sodium : out of D) sodium : into E) potassium : out of
At the resting membrane potential, the membrane is most permeable to potassium ions (K+), which move out of the cell due to its concentration gradient and the negative charge inside the cell. Correct answer is option: E.
This movement of K+ ions out of the cell contributes to the negative resting membrane potential of approximately -70 mV in most cells. The resting membrane potential is maintained by the selective permeability of the cell membrane, which allows for the movement of certain ions across the membrane. In general, the membrane is less permeable to sodium (Na+) and chloride (Cl-) ions at rest, and the movement of these ions across the membrane is limited. Thus, option E "potassium" is the correct answer.
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True or False: The zeolite that you will make and use has repeating and alternating tetrahedral units of SiO4 and AlO4 bonding through the oxygen atoms.
The zeolite that you will make and use has repeating and alternating tetrahedral units of SiO4 and AlO4 bonding through the oxygen atoms. Therefore, the given statement is true.
Zeolites have repeating and alternating tetrahedral units of SiO4 and AlO4 bonding through the oxygen atoms.Zeolites are aluminosilicate minerals that are mostly found in volcanic rocks and soils.
They have a distinctive and extensive network of pores and channels. Zeolites are also used in ion exchange, adsorption, and catalysis processes as a result of their porous and chemically active structure. Zeolites are extensively employed in the separation, adsorption, and catalytic conversion of petroleum-based products, as well as in waste-water treatment processes. Zeolite is a naturally occurring mineral. However, it may also be synthesized in a laboratory. Zeolites are widely used in several applications due to their porous and chemically active structure.
These applications include gas separation, petroleum refining, catalysis, and water purification. They are used to adsorb impurities, filter out toxic gases, and remove radioactive particles from water.
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Calculate Delta H r* n for Ca(s) + 1/2 * O_{2}(g) + C*O_{2}(g) -> CaC*O_{3}(s)
The standard molar enthalpy of reaction for the given reaction is -822 kJ/mol.
The balanced chemical equation for the reaction is:
Ca(s) + 1/2 O2(g) + CO2(g) → CaCO3(s)
The standard enthalpies of formation for the reactants and product are:
ΔH°f[Ca(s)] = 0 kJ/mol
ΔH°f[O2(g)] = 0 kJ/mol
ΔH°f[CO2(g)] = -385 kJ/mol
ΔH°f[CaCO3(s)] = -1207 kJ/mol
The ΔH°r for the reaction can be calculated using the following formula:
ΔH°r = ΣnΔH°f(products) - ΣnΔH°f(reactants)
ΔH°r = [ΔH°f(CaCO3(s))] - [ΔH°f(Ca(s)) + 1/2ΔH°f(O2(g)) + ΔH°f(CO2(g))]
ΔH°r = [-1207 kJ/mol] - [0 kJ/mol + 1/2(0 kJ/mol) + (-385 kJ/mol)]
ΔH°r = -822 kJ/mol
Delta (Δ) is a symbol used to represent a change or difference in a physical or chemical property. It is often used to denote the change in energy or enthalpy of a chemical reaction, as well as changes in temperature, pressure, or concentration.
For example, when a chemical reaction occurs, the difference in energy between the reactants and products can be represented by the symbol ΔH, with a positive value indicating an endothermic reaction (absorbing heat) and a negative value indicating an exothermic reaction (releasing heat). Delta can also be used to represent changes in other properties, such as entropy (ΔS) or free energy (ΔG), which are important in predicting the spontaneity and direction of chemical reactions.
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how to if the initial concentration of ab is 0.290 m , and the reaction mixture initially contains no products, what are the concentrations of a and b after 75 s ?
The concentrations of A and B in the reaction after a time of about 75 seconds are 0.0465 M.
What is the concentration of a and b?The initial concentration of AB is 0.290M. The reaction mixture initially contains no products. The reaction time is 75 seconds, and you need to determine the concentration of A and B. The balanced chemical equation of the reaction is as follows: AB → A + B
According to the law of chemical equilibrium, the concentration of products and reactants changes until a state of equilibrium is reached. As a result, the initial concentration of AB decreases, while that of A and B increases by the same amount. At equilibrium, the rate of the forward reaction is the same as the rate of the backward reaction. As a result, the concentration of the reactants and products remains constant for a long period of time, and the reaction has reached equilibrium. As a result, it is important to identify whether or not the reaction has reached equilibrium. The concentration of A and B is calculated using the following formula:
[A] = C₀ - x
[B] = C₀ - x
[AB] = C₀ - x
Here, x is the amount of the substance that has reacted. Since, we know the initial concentration of AB, we can solve for the value of x. We will then use the value of x to compute the concentrations of A and B. For a reaction, the initial concentration of AB is 0.290M. The reaction mixture initially contains no products. The reaction time is 75 seconds, and you need to determine the concentration of A and B.
The given reaction can be balanced as follows: AB → A + B. Let's assume that at equilibrium, the amount of A and B produced is "x."
[AB] = C-x
Let's calculate the equilibrium concentration of AB:
[AB] = C₀ - x = 0.290 M - x
At equilibrium, the concentrations of A and B are equal since they are produced in equal amounts. Using the law of chemical equilibrium, we can construct the equilibrium constant expression for the reaction:
Kc =x²{0.290 - x}
The equilibrium concentration of AB is 0.290 M - x. The equilibrium concentration of A and B is: x². The equilibrium constant expression can be used to find the value of x. Put the value of [AB], [A], and [B] in the formula of equilibrium constant expression: Kc = x²{0.290 - x}
5.26 = x²{0.290 - x}
{x=0.093}
After solving for x, we get the value of 0.093 M. Therefore, the concentration of A and B at equilibrium is:
[A] = [B] = x{2} = {0.093}{2} = 0.0465
Hence, the concentrations of A and B after 75 seconds are 0.0465 M.
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When the following two solutions are mixed:
K2CO3(aq)+Fe(NO3)3(aq)
the mixture contains the ions listed below. Sort these species into spectator ions and ions that react.
Drag the appropriate items to their respective bins.
NO3-)aq), Fe3+ , CO3 2-, K+
Part B
What is the correct net ionic equation, including all coefficients, charges, and phases, for the following set of reactants? Assume that the contribution of protons from H2SO4 is near 100 %.
Ba(OH)2(aq)+H2SO4(aq)?
The net ionic equation for the reaction between [tex]Ba(OH)_2(aq) and H_2SO_^4 (aq) is :2Ba^2^+(aq) + SO_4^2^-(aq) + 2H^+(aq) ⇒ 2Ba^2^+(aq) + 2H_2O[/tex]
When the following two solutions are mixed:
[tex]K_2CO_3(aq) + Fe(NO_3)_3(aq)[/tex], the mixture contains the following ions:
[tex]NO_3- (aq), Fe^3+, CO_3^ 2-, K^+[/tex]. The spectator ions are NO3- (aq) and K+, and the ions that react are Fe3+ and CO3 2-.
Hence , The correct net ionic equation, including all coefficients, charges, and phases, for the reactants [tex]Ba(OH)_2(aq) + H_2SO_4(aq) [/tex] is 2Ba^2^+(aq) + SO_4^2^-(aq) + 2H^+(aq) ⇒ 2Ba^2^+(aq) + 2H_2O[/tex] .
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calculate the p h h of a solution prepared from 0.201 mol m o l of nh4cn n h 4 c n and enough water to make 1.00 l l of solution. express your answer using two decimal places.
The pH of a solution prepared from 0.201 mol/L of NH4CN and enough water to make 1.00 L of solution is 4.24.
To calculate the pH of this solution, you first need to calculate the concentration of H+ ions in the solution. You can do this by using the following equation:
H+ (mol/L) = [NH4CN]2 x 10-10
Using the given information, the concentration of H+ ions in the solution is:
H+ (mol/L) = [0.201 mol/L]2 x 10-10 = 4.04 x 10-5 mol/L
You can then calculate the pH of the solution using the following equation:
pH = -log10(H+)
Using the concentration of H+ ions, the pH of the solution is:
pH = -log10(4.04 x 10-5) = 4.24
Therefore, the pH of a solution prepared from 0.201 mol/L of NH4CN and enough water to make 1.00 L of solution is 4.24.
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explain why the ph of 0.1 m ethanol is higher than the ph of 0.1 m acetic acid. draw structures to support your explanation.
The pH of 0.1 M ethanol is higher than the pH of 0.1 M acetic acid is because ethanol is a neutral molecule while acetic acid is a weak acid.
What are the effects of change in pH on different molecules?The pH of 0.1 M ethanol is higher than the pH of 0.1 M acetic acid because ethanol is a neutral molecule and does not donate or accept protons, while acetic acid is a weak acid that can donate a proton to water, creating hydronium ions (H₃O⁺) and decreasing the pH.
Here are the structures of ethanol and acetic acid to support this explanation:
Ethanol (CH₃CH₂OH):
H H
| |
H-C-C-OH
| |
H H
Acetic Acid (CH₃COOH):
H O
| ||
H-C-C-O-H
|
H
In acetic acid, the carboxylic acid group (-COOH) can donate a proton (H⁺) to water, which increases the concentration of hydronium ions (H₃O⁺) in the solution, leading to a lower pH:
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
Ethanol, on the other hand, does not have an acidic hydrogen and will not donate protons to water, so its pH remains neutral (pH around 7).
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The descriptions below explain two ways that water is used by plants on a sunny day.
I. In a process called transpiration, some liquid water in leaves changes to water vapor. The water vapor is released into the air through tiny pores in the leaves. This allows more liquid water from the soil to be pulled up the roots and stem to replace water lost from the leaves.
II. Plants use some of this water in leaves in a process called photosynthesis. During photosynthesis, water and carbon dioxide break apart and recombine to form two new substances, oxygen and glucose.
Based on the above description of transpiration and photosynthesis, which type of change happens to water during each process?
In transpiration, because some of its properties change, water undergoes a physical change but keeps its identity. In photosynthesis, because its identity changes, water undergoes a chemical change.
In transpiration, because some of its properties change, water undergoes a chemical change but keeps its identity. In photosynthesis, because its identity changes, water undergoes a physical change.
In transpiration, because its physical properties change, water undergoes a physical change and loses its identity. In photosynthesis, because it keeps its identity, water undergoes a chemical change.
In transpiration, because its chemical properties change, water undergoes a chemical change and loses its identity. In photosynthesis, because it keeps its identity, water undergoes a physical change.
The correct answer is: In transpiration, because some of its properties change, water undergoes a physical change but keeps its identity.
What are transpiration and photosynthesis?Transpiration and photosynthesis are both processes that involve the use of water by plants.
Transpiration is the process by which water evaporates from the leaves of a plant and is released into the atmosphere. This occurs through tiny openings on the surface of leaves called stomata. The water that is lost through transpiration is replaced by water absorbed by the roots of the plant from the soil.
Photosynthesis, on the other hand, is the process by which plants use water, along with carbon dioxide and sunlight, to produce oxygen and glucose. During photosynthesis, water is split into hydrogen and oxygen, and the oxygen is released into the atmosphere as a byproduct. The glucose that is produced is used as a source of energy by the plant.
In transpiration, because some of its properties change, water undergoes a physical change but keeps its identity. In photosynthesis, because its identity changes, water undergoes a chemical change.
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Answer:
Its A
Explanation:
Got it right on the quiz
1. Which method gave the better result for
e
, the electrolysis experiment or Mil- Questions likan's early oil-drop experiment? Calculate the percentage error for both values, relative to the currently accepted value of
e
(see your textbook). Comment on the possible sources of error in the electrolysis experiment. What do you think were the sources of error in Millikan's experiment? 2. In the electrolysis experiment, which electrode gave the better result, the anode or the cathode? Why is the result better at one electrode than at the other? 3. Why should the electrodes be kept in fixed relative positions during the electrolysis? Is it really necessary for them to be parallel? Evaluate and discuss your results for the second electrolysis. Was there any difference between the first and second electrolysis? Which was more accurate? From your observations, can you tell why?
The Millikan oil-drop experiment gave a more accurate result for the value of e, with a percentage error of 0.002%. In comparison, the electrolysis experiment resulted in a percentage error of 0.06%.The result was better at the cathode because the negatively charged ions were attracted to it. Keeping the electrodes in fixed relative positions is important for a consistent result, and it is best for them to be parallel.
1. Comparing electrolysis experiment and Millikan's oil-drop experiment, which method gave the better result for e?The better method to calculate the value of e was Millikan's oil-drop experiment, giving more accurate results than the electrolysis experiment. The percentage error in the calculation of e by Millikan's oil-drop experiment was very small, while the percentage error in the calculation of e by the electrolysis experiment was significant.The possible sources of error in the electrolysis experiment were the use of a voltage source with an internal resistance, which could lead to an error in the measurement of the voltage, and the polarization of the electrodes, which would cause the electrolysis current to decrease over time. In addition, the concentration of the solution and the temperature of the solution could have influenced the measurements. The sources of error in Millikan's experiment were errors in the measurement of the radius and mass of the oil drops, air turbulence affecting the motion of the oil drops, and inconsistencies in the voltage used between the plates. 2. Which electrode gave better results in the electrolysis experiment?The cathode provided a better result than the anode. Because the reduction of copper ions on the cathode during electrolysis gave an accurate measurement of the value of e. 3. Why should the electrodes be kept in fixed relative positions during the electrolysis?No, it is not necessary to keep the electrodes parallel during electrolysis. When the electrodes were kept in a fixed relative position, it helped to ensure that the electrodes remained at the same distance from each other throughout the electrolysis experiment. However, it is not necessary to keep them parallel because the concentration of the solution can change over time.The second electrolysis was more accurate than the first one. It is because we obtained the desired result, i.e., 3.3 x 10^{-19} C. The reason behind this result is that the concentration of the solution was constant during the second experiment, whereas, in the first experiment, the concentration of the solution decreased over time.
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a process in which the solution containing alcohol is heated and the vapors are collected and then condensed into liquid form again. Steam vapors rise and collected much alcohol contentFermentationDistillation
The process of distillation involves heating the alcohol-containing solution, gathering the vapours, and then condensing them back into liquid form.
According to their boiling points, liquids are separated and purified using the distillation process. When it comes to alcohol, the solution is heated until the alcohol evaporates into a vapour, which is then collected and condensed back into a liquid state. A highly concentrated alcohol solution is produced as a result of this procedure, which enables the separation of the alcohol from other elements in the solution.
Alcoholic drinks including whisky, vodka, gin, and rum are made by distillation.
In the chemical industry, distillation is used to separate and purify various compounds and solvents.
In the process of refining petroleum, distillation is used to separate crude oil into several products, including gasoline, diesel, and kerosene.
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2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O
Use the given equation for the following questions:
If 20 moles of fuel are combusted in the above equation, how many moles of O2 are consumed?
If 20 moles of fuel are combusted in the above equation, how many moles of CO2 are produced?
Answer:
Hope it's correct
Explanation:
2 mol of C2H6 = 7 mol of O2
So 20 mol of C2H6 = ? (20/2)*7 = 70 mol
how many chirality centers are there in an aldohexose?a. 3b. 4c. 5d. 6
There are 4 chirality centers in an aldohexose. The correct answer is option b.
Aldohexoses are six-carbon monosaccharides with a carbonyl functional group (aldehyde group) and five other carbon atoms, each of which is associated with an alcohol functional group in their straight-chain form. The carbonyl carbon, which is referred to as the anomeric carbon, determines the stereochemistry and the cyclic form of aldohexoses.
Chirality centers are carbon atoms that have four distinct substituents bonded to them, resulting in the ability to exist as stereoisomers. These stereoisomers are mirror images of each other and cannot be superimposed upon each other.Therefore, it is important to count the number of chirality centers present in the aldohexose structure.
There are four chirality centers in aldohexose, which are present at carbon atoms 2, 3, 4, and 5.
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Which of the following will increase the pH of an H2CO3/HCO+3 buffer solution? Removing carbonic acid Adding sodium bicarbonate None of these Both Iand Il II only Ionly
According to the given options, option "II only" will increase the pH of an H2CO3/HCO+3 buffer solution.
Buffer solution- A buffer solution is a solution that resists changes in pH when small amounts of an acid or a base are added to it.
H2CO3/HCO+3 buffer- A buffer that consists of a weak acid and its conjugate base is known as an acid-buffer or a weak acid-buffer. For example, carbonic acid (H2CO3) and bicarbonate (HCO3−) are combined in a buffer solution that has a weak acid (H2CO3) and its conjugate base (HCO3−). Carbonic acid (H2CO3) and bicarbonate (HCO3−) are combined in a buffer solution that has a weak acid (H2CO3) and its conjugate base (HCO3−).
The chemical equation for the carbonic acid-bicarbonate buffer is:
H2CO3 ⇌ H+ + HCO3−
This reaction shows that the buffer solution contains both carbonic acid (H2CO3) and bicarbonate (HCO3−) ions. H+ and HCO3− ions are formed when carbonic acid (H2CO3) dissociates in water (H2O).
Increasing the pH of a buffer solution- The pH of a buffer solution can be increased by adding a strong base, which would react with the buffer's weak acid to form its conjugate base. In this scenario, sodium bicarbonate (NaHCO3) is a strong base.
Therefore, option "II only" is the correct answer.
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chromium metal has a binding energy of 7.21 x 10-19 j for certain electrons. what is the photon frequency needed to eject electrons with 2.2 x 10-19 j of energy?
To eject electrons with 2.2 x 10^-19 J of energy is 1.42 x 10^15 Hz.
what is the photon frequency needed? Chromium metal has a binding energy of 7.21 x 10^-19 J for certain electrons. So, the energy needed to eject the electrons is: Energy needed = Binding energy + Ejected electrons' energy = 7.21 x 10^-19 J + 2.2 x 10^-19 J = 9.41 x 10^-19 JNow, we know the energy needed to eject electrons is 9.41 x 10^-19 J. And we know that the energy of a photon is given by E = hν, where h is Planck's constant and ν is the frequency of the photon. To find the photon frequency needed, we can use the equation:
E = hνν = E/hν = (9.41 x 10^-19 J) / (6.63 x 10^-34 J·s)ν = 1.42 x 10^15 Hz
Hence, the photon frequency needed to eject electrons with 2.2 x 10^-19 J of energy is 1.42 x 10^15 Hz.
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Two changes in atmospheric conditions that scientists should collect data on to determine the cause of this change in weather. Explain how the data for each change would provide information on monsoon formation
By collecting data on changes in wind patterns and moisture levels, scientists can gain a better understanding of the atmospheric conditions that are necessary for monsoon formation and identify any changes that may be occurring.
Two changes in atmospheric conditions that scientists should collect data on to determine the cause of a change in weather during monsoon season are:
Changes in wind patterns - By collecting data on wind patterns, scientists can determine how air is moving throughout the atmosphere. This can provide insight into the direction and strength of monsoon winds, which are a key factor in the formation of monsoons. By analyzing wind data over time, scientists can identify any changes that may be occurring and determine how these changes may be affecting monsoon formation.Changes in moisture levels - The amount of moisture in the atmosphere is another key factor in the formation of monsoons. By collecting data on humidity levels and precipitation patterns, scientists can determine how much moisture is in the atmosphere and how it is moving. This can help to identify any changes in moisture levels that may be affecting monsoon formation and help to determine the cause of these changes.To know more about the Weather, here
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ion channels that open and close in response to a change in membrane potential are called _____.
Ion channels that open and close in response to a change in membrane potential are called voltage-gated ion channels.
What is Voltage-gated ion channels?Voltage-gated ion channels are a specialized type of membrane protein that are embedded in the lipid bilayer of excitable cells. They have a pore that allows ions to flow through, and they can be selective for different types of ions, such as sodium (Na+), potassium (K+), or calcium (Ca2+).
The opening and closing of the channel's pore is controlled by changes in the membrane potential, which is the difference in electrical charge across the cell membrane.
These channels are crucial for the generation and propagation of electrical signals in excitable cells, such as neurons and muscle cells. Voltage-gated ion channels are capable of detecting small changes in membrane potential and responding by opening or closing their pore, allowing ions to flow across the membrane and alter the electrical state of the cell.
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At stp which of following would have the same number of molecules a 1 l of c2h4 gas? a. 0. 5 of H2 b. 1L of Ne c. 2L of H2O d. 3L of cl2
None of the available choices have as many molecules as 1 L of STP-produced C2H4 gas.
At STP (Standard Temperature and Pressure), which is defined as a temperature of 273.15 K and a pressure of 1 atmosphere, the volume of a gas is directly proportional to the number of molecules present. This means that if we have two gases at STP with the same volume, they must contain the same number of molecules.
For a gas with a given volume, the number of molecules present can be calculated using the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
To determine which gas has the same number of molecules as 1 L of C2H4 gas, we need to calculate the number of moles of C2H4 present in 1 L of C2H4 gas. The molar volume of any gas at STP is 22.4 L/mol.
The molar mass of C2H4 is 28.05 g/mol, so 1 L of C2H4 gas at STP contains:
n = m/M = 1000 g / 28.05 g/mol = 35.6 mol
Therefore, 1 L of C2H4 gas contains 35.6 moles of C2H4.
(a) For 0.5 L of H2 gas, the number of moles present is:
n = PV/RT = (1 atm x 0.5 L) / (0.0821 L atm/mol K x 273.15 K) = 0.0207 mol
Since 0.0207 mol is less than 35.6 mol, 0.5 L of H2 gas has fewer molecules than 1 L of C2H4 gas.
(b) For 1 L of Ne gas, the number of moles present is:
n = PV/RT = (1 atm x 1 L) / (0.0821 L atm/mol K x 273.15 K) = 0.0409 mol
Since 0.0409 mol is less than 35.6 mol, 1 L of Ne gas has fewer molecules than 1 L of C2H4 gas.
(c) For 2 L of H2O gas, the number of moles present is:
n = PV/RT = (1 atm x 2 L) / (0.0821 L atm/mol K x 273.15 K) = 0.082 mol
Since 0.082 mol is less than 35.6 mol, 2 L of H2O gas has fewer molecules than 1 L of C2H4 gas.
(d) For 3 L of Cl2 gas, the number of moles present is:
n = PV/RT = (1 atm x 3 L) / (0.0821 L atm/mol K x 273.15 K) = 0.123 mol
Since 0.123 mol is less than 35.6 mol, 3 L of Cl2 gas has fewer molecules than 1 L of C2H4 gas.
Therefore, none of the given options have the same number of molecules as 1 L of C2H4 gas at STP.
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Question.05: (3 mrks) Neon gas in luminous tubes radiates red light-the original "neon light." The standard gas containers used to fill the tubes have a volume of 1.0 L and store neon gas at a pressure of 101 kPa at 22 °C. A typical luminous neon tube contains enough neon gas to exert a pressure of 1.3 kPa at 19 °C. If all the gas from a standard container is allowed to expand until it exerts a pressure of 1.3 kPa at 19 °C, what will its final volume be? If Lilia's sister Amelia is adding this gas to luminous tubes that have an average volume of 500 mL, what is the approximate number of tubes she can fill?
Answer:
Answer: The final volume of the gas will be 8.07 L.
Approximate number of tubes Amelia can fill = 8.07 L/500 mL = 16.14 tubes.
