Answer:
e. Na, Fe, K
Explanation:
The group of elements that will be positive charge ions is metal. You can find metal in the first 2 columns of the periodic table and in the transition area. Natrium/sodium (Na), iron (Fe), and kalium/potassium(K) categorized as metal and they will form positive charge ions.
On the other hand carbon(C), oxygen (O), arsenic(As) and bromine(Br) is gas and will form negative charge ions. Gas located on the right side of the periodic table.
The elements that readily form positively charged ions, or cations, are more often metals like Sodium (Na), Iron (Fe), and Potassium (K). Therefore, the correct answer from the options given is 'Na, Fe, K'. option e.
Explanation:The elements that form cations, or positively charged ions, readily are elements that tend to lose electrons. This characteristic is typically associated with metals. In the options given, Na (Sodium), Fe (Iron), and K (Potassium) are the ones which are more likely to form cations. So, the correct answer to your question: 'Which of the following elements form cations (positively charged ions) readily: C, O, Na, Fe, As, Br, K?' would be option e. Na, Fe, K.
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Be sure to answer all parts. In winemaking, the sugars in grapes undergo fermentation by yeast to yield CH3CH2OH (ethanol) and CO2. During cellular respiration, sugar and ethanol are "burned" to water vapor and CO2. (a) Using C6H12O6 for sugar, calculate ΔH o rxn of fermentation and of respiration (combustion). Fermentation = kJ Respiration = kJ (b) Write a combustion reaction for ethanol. Include the physical states of each reactant and product. (c) Which releases more heat from combustion per mole of C, sugar or ethanol?
a) ΔH of Fermentation = - 2816 kJ/mol
ΔH of Respiration = - 1409.2 kJ/mol
b)C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)
c) Combustion per mole of sugar
Let's solve for each part one by one:
a)
The fermentation of sugar is given as follows:C₆H₁₂O₆ (s) + 6O₂ (g) → 6O₂ (g) + 6H₂O (l)
ΔHrxn = ΔHformation (products) - ΔHformation (reactants)
[tex]= (6 mol * -393.5kJ/mol)+ (6mol * -285.8kJ/mol) - (1 mol * -1260kJ/mol + 6mol * 0)\\\\= -2 816 kJ/mol[/tex]
The heat of combustion of ethanol is given as follows:C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)
Let's assume that 1 mole of ethanol is burnt. The heat of reaction, ΔHrxn is given by this equation:
ΔHrxn = ΔHformation (products) - ΔHformation (reactants)
[tex]= (2 mol* - 393.5kJ/mol) + ( 3mol * -285.8kJ/mol) - [ (1mol * -235 kJ/mol) + ( 3mol * 0.0000kJ)]\\\\= -1644 - (-235.2)\\\\= - 1409.2 kJ/mol[/tex]
The negative sign indicates that the process of combustion is exothermic.
b) The combustion reaction for ethanol can be given as:
C₂H₅OH (l) + 3O₂ (g) → 3H₂O(g) + 2CO₂(g)
c) From the comparisons of the ΔHrxn, sugar produces more energy.
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Final answer:
The enthalpy change for fermentation and respiration (combustion) can't be calculated without additional data on standard enthalpies of formation. The combustion reaction for ethanol is balanced, and to compare the heat released from sugar or ethanol combustion, we would need their specific heats of combustion in kJ/mol.
Explanation:
The question asks us to calculate the change in enthalpy (ΔHorxn) for both fermentation and respiration (combustion) processes and write a balanced chemical equation for the combustion of ethanol. To answer these, we start by looking at each process.
Fermentation
The balanced chemical equation for the fermentation of glucose to ethanol (CH3CH2OH) and carbon dioxide (CO2) is given by:
C6H12O6 → 2 CH3CH2OH + 2 CO2
We cannot calculate ΔHorxn for fermentation without the standard enthalpy of formation for each species.
Respiration (Combustion)
The combustion of glucose can be represented as:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
Again, to calculate ΔHorxn, we need the standard enthalpies of formation for each compound.
Combustion Reaction for Ethanol
The balanced combustion reaction of ethanol is:
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)+ 29.7 kJ/g
This reaction shows that for each mole of ethanol burned, carbon dioxide and water are produced, and heat is released.
Comparison of Heat Released
To compare which releases more heat per mole of carbon, sugar or ethanol, we would need to know the specific heat of combustion for each substance in kJ/mol.
What is the density of a piece of aluminum that has a mass of 270 grams and a volume of 100.0 cubic centimeters? If this piece was cut in half what would be the density of one of the pieces?
The density of the original piece of aluminum is 2.7 grams per cubic centimeter. If one piece is cut in half, the density of each piece would remain the same.
Explanation:In order to find the density of an object, you divide the mass of the object by its volume. The mass of the aluminum piece is 270 grams and its volume is 100.0 cubic centimeters. So, the density would be 270 grams divided by 100.0 cubic centimeters, which equals 2.7 grams per cubic centimeter.
If the piece of aluminum is cut in half, the mass of each piece would be half of the original mass, so each piece would have a mass of 135 grams. Since the volume of each piece remains the same, 100.0 cubic centimeters, the density of one of the pieces would still be 2.7 grams per cubic centimeter.
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A sample of potassium nitrate (49.0 g) is dissolved in 101 g of water at 100 °c with precautions taken to avoid evaporation of any water. The solution is cooled to 30 °c and a small amount of precipitate is observed. This solution is __________.a. hydrated b. saturated c. unsaturated d. supersaturated e. placated
Final answer:
The solution in question is saturated because the amount of KNO3 dissolved is less than the amount initially added at the given temperature.
Explanation:
Based on the information provided, the solution in question is a saturated solution. A saturated solution is one in which the maximum amount of solute has been dissolved in a given amount of solvent at a specific temperature. In this case, the solubility of potassium nitrate (KNO3) at 30 °C is approximately 48 g, which is less than the 80 g of KNO3 that was initially added to the solution. Therefore, the solution is saturated.
All methods of chromatography operate on the same basic principle that Select one: a. one component of the mixture will chemically react with the mobile phase b. the one component of the mixture will be completely insoluble in the mobile phase c. the components of the mixture will destribute unequally between mobile and stationary phase d. one component of the mixture will chemically react with the stationary phase
Answer:C
Explanation:
Chromatography separates compounds by taking advantage of their polarity. The stationary phase is generally very polar. The mobile phase can be pure hexane or various ratios of hexane with a polar eluent added. The more polar the compound, the more it interacts with the stationary phase and won’t move very far up the plate compared to the non-polar or less polar compounds that interact more with the non-polar hexane.
Final answer:
All methods of chromatography operate based on the principle that components of a mixture will distribute unequally between the mobile and stationary phase, due to differences in intermolecular forces and affinities.
Explanation:
The question asks about the basic principle on which all methods of chromatography operate. The correct answer is c. The components of the mixture will distribute unequally between the mobile and stationary phases. In chromatography, the separation of components is achieved because different components have different affinities for the stationary and mobile phases. A component with a higher affinity for the stationary phase will move more slowly through the chromatography system compared to a component with a higher affinity for the mobile phase. This differing affinity is often due to differences in the strength of intermolecular forces between the components and the phases. For example, in liquid chromatography, if a compound has strong intermolecular forces with the stationary phase (e.g., through hydrogen bonding or van der Waals forces), it will tend to be 'adsorbed' onto the stationary phase and thus move through the system more slowly compared to compounds that have weaker interactions and therefore travel with the mobile phase.
Water containing large amounts of Mg2+and Ca2+ions is said to be hard because it is hard to make soap lather in the water. True False
Answer:True
Explanation:
Water is said to be hard when it contains calcium ions or magnesium ions dissolved in it. These ions are able to react with soap in such a way that the soap is prevented from forming lather with the water. Hard water occurs when water passes over calcium or magnesium bearing minerals and dissolves some of it. Hardness due to the presence of calcium ions can easily be removed by boiling the water.
In most chemical reactions the amount of product obtained is
Answer:
called theoretic yield
In most chemical reactions, the amount of product obtained is typically less than the theoretical yield due to various reasons such as incomplete reactions, loss of product during isolation, or the reversible nature of some reactions.
Explanation:In most chemical reactions, the amount of product obtained is usually less than the theoretical yield which is the amount predicted by a stoichiometric calculation based on the number of moles of all reactants present. This is because some reactants may not react to form the product, some products may be lost during the isolation step, or the reaction may not go to completion because it is reversible. For instance, if you mix 10 grams of hydrogen with 10 grams of oxygen in a sealed container, you won't get 20 grams of water, but less, because not all the hydrogen and oxygen will react.
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Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25∘C diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Determine △ Hrxn for C(diamond) → C(graphite) with equations from the following list: (1) C(dianond)+O2(g)⟶CO2(g)ΔH=−395.4kJ (2) 2CO2(g)⟶2CO(g)+O2(g)ΔH=566.0kJ (3) C(graphite)+O2(g)→CO2(g)ΔH=−393.5kJ (4) 2CO(g)⟶C(graphite)+CO2(g)ΔH=−172.5kJ
Answer:
-1.9 KJ/mol
Explanation:
In order to solve the problem, we have to rearrange the equations in a way in which all molecules of O₂ and CO₂ are eliminated:
2C(diamond) + 2O₂(g) → 2CO₂(g) ΔH₁= 2 x (-395.4 KJ) ------> we multiply by 2 both reactants and products
2 CO₂(g) → 2CO(g) + O₂(g) ΔH₂= 566.0 KJ
CO₂(g) → C(graphite) + O₂(g) ΔH₃= -1 x (-393.5 KJ) ------> we use reverse rxn
2CO(g) → C(graphite) + CO₂(g) ΔH₄= -172.5 KJ
When we cancel the molecules that appear both in reactants and products, the total reaction is the following:
2C(diamond) → 2C(graphite)
ΔHt= ΔH₁ + ΔH₂ + ΔH₃ + ΔH₄ = 2 x (-395.4 KJ) + 566.0 KJ + (-1 x (-393.5 KJ)) - 172.5 KJ
ΔHt= 347.2 KJ
This is for 2 mol of C(diamond) which are converted in 2 mol of C(graphite). To obtain ΔH for the reaction of 1 mol C(diamond) to 1 mol (graphite) we have to divide into 2:
ΔH= -3.8 KJ/2mol= -1.9 KJ/mol
To determine ΔHrxn for the reaction C(diamond) → C(graphite), we can use Hess's law and the given equations. By canceling out common compounds/products and summing the remaining equations, we can calculate ΔHrxn. The enthalpy change can be determined by summing the enthalpy changes of the canceled equations.
Explanation:This type of calculation usually involves the use of Hess's law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. For this specific reaction, we can use the given equations to determine the enthalpy change ΔHrxn for C(diamond) → C(graphite).
Step 1: C(diamond) + O2(g) → CO2(g) (ΔH = -395.4 kJ)Step 2: 2CO2(g) → 2CO(g) + O2(g) (ΔH = 566.0 kJ)Step 3: C(graphite) + O2(g) → CO2(g) (ΔH = -393.5 kJ)Step 4: 2CO(g) → C(graphite) + CO2(g) (ΔH = -172.5 kJ)To obtain the ΔHrxn for C(diamond) → C(graphite), we need to cancel out common compounds/products in these equations. From the given equations, we can see that CO2(g) is a common compound in steps 1, 2, and 3, and thus we can cancel it out. Likewise, O2(g) is present in steps 1, 2, and 3, so we can also cancel it out. By summing the canceled equations, we get the final equation:
C(diamond) → C(graphite)
The enthalpy change for this reaction is the sum of the enthalpy changes of the canceled equations:
ΔHrxn = ΔH1 + ΔH2 + ΔH3 + ΔH4 = -395.4 kJ + 566.0 kJ + (-393.5 kJ) + (-172.5 kJ) = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + (-393.5 kJ) + 566.0 kJ + (-172.5 kJ) = -395.4 kJ - 393.5 kJ + 566.0 kJ - 172.5 kJ = -395.4 kJ - 393.5 kJ + 566.0 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ = -395.4 kJ + 566.0 kJ - 393.5 kJ - 172.5 kJ
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The specific branch of chemistry that focuses on molecules such as salts and water that constitute non-living matter, but are still important to living things, is termed ____________chemistry.
Answer:
Inorganic chemistry
Explanation:
Inorganic chemistry can be defined as the study of the composition and constituent of materials from non-biological origins, materials without carbon-hydrogen bonds such as: metals, salts, water and minerals.
The branch of chemistry dealing with the nonliving constituent, but important chemical to living matter, is inorganic chemistry.
Chemistry has been the branch of science that has been dealing with the chemical composition and structure of the compounds.
Branch of chemistryThe chemistry has been divided into various branch, based on the application of the studied in various field.
The branch of chemistry that has been dealing with the molecules constituent of the non-living matter has been inorganic chemistry.
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A compound contains carbon, hydrogen, and chlorine. It has a molar mass of 98.95 g/mol. Analysis of a sample shows that it contains 24.27% carbon and 4.07% hydrogen. What is its molecular formula?
Answer:
C2H4Cl2
Explanation:
Firstly, we know that the compound contains only three elements. These are carbon, hydrogen and oxygen. We have the percentage compositions of carbon and hydrogen, thus we need the one for chlorine. To get the one for chlorine, we simply subtract that of carbon and hydrogen from a total of 100%.
Hence percentage composition of chlorine = 100 - 24.27 - 4.07 = 71.66%
Now, we divide the percentage compositions by the atomic masses. The atomic masses of carbon, hydrogen and chlorine are 12, 35.5 and 1 respectively. We go on to the divisions as follows.
C = 24.27/12 = 2.0225
H = 4.07/1 = 4.07
Cl = 71.66/35.5 = 2.02
We then go on to divide each by the smallest which is 2.02
C = 2.0225/2.02 = 1
H = 4.07/2.02 = 2
Cl = 2.02/2.02 = 1
Hence the empirical formula is CH2Cl
Now, since the molecular mass is 98.95, we need to calculate the molecular formula
Hence, [CH2Cl]n = 98.95
[12 + 2(1) + 35.5]n = 98.95
[12 + 2 + 35.5]n = 98.95
49.5n = 98.95
n = 98.95/49.5 = 2
The molecular formula is thus [CH2Cl]2 = C2H4Cl2
The amino acid substitution of Val for Glu in Hemoglobin S results in aggregation of the protein because of __________ interactions between subunits.
Answer: Hydrophobic
Explanation:
The glutamic acid in the 6th position of beta chain of HbA is changed to valine in HbS. The substitution of hydrophilic glutamic acid by hydrophobic valine causes a sickness on the surface of the molecule.
This single nucleotide change polymerises hemoglobin molecules in the red blood cell.
So the hydrophobic nature of valine causes the aggregation of hemoglobin protein.
Which of the following are true about balanced chemical reactions? (Select all that apply)
a
Atoms of each element are equal on both sides.
b
All chemical formulas have a coefficient greater than one.
c
It is better to change subscripts to balance equations.
d
Coefficients are added to balance the equation.
e
Chemical equations must be balanced to satisfy the Law of Conservation of Mass.
f
Balanced chemical equations help to determine the number of moles needed in the reaction.
Answer: D. There must be as equal number of atoms of each element on both sides of the equation.
Explanation: I just took the test on Plato and this is correct
1. Where can you find safety data sheets (MSDS) for the chemicals used in the lab ?
2. Which region in the IR spectrum could be used to distinguish between butanoic acid and 2-butanone?
(a) 3200-3600 cm⁻¹
(b) 1600 cm⁻¹
(c) 1680-1750 cm⁻¹
(d) 2500-3300 cm⁻¹
Answer:
Option d
Explanation:
Safety data sheets (MSDS) contains information regarding safe handling practices of hazardous properties of chemicals used, remedies on exposure.
By rule, MSDS should be provided by suppliers of the chemicals.
physical and chemical properties, technical data, trade and common names, hazards, remedies on exposure and safe handling practices on chemicals to users and those involved in their handling and transportation.
2-butanone is a ketone having carbonyl group (C=O) whereas butanoic acid is a carboxylic acid.
IR frequency of ketonic C=O is 1750 – 1680 cm⁻¹.
Butanoic acid also has C=O group but peak is observed at 3000 – 2500 cm⁻¹.which corresponds to Carboxylic Acid O-H Stretch. Therefore, if frequency 3000 – 2500 cm⁻¹. frequency can be used to distinguish between between butanoic acid and 2-butanone.
Therefore, among given, option d is correct.
Safety Data Sheets (SDS) can be found at http://www.msds.com. The IR spectrum region 1680-1750 cm⁻¹ distinguishes between butanoic acid and 2-butanone through different behavior of their carbonyl groups.
Explanation:Locating Safety Data Sheets and Distinguishing IR Spectrum Regions
Safety Data Sheets (SDS), formerly known as Material Safety Data Sheets (MSDS), for chemicals used in the lab can often be found on dedicated websites like http://www.msds.com or directly from the chemical manufacturer's website. These sheets are crucial for understanding the handling, risks, and disposal procedures of chemicals.
To distinguish between butanoic acid and 2-butanone using infrared (IR) spectroscopy, you would look at differing absorbance regions that are characteristic of their functional groups. Butanoic acid has a carboxyl group that typically shows absorbance in the O-H stretch region (2500-3300 cm⁻¹) and the C=O stretch of acids (1700-1750 cm⁻¹). In contrast, 2-butanone has a carbonyl group that absorbs in the C=O stretch region (1680-1750 cm⁻¹), but without the broad O-H stretching band. Thus, the region that could be used to distinguish between butanoic acid and 2-butanone is option (c) 1680-1750 cm⁻¹, which captures the different behavior of the carbonyl group in each compound.
What is the mass of water necessary to generate 11.2 L of hydrogen gas if calcium metal reacts with water at standard temperature and pressure (STP)?
Answer:
The mass of water is 18 g
Explanation:
The reaction of calcium with water can be represented in the equation below:
Ca + 2H₂O --------->Ca(OH)₂ +H₂
1 Mole of gas at STP = 22.4L
From the displacement reaction above, calculate the mass of water that will produce 22.4L of hydrogen gas at STP.
Mass of 2H₂O = 2(2x1 + 16) = 2X18 = 36 g/mol
Using proportional analysis;
36 g of 2H₂O produced 22.4 L of H₂, then
what mass of 2H₂O will produce 11.2L of H₂ ?
Mathematically,
22.4 L ----------------------------------> 36g
11.2 L -----------------------------------> ?
Cross and multiply, to obtain the expression below
= (11.2 X 36)/22.4
= 18 g
Therefore, the mass of water is 18 g
Note that 5° of longitude or latitude equals about 563 km. Use this information and a ruler to create a scale to use with this map. Draw and label your scale in the legend of the map.
Answer:
A suitable scale, say 1 cm: 100 km can be used.
Explanation:
Thinking process:
The best way to approach the question will be to consider the requirements. This is a simple case of scaling. In order to achieve the objective, you need to choose a scale that does not consume space and that presents more details at the same time.
For instance, a scale of 1 cm to 100 km will give me lines which are a little more then 5 cm. This can be presented as:
1: 500
This is appreciable for the paper size.
Which results when an atom has such a strong attraction for electrons that it pulls one or more electrons completely away from another atom?
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Heavy elements, like uranium, were/are created _______________. a. in supernovae and star collisions b. all of the above c. in the Big Bang. d. through nuclear fusion in mid-sized suns, like ours.
Answer:
a. in supernovae and star collisions
Explanation:
The periodical table contains some heavier elements, which are formed as neutron stars pairs hit eachother and erupt cataclysmically.
The star emitts very large quantities of energy and neutrons during supernova, which allow for the production of heavier elements than iron, such as uranium and gold. All these elements are ejected into space during the supernova explosion.
Pick the correct statement for the following isotope: a. 42Ca 42 is the mass number and 20 is the atomic number. b. 42 is the number of neutrons and 20 is the number of protons. c. 42 is the number of protons and 20 is the number of electrons. d. 42 is the atomic number and 20 is the number of neutrons.
Answer:
A
Explanation:
To label an element correctly using a combination of the symbol, mass number and atomic number furnishes some important information about the element.
We can obtain these information from the element provided that correct labeling of the element is presented. Firstly, after writing the symbol of the element, the atomic number is placed as a subscript on the left while the mass number of the atomic mass is placed as a superscript on the same left.
Looking at the question asked, we have the element symbol in the correct position as Ca, with 42 also in the correct position which is the mass number. The third number which is 20 is thus the atomic number of the element.
A 700 g peregrine falcon dives toward the ground from a height of 80 m and has a kinetic energy of 2,835 J. What is its speed?
Answer:
Kinetic energy is gven by
[tex]Kinetic Energy = \frac{1}{2} mv^{2}[/tex]
Hence the velocity can be found by [tex]v^{2} = \frac{2KE}{m}[/tex]
or [tex]\sqrt{\frac{2(2835J)}{0.7Kg} }[/tex] = [tex]\sqrt{8100\frac{m^{2} }{s^{2} } }[/tex] = 90m/s
Explanation:
The kinetic energy is the energy due to motion and is given by the kinetic energy equation KE = 1/2mv^2
It is the energy required to stop a body in motion and as seen from the equation, objects with large speed or mass have larger amount of kinetic energy or a large object effect can be made milder by lowering its speed so also a small object can increase its effect, for example a bullet by increasing its speed
The concentrations of Fe and K in a sample of riverwater are 0.0400 mg/kg and 1.30 mg/kg, respectively. Express the concentration in molality.
Answer :
The concentration in molality of Fe is, [tex]7.1\times 10^{-7}mol/kg[/tex]
The concentration in molality of K is, [tex]3.3\times 10^{-5}mol/kg[/tex]
Explanation:
First we have to calculate concentration in molality of Fe.
Molar mass of Fe = 56 g/mol
Concentration of Fe = 0.0400 mg/kg = [tex]4\times 10^{-5}g/kg[/tex]
Conversion used : 1 g = 1000 mg
[tex]\text{Concentration in molality}=7.1\times 10^{-7}mol/kg[/tex]
Thus, the concentration in molality of Fe is, [tex]7.1\times 10^{-7}mol/kg[/tex]
Now we have to calculate concentration in molality of K.
[tex]\text{Concentration in molality}=\frac{\text{Concentration of K}}{\text{Molar mass of K}}[/tex]
Molar mass of K = 39 g/mol
Concentration of K = 1.30 mg/kg = [tex]1.3\times 10^{-3}g/kg[/tex]
Conversion used : 1 g = 1000 mg
[tex]\text{Concentration in molality}=\frac{1.3\times 10^{-3}g/kg}{39g/mol}[/tex]
[tex]\text{Concentration in molality}=7.1\times 10^{-7}mol/kg[/tex]
Thus, the concentration in molality of K is, [tex]3.3\times 10^{-5}mol/kg[/tex]
To express the concentration of Fe and K from mg/kg to molality, convert their masses to grams, calculate the number of moles using their respective molar masses, and divide by the solvent's mass in kilograms.
Explanation:The question asks to convert the concentration of Fe and K from mg/kg to molality. Molality is a measure of the concentration of a solute in a solution in terms of the amount of substance in moles per kilogram of solvent. First, one needs to convert the mass of Fe and K from mg to g, which is simply dividing by 1000 as there are 1000 mg in a gram. Then, we find the molar mass of Fe (approximately 55.845 g/mol) and K (approximately 39.0983 g/mol) and calculate the number of moles for each.
Molality is then determined by dividing the number of moles of solute by the mass of the solvent in kilograms. Since water is the solvent and its density is typically close to 1 kg/L, for practical purposes, the mass of solvent is usually approximated as equal to the volume of the solution in liters (provided the solution concentration is relatively low, which it is in this case).
What potential difference is required to bring the proton to rest?
Answer:
0
Explanation:
Δk = (kf - Ki). Kf is 0 because your stopping it [m(v_i)^2 ]/2q = V
The potential difference required to bring a proton to rest can be calculated by knowing the initial kinetic energy of the proton. This energy, when applied in the opposite direction, would act to decelerate the proton and bring it to rest.
Explanation:The potential difference required to stop a moving proton can be calculated using the equation PEelec = qV, where q is the proton's charge and V is the potential difference (also known as voltage). Given that the proton's charge, q = qe = 1.6×10−¹⁹ Coulombs, we can rearrange the equation to solve for V: V = PEelec/q. If we know the initial kinetic energy of the proton (its energy due to motion, which can be thought of as the electrical energy it gained from accelerating through the potential difference), then the same voltage (potential difference) applied in the opposite direction will bring the proton to rest.
In other words, if the proton gained a certain amount of energy (in electron volts, eV) while accelerating through a potential difference, it will require the same amount of energy in the opposite direction to decelerate and come to rest.
For example, if the proton gained 1 electron Volt (1 eV) of energy from the potential difference, it would require a potential difference of -1 V to bring it to rest. This is because 1 eV = (1 V)(1.6×10−¹⁹ C).
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What is the density, in g/mL, of a cube of lead (Pb) that weighs 0.371 kg and has a volume of 2.00 in.3
Answer:
11.32 g/mL
Explanation:
Given that:-
Mass = 0.371 kg = 371 g ( 1 kg = 1000 g)
Volume = 2.00 in³
The conversion of in³ to mL is as shown below:-
1 in³ = 16.3871 mL
So, Volume = [tex]2.00\times 16.3871\ mL[/tex] = 32.7741 mL
The expression for the calculation of density is shown below as:-
[tex]\rho=\frac{m}{V}[/tex]
Applying the values as:-
[tex]\rho=\frac{371\ g}{32.7741\ mL}=11.32\ g/mL[/tex]
Final answer:
To find the density of a lead cube weighing 0.371 kg and having a volume of 2.00 in.3, first convert the mass to grams and the volume to cm3, resulting in a density of approximately 11.3 g/cm3.
Explanation:
The question is asking for the density of a cube of lead (Pb) that has a mass of 0.371 kg and a volume of 2.00 in.3. First, it is necessary to convert the mass from kilograms to grams (since density is often expressed in g/mL) by multiplying the mass by 1000 (0.371 kg × 1000 = 371 g). Then, we need to convert the volume from cubic inches to cubic centimeters (cm3), as the requested density unit is g/mL and 1 cm3 is equivalent to 1 mL. Knowing that 1 in.3 = 16.387 cm3, we convert the volume of the lead cube to cm3 (2.00 in.3 × 16.387 = 32.774 cm3). Finally, to find the density in g/mL, we divide the mass in grams by the volume in mL (371 g / 32.774 cm3 = 11.3 g/cm3).
Arrange the following alkyl bromides in order from most reactive to least reactive in an SN₂ reaction: 1-bromo-2-methylbutane, 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, and 1-bromopentane.
Answer:
1- bromopentane 1- bromo-3-methylbutane 1-bromo2-methylbutane 2-bromo-2-methylbutane
Explanation:
SN square reaction is a concentrated reaction. All the bond making and bond braking occur in one step
The nucleophile attack the back side of the carbon that bears the halide and replace it
With respect to the categories of assets, liabilities, and stockholders' equity presented on the balance sheet (statement of financial position), what are U.S. GAAP and IFRS differences?
Explanation:
GAAP is a generally accepted accounting principle in U.S. it refers to common sets of accepted accounting principle, standards, procedures that the companies and its accountants must follow in order to compile their financial statement.
IFRS are sets of international accounting standards That specify how the financial statements will disclose different types of transactions and other activities. The International Accounting Standards Board (IASB) issues IFRS which defines precisely how accountants are required to maintain and record their accounts. In an attempt to have an universal accounting system, IFRS was developed so that business and accounts can be interpreted from industry to industry, and country to country.
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Chromium reacts with oxygen according to the equation: 4Cr + 3O2 Cr2O3(s). Determine the moles of chromium(III) oxide produced when 4.58 mol of chromium is allowed to react.
Answer:
2.29 moles of Cr₂O₃ are produced
Explanation:
This is the reaction:
4 Cr + 3O₂ → 2Cr₂O₃
Ratio for this equation is 4:2, so 4 moles of chromium can produce the half of moles of chromium(III) oxide
4.58 mol of Cr may produce (4.58 .2)/4 = 2.29 moles of Cr₂O₃
The moles of chromium(III) oxide produced from the reaction of 4.58 mol of chromium with excess oxygen is 2.29 mol, based on the stoichiometry of the chemical equation.
Explanation:To determine the moles of chromium(III) oxide produced from 4.58 moles of chromium reacting with oxygen, we use the stoichiometry of the balanced chemical equation:
4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)
According to this equation, 4 moles of chromium react completely with 3 moles of oxygen to produce 2 moles of chromium(III) oxide. We use this ratio to calculate the amount of chromium(III) oxide produced:
Moles of chromium(III) oxide = (Moles of chromium × Moles of chromium(III) oxide produced) / Moles of chromium reacted
= (4.58 mol Cr × 2 mol Cr2O3) / 4 mol Cr
= 2.29 mol Cr2O3
Therefore, when 4.58 mol of chromium are allowed to react with excess oxygen, 2.29 mol of chromium(III) oxide are produced.
What volume (in mL) of 6 M acetic acid would have to be added to 500mL of a solution of 0.20M sodium acetate in order to achieve a pH = 5.0? The pKa of acetic acid is 4.75.
Answer:
9.3 mL of 6 M acetic acid needs to be added to 500 mL of a solution of 0.20 M sodium acetate to a achieve a pH of 5.0
Explanation:
This problem can be solved by the Henderson-Hasselbalch equation. It's formula is:
[tex]pH=pKa+log(\frac{[CH_3COO^-]}{[CH_3COOH]})[/tex]
The molar concentration can be replaced by the moles of the solute as the volume of the buffer will be the same for both species
[tex]pH=pKa+log(\frac{n_{CH_3COO^-}}{n_{CH_3COOH}})[/tex]
Placing the given data:
[tex]5.0=4.75+log(\frac{0.1}{n_{CH_3COOH}})[/tex]
[tex]n_{CH_3COOH}=(\frac{0.1}{10^{5.0-4.75}})\\\\n_{CH_3COOH}=(\frac{0.10}{1.778})\\\\ n_{CH_3COOH}=0.056moles[/tex]
The volume required to obtain the above-calculated moles can be determined by the molarity of acetic acid
[tex]M_{CH_3COOH}=\frac{n_{CH_3COOH}}{V_{CH_3COOH}(L)}\\\\V_{CH_3COOH}=\frac{n_{CH_3COOH}}{M_{CH_3COOH}}\\\\V_{CH_3COOH}=\frac{0.056}{6.0}\\\\V_{CH_3COOH}=0.0093L\\\\or\\\\V_{CH_3COOH}=9.3mL[/tex]
The above volume of acetic acid can be added to 500 mL of acetate to form 5.0 pH buffer
The volume of 6 M acetic acid needed is (0.356/6)x(500+V(A-)).
Explanation:To calculate the volume of 6 M acetic acid needed to achieve a pH of 5.0, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, the ratio of sodium acetate ([A-]) to acetic acid ([HA]) should be 1:1 to achieve a pH of 5.0.
Using the equation, we can rearrange it to find the concentration of acetic acid:
pH = pKa + log([A-]/[HA])
[HA] = [A-] x 10^(pH - pKa)
Plugging in the values, we get:
[HA] = 0.20 M x 10^(5.0 - 4.75) = 0.20 M x 10^0.25 = 0.20 M x 1.78 = 0.356 M
Now, we can calculate the volume of 6 M acetic acid needed:
[HA] x V(HA) = [HA] x V(A-)
0.356 M x V(HA) = 0.356 M x (500 mL + V(A-))
V(HA) = 0.356 M x (500 mL + V(A-)) / 6 M
V(HA) = (0.356/6)x(500+V(A-)) mL
Since the ratio of sodium acetate to acetic acid is 1:1, the volume of 6 M acetic acid needed is equal to the volume of 0.20 M sodium acetate added to the solution. Therefore, the volume of 6 M acetic acid needed is (0.356/6)x(500+V(A-)).
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What hybridization is required for central atoms that have a trigonal planar arrangement of electron pairs? sp sp2 sp3
How many unhybridized p atomic orbitals are present when a central atom exhibits trigonal planar geometry?
Answer:
The required hybridization for a trigonal planar arrangement is sp2.There is one p atomic orbital unhybridized.Explanation:
In the sp2 hybridization, only two p atomic orbitals (out of three) are hybridized with the s orbital, thus forming a total of three sp2 orbitals.
These three orbitals will point in different directions to minimize the electron repulsion between them. When there are three such orbitals, the geometry that allows such minimization is the trigonal planar.
The unhybridized p orbital is found perpendicular to the plane of the 3 sp2 orbitals.
Hard water often contains dissolved Ca2 and Mg2 ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. A solution is 0.050 M in calcium chloride and 0.085 M in magnesium nitrate. What mass of sodium phosphate would you add to 1.5 L of this solution to completely eliminate the hard water ions.
The mass of sodium phosphate needed to completely eliminate the hard water ions from the solution is approximately 11.06 grams.
First, we need to determine the moles of calcium and magnesium ions present in the solution:
1. Moles of [tex]Ca^2+ ions[/tex] = Molarity of CaCl2 * Volume of solution
Moles of [tex]Ca^2+ ions[/tex] = 0.050 mol/L * 1.5 L = 0.075 mol
2. Moles of [tex]Mg^2+ ions[/tex] = Molarity of Mg(NO3)2 * Volume of solution
Moles of [tex]Mg^2+ ions[/tex] = 0.085 mol/L * 1.5 L = 0.1275 mol
Next, we need to find out the mole ratio of phosphate ions required to precipitate these ions. Since both calcium and magnesium ions form insoluble precipitates with phosphate ions in a 1:1 ratio, the moles of phosphate ions required will be equal to the sum of moles of calcium and magnesium ions:
Moles of phosphate ions = [tex]Moles of Ca^2+ ions + Moles of Mg^2+ ions[/tex]
Moles of phosphate ions = 0.075 mol + 0.1275 mol = 0.2025 mol
Now, we can calculate the mass of sodium phosphate required using its molar mass and the moles of phosphate ions:
Mass of sodium phosphate = Moles of phosphate ions * Molar mass of Na3PO4
Mass of sodium phosphate = 0.2025 mol * 163.94 g/mol = 33.18 g
However, sodium phosphate (Na3PO4) dissociates into three sodium ions [tex](Na^+)[/tex] and one phosphate ion[tex](PO4^3-)[/tex] . Therefore, to find the mass of the compound that would provide the required moles of phosphate ions, we need to adjust for this:
Mass of sodium phosphate = (33.18 g / 3) * 1 = 11.06 g
So, you would need approximately 11.06 grams of sodium phosphate to completely eliminate the hard water ions from the solution.
- Calculate the moles of [tex]Ca^2+ and Mg^2+[/tex] ions using their respective molarities and the volume of the solution.
- Determine the moles of phosphate ions required by summing the moles of [tex]Ca^2+ and Mg^2+[/tex]ions.
- Calculate the mass of sodium phosphate required using its molar mass and the moles of phosphate ions.
- Adjust the calculated mass for the fact that sodium phosphate dissociates into three sodium ions and one phosphate ion.
Complete Question:
Hard water often contains dissolved Ca2 and Mg2 ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. A solution is 0.050 M in calcium chloride and 0.085 M in magnesium nitrate. What mass of sodium phosphate would you add to 1.5 L of this solution to completely eliminate the hard water ions.
A water bath in a physical chemistry lab is 1.85m long, 0.810m wide and 0.740m deep. If it is filled to within 2.57 in from the top, how many liters of water are in it
Final answer:
The water bath, filled to within 2.57 inches of the top, contains approximately 1014.48 liters of water. This calculation involves converting inches to meters, calculating the volume in cubic meters, and then converting that volume to liters.
Explanation:
To calculate the volume of water in the water bath, we need to take into account the dimensions given and the fact that it is filled to within 2.57 inches from the top. First, we need to convert the depth from which it's filled into meters, as the other dimensions are given in meters.
2.57 inches = 0.06533 meters (since 1 inch = 0.0254 meters)
The actual depth filled with water will be:
0.740 m (total depth) - 0.06533 m = 0.67467 m
Next, we multiply the length, the width, and the newly calculated depth to get the volume in cubic meters:
Volume = 1.85 m × 0.810 m × 0.67467 m = 1.01448 cubic meters
To convert the volume from cubic meters to liters, we use the conversion factor 1 cubic meter = 1000 liters:
Volume = 1.01448 m3 × 1000 L/m3 = 1014.48 liters
Therefore, there are approximately 1014.48 liters of water in the water bath.
We create an electron with wavefunction ψ = [ψ1s(r)+3ψ3s(r)]/ √ 10. Find the expected value hEi of the energy in that state. Use the energies of the hydrogen atom to extract your answer in electron volts.
Answer: r=132pm
Explanation:
r = ⟨r⟩=5a0/Z
r=(5(5.29177*〖10〗^(-11) m) x^2)/2
r=132pm
Flammable materials, like alcohol, should never be
dispensed or used near
A. an open door.
B. an open flame.
C. another student.
D. a sink
Answer:
b.open flame because it is fundamental end of the alcohol mixes in with the flame then it will become a bigger fire
The safest practice is to never use flammable materials like alcohol near an open flame as it can easily ignite and cause a fire.
Explanation:Flammable materials such as alcohol should never be dispensed or used near an open flame (option B). These substances are highly reactive and can easily ignite, potentially causing a fire. It's crucial for safety purposes to avoid using flammable materials around direct sources of heat or open flames. This rule is applicable not only in chemistry laboratories, but also in any other scenario where flammable substances are present.
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