Answer:
Decreases
Explanation:
Fatty acid which have the double bond or triple bond are called unsaturated fatty acids. Because of the double or triple bond, unsaturated fatty acids are loosely packed and form some distance among molecules which lowers the melting point of unsaturated fatty acids.
So, if the unsaturation of fatty acid will increase, it leads to more branched and loosely packed molecules and decreases the melting temperature accordingly.
What does the period number tell about the energy levels occupied by
electrons in an atom?
A. The period number tells how many electrons are in the highest
energy level of the atom.
B. The period number tells which is the highest energy level occupied
by the electrons.
C. The period number tells how many electrons are in each sublevel
of the atom.
D. The period number tells how many energy sublevels are occupied
in the atom.
Answer: B. The period number tells which is the highest energy level occupied by the electrons
Explanation:
The period number ( denoted by 'n' ) is the outer energy level that is occupied by electrons in an atom. The period number that an element is in, is the number of energy levels that the element has.When we move across a period from left to right in a periodic table the number of electrons in atoms increases within the same orbit.Thus, we can say that the period number tells which is the highest energy level occupied by the electrons in an atom.
hence, the correct option is B. The period number tells which is the highest energy level occupied by the electrons.
The period number tell about the energy levels occupied by electrons in an atom B. The period number tells which is the highest energy level occupied by the electrons. option B , second option is correct.
What are energy levels ?The fixed distances from an atom's nucleus where electrons may be found are referred to as energy levels (also known as electron shells). Higher energy electrons have greater energy as you move out from the nucleus. A region of space within an energy level known as an orbital is where an electron is most likely to be found.
When a quantum mechanical system or particle is bound, or spatially constrained, it can only take on specific discrete energy values, or energy levels. Classical particles, on the other hand, can have any energy level.
Therefore, option B , second option is correct.
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What is the percent composition of sulfur in SO2?
OA
32.1 g
(32.1 g + 16.0 g + 16.0 g)
x 100%
32.1 g
O B.
x 100%
(16.0 g + 16.0 g)
O c.
(32.1 g + 16.0 g + 16.0 g)
X 100%
32.1 g
(32.1 g)(6.02 x 102)
O D.
100 g
Answer:
5 .04 percent.
Explanation:
it is also known as sulfur dioxide. so it's 5.0.4 percent.
Your question is incomplete. Read below to find the content.
0.4 % is the percent composition of sulfur in SO2.
How is SO2 prepared from sulfur?In the laboratory, sulfur dioxide is prepared by the reaction of metallic sulfite or a metallic bisulfite with dilute acid. For example, a reaction between the dilute sulphuric acid and sodium sulfite will result in the formation of SO2.
What do you mean by percentage composition?The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100. Here, the quantity is measured in terms of grams of the elements present.
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Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for this reaction at 298.15K if the pressure of each gas is 22.20 mm Hg.
Answer:
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
ΔG ≅ 199.91 kJ
Explanation:
Consider the reaction:
[tex]2N_{2(g)} + O_{2(g)} \to 2N_2O_{(g)}[/tex]
temperature = 298.15K
pressure = 22.20 mmHg
From, The standard Thermodynamic Tables; the following data were obtained
[tex]\Delta G_f^0 \ \ \ N_2O_{(g)} = 103 .8 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times \Delta G_f^0 \ N_2O_{(g)} - ( 2 \times \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_{2(g)})[/tex]
[tex]\Delta G^0 _{rxn} = 2 \times 103.8 \ kJ/mol - ( 2 \times 0 + 0)[/tex]
[tex]\Delta G^0 _{rxn} = 207.6\ kJ/mol[/tex]
The equilibrium constant determined from the partial pressure denoted as [tex]K_p[/tex] can be expressed as :
[tex]K_p = \dfrac{(22.20)^2}{(22.20)^2 \times (22.20)}[/tex]
[tex]K_p = \dfrac{1}{ (22.20)}[/tex]
[tex]K_p[/tex] = 0.045
[tex]\Delta G = \Delta G^0 _{rxn} + RT \ lnK[/tex]
where;
R = gas constant = 8.314 × 10⁻³ kJ
[tex]\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15 \ ln(0.045)[/tex]
[tex]\Delta G =207.6 + 2.4788191 \times \ ln(0.045)[/tex]
[tex]\Delta G =207.6+ (-7.687048037)[/tex]
[tex]\Delta G =[/tex] 199.912952 kJ
ΔG ≅ 199.91 kJ
Which best describes the total mass of a sample of water when it condenses
from a liquid to a gas?
A. The mass is less because the water molecules get closer together
and take up more space.
B. The mass is the same because the decrease in energy equals the
increase in the number of molecules.
C. The mass is the same because water molecules are not created or
destroyed during a phase change.
D. The mass is greater after water condenses because the mass of
the molecules increases.
Answer:
Its C I hopefully help you
1. Methanol is a high-octane fuel used in high performance racing engines. 2 CH3OH(l) + 3O2(g) → 2CO2(g) + 4 H20(g) a) Calculate ∆H० and ∆S० using thermodynamic data, and then ∆G
Answer:
The reaction given in the question is:
2CH₃OH (l) + 3O₂ (g) ⇒ 2CO₂ (g) + 4H₂O (g)
The values of ΔH°formation and ΔS° of the reactants and products given in the reaction based on the thermodynamics data is:
ΔH°formation values of CH3OH (l) is -238.4 kJ/mol, CO2(g) is -393.52 kJ/mol, H2O (g) is -241.83 kJ/mol and O2 (g) is 0.
The S° values of CH3OH (l) is 127.19 J/molK, CO2(g) is 213.79 J/molK, H2O (g) is 188.84 J/moleK, and O2 (g) is 205.15 J/molK.
Now the values of ΔH° and ΔS° are,
ΔH°rxn = 2 * ΔH°formation CO2 (g) + 4 * ΔH°formation H2O (g) - 2*ΔH°formation CH3OH (l)
ΔH°rxn = 2 * (-393.52) + 4 (-241.83) -2 * (-238.4)
ΔH°rxn = -1277.56 kJ/mole
ΔS°rxn = 2 * S° CO2 (g) + 4 * S° H2O (g) - 2*S° CH3OH (l) - 3 * S° O2 (g)
ΔS°rxn = 2 * 213.79 + 4 * 188.84 - 2 * 127.19 - 3*205.15
ΔS°rxn = 313.11 J/mole/K
Now the formula for calculating ΔG°rxn is,
ΔG°rxn = ΔH°rxn - TΔS°rxn
ΔG°rxn = -1277.56 * 1000 J/mole - 298 * 313.11 J/mole
ΔG°rxn = -1370.86 kJ/mol
If the sugar concentration in a cell is 3% and the concentration in the cell’s environment is 5%, how can the cell obtain more sugar? {Hint: Sugar is polar and does not pass through the cell membrane easily.}
Select one:
a. Sugar can undergo facilitated diffusion through a channel protein, and no energy is required.
b. The cell must use active transport to pump the sugar in. Energy is required.
c. Sugar can diffuse straight through the phospholipid bilayer.
d. The cell CANNOT obtain more sugar. It is doomed.
The cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.
FACILITATED DIFFUSION:
Facilitated diffusion is the process whereby molecules move across a cell through the help of carrier/channel proteins. Facilitated diffusion is a type of passive transport and hence, does not require energy to occur. This is because movement occurs down a concentration gradient (high to low). According to this question, the sugar concentration in a cell is 3% and the concentration in the cell’s environment is 5%. This means that the sugar can travel down a concentration gradient across the cell membrane. However, because sugar is a polar molecule and does not pass through the cell membrane easily, a carrier proteins is needed to aid its movement. Therefore, the cell can obtain more sugar when the sugar undergo facilitated diffusion through a channel protein, and no energy is required.Learn more at: https://brainly.com/question/18122054?referrer=searchResults
A gas mixture containing N2 and O2 was kept inside a 2.00 L container at a temperature of 23.0°C and a total pressure of 1.00 ATM the partial pressure of oxygen was 0.722 ATM how many grams of nitrogen are present in the gas mixture
Answer:
0.641 g of Nitrogen are present in the mixture.
Explanation:
We use the Ideal Gases Law, to solve this question.
For the mixture:
P mixture . V mixture = mol mixture . R . T
We convert the T° to K → 23°C + 273 = 296 K
R = Ideal gases constant → 0.082 L.atm/mol.K
1 atm . 2L = mol mixture . 0.082 L.atm/mol.K . 296K
2 atm.L / ( 0.082 mol /L.atm) . 296 = 0.0824 moles
We know that sum of partial pressure = 1
Partial pressure N₂ + Partial pressure O₂ = 1
1 - 0.722 atm = Partial pressure N₂ → 0.278 atm
We apply the mole fraction concept:
Partial pressure N₂ / Total pressure = Moles N₂ / Total moles
Moles N₂ = (Partial pressure N₂ / Total pressure) . Total moles
Moles N₂ = (0.278 atm / 1 atm) . 0.0824 mol → 0.0229 moles
We convert the moles to mass → 0.0229 mol . 28 g/mol = 0.641 g
641 mg
18. Sucralose contains which two functional groups: (2 points)
A) benzene
B) halogen
C) carboxyl
D) hydroxy!
Answer:
The correct answer is option B and D, that is, halogen (chlorine) and hydroxyl.
Explanation:
An artificial sweetener and sugar substitute is sucralose. It is noncaloric as the majority of the sucralose ingested does not get dissociated within the body. The generation of sucralose takes place by the chlorination of sucrose. It is about 300 to 1000 times sweeter in comparison to sucrose.
The consumption of sucralose is safe for both nondiabetics and diabetics, it is used in various food and beverage components due to non-caloric sweetener characteristics. It does not affect the levels of insulin and does not affect dental health. As it is produced by chlorination of sucrose, thus, the functional groups present in it are a halogen (chlorine) and a hydroxyl.
Which of the following elements is in the same family as fluorine?
a. silicon
b. antimony
O c. iodine
O d. arsenic
e. None of these.
Answer:
c iodine
Explanation:
fluorine is a halogen group element like Bromine, Iodine,Astatine,Chloride
A 40.80 gram sample of copper is heated in the presence of excess sulfur. A metal sulfide is formed with a mass of 51.09 g. Determine the empirical formula of the metal sulfide.
Answer:
Cu₂S
Explanation:
From the question,
Cu S
Mass: 40.80 g 51.09-40.80 = 10.29 g
Mole ratio: 40.80/63.5 10.29/32.1
0.64 : 0.32
Divide by the smallest,
0.64/0.32 : 0.32/0.32
2 : 1
Therefore,
Empirical formula = Cu₂S.
This pluton occurs deep in Earth and does not cause any changes to the surface of Earth . True or False
Answer:
The given statement is false.
Explanation:
However, if the pluton exists beneath the ground, this could be conveniently shown in the illustration something from the peak such pluton appears convex in form resembling a lopolith and perhaps diapir, which would be a particular form of statistically significant pluton recognized as the sill.Mostly from the figure it could also be shown that subsurface sheets are lined or curved, throughout the pluton mold. And therefore it is inferred that such a pluton creates adjustment to something like the ground atmosphere by altering the form of the levels above it.So that the given is incorrect.
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties.HCl(g)+NaOH(s)⟶NaCl(s)+H2O(l)
Answer:
-179.06 kJ
Explanation:
Let's consider the following balanced reaction.
HCl(g) + NaOH(s) ⟶ NaCl(s) + H₂O(l)
We can calculate the standard enthalpy change for the reaction (ΔH°r) using the following expression.
ΔH°r = 1 mol × ΔH°f(NaCl(s)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(HCl(g)) - 1 mol × ΔH°f(NaOH(s))
ΔH°r = 1 mol × (-411.15 kJ/mol) + 1 mol × (-285.83 kJ/mol) - 1 mol × (-92.31 kJ/mol) - 1 mol × (-425.61 kJ/mol)
ΔH°r = -179.06 kJ
Diluting sulfuric acid with water is highly exothermic:
(Use data from the Appendix to find for diluting 1.00 mol of H2SO4(l) (d = 1.83 g/mL) to 1 L of 1.00 MH2SO4(aq) (d = 1.060 g/mL). )
Suppose you carry out the dilution in a calorimeter. The initial T is 25.2°C, and the specific heat capacity of the final solution is 3.458 J/gK. What is the final T in °C ?
Answer:
The correct answer is 51.2 degree C.
Explanation:
The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.
Now the dHreaction = dHf (product) - dHf (reactant)
= -909.3 - (-814)
dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.
The heat change in calorimetry can be determined by using the formula,
q = mass * specific heat capacity * change in temperature -----------(i)
Based on the given information, the density of H₂SO₄ is 1.060 g/ml
The volume of H₂SO₄ is 1 Liter
Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams
The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.
ΔT = T₂ -T₁ = T₂ - 298.2 K
Now putting the values in equation (i) we get,
95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)
(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K
T₂ = 298.2 K + 26 K
T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.
In this experiment, you will analyze your sample by TLC by spotting pure benzophenone and your product and eluting with 5:1 hexanes/ethyl acetate. Choose the statement that BEST describes what you should observe for a successful experiment.
A) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value.
B) The benzophenone in the standard and reaction samples should travel the least and will have the same Rf value. No unreacted benzhydrol should be observed at a higher Rf value.
C) The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. Unreacted benzhydrol should be observed at a lower Rf value.
D) The benzophenone in the standard and reaction samples should travel the farthest and will have the different Rf values. No unreacted benzhydrol should be observed at a lower Rf value.
Answer:
The correct answer is statement A.
Explanation:
In a media comprising 20 percent ethyl acetate/hexane, as the benzophenone is non-polar, so it will travel farther with high Rf value. On the other hand, as benzohydrol is a polar molecule, therefore, it should be at lower Rf value and will not rise in the given media.
For an experiment to be successful, there should not be any unreacted benzohydrol to be left in the experimental system. The benzophenone in the reaction as well as the standard samples should exhibit similar Rf value.
The benzophenone in the standard and reaction samples should travel the farthest and will have the same Rf value. No unreacted benzhydrol should be observed at a lower Rf value to analyze by TLC.
What is TLC ?In synthetic chemistry, thin-layer chromatography (TLC) is a method that is frequently used to identify compounds, assess their purity, and monitor the progress of a reaction. Additionally, it enables the solvent system for a specific separation problem to be optimized.
The foundation of TCL is the adsorption-based separation theory. The separation depends on how sensitive different chemicals are to the stationary and mobile phases.
TLC, or thin layer chromatography, is a technique for isolating the components of mixtures before analysis. TLC can be used to identify compounds, ascertain their identities, and ascertain the purity of a compound.
Thus, option A is correct.
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A molecule of aluminum fluoride has one aluminum atom. How many fluorine atoms are present?
Answer:
3 fluorine atoms will be present
Answer:
3
Explanation:
The chemical formula of aluminum fluoride is AlF3. As you can see, there is a 1:3 ratio of aluminum atoms to fluorine atoms. Therefore, if a molecule of AlF3 has one aluminum atom, you know there must be 3 fluorine atoms present.
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If phosphorus (P) has 4 naturally occurring isotopes, phosphorus-29(32.7.%), phosphorus-30(48.03%), phosphorus-31(18.4%), and phosphorus-33 (0.87%), what is its average r.a.m.?
The Average atomic mass of phosphorus is 29.9.
What is Average atomic mass ?The average atomic mass (sometimes called atomic weight) of an element is the weighted average mass of the atoms in a naturally occurring sample of the element.
Average masses are generally expressed in unified atomic mass units (u), where 1 u is equal to exactly one-twelfth the mass of a neutral atom of carbon-12.
An element can have differing numbers of neutrons in its nucleus, but it always has the same number of protons.
The versions of an element with different neutrons have different masses and are called isotopes.
The average atomic mass for an element is calculated by summing the masses of the element’s isotopes, each multiplied by its natural abundance on Earth i.e,
Average atomic mass of P = ∑(Isotope mass) (its abundance)
∴ Average atomic mass of P = (P-29 mass) (its abundance) + (P-30 mass)(its abundance) + (P-31 mass) (its abundance) + (P-33 mass) (its abundance)
Abundance of isotope = % of the isotope / 100.
∴ Average atomic mass of P = (29)(0.327) + (30)(0.4803) + (31)(0.184) + (33)(0.0087) = 29.88 a.m.u ≅ 29.9 a.m.u.
Hence , The Average atomic mass of phosphorus is 29.9.
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The NMR spectrum of your final compound will contain extra peaks that were not present in your starting material. For what hydrogen nuclei do those peaks occur?
Answer:
The peaks are registered from tetramethyl silane (TMS)
Explanation:
Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.
Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.
I hope this helped.
Which is the electron configuration for bromine?
Answer:
The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.
Explanation:
1s^2
2s^2
2p^6
3s^2
3p^6
4s^2
3d^10
4p^5
1)The average lethal dose of Valium is 1.52 mg/kg of body weight. Estimate how many grams of Valium would be lethal for a 200.-lb woman. Show all your calculations. (1lb = 453.6 g)
2) A patient in hospital is receiving the antibiotic amoxcillin IV at the rate of 50. mL/h. The IV contains 1.5 g of the antibiotic in 1000. mL. (IV stands for intravenous). Calculate the mg/min of the drip. Show all your calculations
Answer:
1. 0.138g of valium would be lethel in the woman
2. 125mg/min is the drip of the patient
Explanation:
1. In a body, an amount of Valium > 1.52mg / kg of body weight would be lethal.
A person that weighs 200lb requires:
200lb × (453.6g / 1lb) × (1kg / 1000g) = 90.72kg (Weight of the woman in kg)
90.72kg × (1.52mg / kg) =
137.9mg ≡
0.138g of valium would be lethel in the woman2. The IV contains 1.5g = 1500mg/mL.
If the patient is receiving 5.0mL/h, its rate in mg/h is:
5.0mL/h × (1500mg/mL) = 7500mg/h
Now as 1h = 60min:
7500mg/h × (1h / 60min) =
125mg/min is the drip of the patientFructose-2,6-bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose-1,6-bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose-2,6-bisphosphate is regulated by many hormones, second messengers, and enzymes.
How do the following affect glycolysis and gluconeogenesis?
Activate glycolysis Inhibit gluconeogenesis Activate gluconeogenesis Inhibit glycolysis
1. increased levels of fructose-2,6-bisphosphatase
2. activation of fructose-2,6-bisphosphate (FBPase-2)
3. increased glucagon levels
4. activation of PFK-2
5. increased levels of CAMP
Answer:
1. Increased levels of fructose-2,6-bisphosphatase : Activate gluconeogenesis Inhibit glycolysis
2. Activation of fructose-2,6-bisphosphate (FBPase-2) : Activate glycolysis Inhibit gluconeogenesis
3. Increased glucagon levels : Activate gluconeogenesis Inhibit glycolysis
4. Activation of PFK-2 : Activate glycolysis Inhibit gluconeogenesis
5. Increased levels of CAMP : Activate gluconeogenesis Inhibit glycolysis
Explanation:
Glycolysis is the breakdown of glucose molecules in order to release energy in the form of ATP in response to the energy needs of the cells of an organism.
Gluconeogenesis is the process by which cells make glucose from other molecules for other metabolic needs of the cell other than energy production.
Glycolysis and gluconeogenesis are metabolically regulated in the cell by various enzymes and molecules.
The following shows the various regulatory methods and their effects on both processes:
1. The enzyme fructose-2,6-bisphosphatase functions in the regulation of both processes. It catalyzes the breakdown of the molecule fructose-2,6-bisphosphate which is an allosteric effector of two enzymes phosphofructokinasse-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase-1 which fuction in glycolysis and gluconeogenesis respectively.
Increased levels of fructose-2,6-bisphosphatase activates gluconeogenesis and inhibits glycolysis by its breakdown of fructose-2,6-bisphosphate.
2. Fructose-2,6-bisphosphate increases the activity of PFK-1 and inhibits the the activity of FBPase-1. The effect is that glycolysis is activated while gluconeogenesis is inhibited.
3. Glucagon is a hormone that stimulates the synthesis of cAMP. It fuctions to activate gluconeogenesis and inhibit glycolysis.
4. Phosphosfructikinase-2, PFK-2 is an enzyme that catalyzes the formation of fructose-2,6-bisphosphate. Activation of PFK-2 results the activation of glycolysis and inhibition of gluconeogenesis.
5. Cyclic-AMP (cAMP) synthesis in response to glucagon release serves to activate a cAMP-dependent protein kinase which phosphorylates the bifunctional protein PFK-2/FBPase-2. This phosphorylation enhances the activity of FBPase-2 while inhibiting the activity of PFK-2, resulting in the activation of gluconeogenesis and inhibition of glycolysis.
Suppose you have a container filled with air at 212 oF. The volume of the container 1.00 L, the pressure of air is 1.00 atm. The molecular composition of air is 79% N2 and 21% O2 for simplification. Calculate the mass of air and moles of O2 in the container.
Answer:
[tex]m_{air}=0.947g[/tex]
[tex]n_{O_2} =0.00686molO_2[/tex]
Explanation:
Hello,
In this case, we can firstly use the ideal gas equation to compute the total moles of the gaseous mixture (air) with the temperature in Kelvins:
[tex]T=212\°F=100\°C=373.15K\\\\n=\frac{PV}{RT}=\frac{1.00atm*1.00L}{0.082\frac{atm*L}{mol*K}*373.15K}\\ \\n=0.0327mol[/tex]
In such a way, since the molar mass of air is 28.97 g/mol, we can compute the mass of air with a single mass-mole relationship:
[tex]m_{air}=0.0327mol*\frac{28.97g}{1mol} =0.947g[/tex]
Finally, knowing that the 21% of the 0.0327 moles of air is oxygen, its moles turn out:
[tex]n_{O_2}=0.0327mol*\frac{0.21molO_2}{1mol} =0.00686molO_2[/tex]
Best regards.
Draw the structure of 1,4-hexanediamine.
Draw the molecule on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced TemplateTowbars. The single bond is active by default. Include all hydrogen atoms.
View Available Hint(s)
Answer:
1,4-hexanediamine contains two [tex]-NH_{2}[/tex] functional groups.
Explanation:
1,4-hexanediamine is an organic molecule which contains two [tex]-NH_{2}[/tex] functional groups at C-1 and C-4 position.
The longest carbon chain in 1,4-hexanediamine contains six carbon atoms.
Molecular formula of 1,4-hexanediamine is [tex]C_{6}H_{16}N_{2}[/tex].
1,4-hexanediamine used as a bidentate ligand in organometallic chemistry.
The structure of 1,4-hexanediamine is shown below.
A. Identify the structure drawn below.
Answer:
Hexane
Explanation:
You have a carbon structure with only single bonds. This means that the name will end in -ane.
There are 6 carbon atoms. This means that the name will begin with hex-.
The structure is hexane.
6. Potassium hydrogen phthalate (KHP, KHC8H4O4) is also a good primary standard. 20 mL of NaOH was titrated with 0.600 M KHC8H4O4 solution. The data was graphed and the equivalence point was found when 15.5 mL of the standard 0.600 M KHP solution was added. The reaction equation is: a. What is the molar ratio of NaOH:KHC8H4O4? b. What is the molarity of the NaOH solution?
Answer:
a. 1
b. 0.465M NaOH
Explanation:
KHP reacts with NaOH as follows:
KHP + NaOH → KP⁻ + Na⁺ + H₂O
a. Molar ratio represents how many moles of NaOH reacts per mole of KHP. As you can see in the reaction, 1 mole of NaOH reacts with 1 mole of KHP. Molar ratio is:
1/1 = 1
b. With volume and molar concentration of the KHP solution you can find how many moles of KHP were added until equivalence point, thus:
15.5mL = 0.0155L ₓ (0.600 moles KHP / L) = 0.0093 moles of KHP
In equivalence point, moles of NaOH = Moles KHP. That means moles of NaOH titrated are 0.0093 moles NaOH.
The volume of the NaOH solution was 20mL = 0.020L. Molarity of the solution is:
0.0093 moles NaOH / 0.020L =
0.465M NaOHa. The balanced equation shows a 1:1 molar ratio between NaOH and KHC₈H₄O₄. This means that for every 1 mole of NaOH, we require 1 mole of KHC₈H₄O₄. Therefore, the molar ratio of NaOH:KHC₈H₄O₄ is 1:1.
The balanced equation for the reaction:
NaOH + KHC₈H₄O₄ → NaKC₈H₄O₄ + H₂O
b. Molarity of KHP solution × volume of KHP solution = Molarity of NaOH solution × volume of NaOH solution at the equivalence point
Molarity of KHP solution = 0.600 M
Volume of KHP solution = 15.5 mL = 0.0155 L
Volume of NaOH solution at the equivalence point = 20 mL = 0.0200 L
Molarity of NaOH solution = (Molarity of KHP solution × volume of KHP solution) / volume of NaOH solution at the equivalence point
Molarity of NaOH solution = (0.600 M × 0.0155 L) / 0.0200 L
Molarity of NaOH solution ≈ 0.465 M
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Write electron configurations for the following ion: Cd2 Cd2 . Express your answer in order of increasing orbital energy. For example, the electron configuration of LiLi would be entered in complete form as 1s^22s^1 or in condensed form as [He]2s^1.
Answer:
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
Explanation:
Before proceeding to write out the electron configuration of Cd2+, we have to obtain the electron configuration of Cadmium (Cd),
Cadmium has an atomic number of 48, this means that a neutral cadmium atom will have a total of 48 electrons surrounding its nucleus.
The electronic configuration of Cadmium is;
Cd: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
The shorthand notation is given as;
Cd: [Kr] 4d¹⁰5s²
Cd2+ means that it has two less electrons, hence it's electron configuration is given as;
Cd2+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 4d10 or [Kr] 4d¹⁰
Solid sodium oxide and gaseous water are formed by the decomposition of solid sodium hydroxide (NaOH) .
Write a balanced chemical equation for this reaction.
Answer:
2NaOH(s) → Na₂O(s) + H₂O(g)
Hope that helps.
Mass of the condensed unknown liquid: 0.3175 g Temperature of the water bath: 99.00 oC Pressure of the gas: 748.2 mmHg Volume of the flask (volume of the gas): 145.0 mL Given : Kelvin = t oC + 273.15 1 L = 1000 mL 1 atm = 760 mmHg Gas constant: R = 0.08206 atm L / mole K; Ideal Gas Law: PV = nRT 1. What is the pressure of the gas in atm? (1 points) 2.
Answer:
1. 0.98 atm
Explanation:
The following data were obtained from the question:
Mass of unknown liquid (m) = 0.3175 g
Temperature (T) = 99 °C
Pressure (P) = 748.2 mmHg
Volume (V) = 145.0 mL
Gas constant (R) = 0.08206 atm.L/Kmol
1. Determination of the pressure in atm.
760 mmHg = 1 atm
Therefore,
748.2 mmHg = 748.2/760 = 0.98 atm
Therefore, the pressure in atm is 0.98 atm.
A compound is found to contain 18.28 % phosphorus , 18.93 % sulfur , and 62.78 % chlorine by mass. To answer the question, enter the elements in the order presented above. QUESTION 1: The empirical formula for this compound is . QUESTION 2: The molar mass for this compound is 169.4 g/mol. The molecular formula for this compound is
Answer:
1. EF = PSCl₃; 2. MF = PSCl₃
Explanation:
1. Empirical formula
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our first job is to calculate the molar ratio of P:S:Cl.
Assume 100 g of the compound.
(a) Calculate the mass of each element.
Then we have 18.28 g P, 18.93 g S, and 67.28 g Cl.
(b) Calculate the moles of each element
[tex]\text{Moles of P} = \text{18.28 g C} \times \dfrac{\text{1 mol P}}{\text{30.97 g P}} = \text{0.5902 mol P}\\\\\text{Moles of S} = \text{18.93 g S} \times \dfrac{\text{1 mol S}}{\text{32.06 g S }} = \text{0.5905 mol S}\\\\\text{Moles of Cl} = \text{62.78 g Cl} \times \dfrac{\text{1 mol Cl}}{\text{35.45 g Cl }} = \text{1.771 mol Cl}[/tex]
(c) Calculate the molar ratio of the elements
Divide each number by the smallest number of moles
P:S:Cl = 0.5902:0.5905:1.898 = 1:1.000:3.000 ≈ 1:1:3
(d) Write the empirical formula
EF = PSCl₃
The empirical formula for this compound is PSCl₃.
2. Molecular formula
(a) Calculate the ratio of the molecular and empirical formula masses
n = (169.4 u)/(169.40 u) = 1.000 ≈ 1
(b) Calculate the molecular formula
MF = (EF)ₙ = (EF)₁ = PSCl₃
The molecular formula for this compound is PSCl₃.
Draw the Lewis structure of acetaldehyde (CH₃CHO) and then choose the appropriate pair of molecular geometries of the two central atoms.
a) tetrahedral/trigonal pyramidal.
b) trigonal planar / linear.
c) tetrahedral / trigonal planar.
d) trigonal/tetrahedral.
e) planar /trigonal planar.
Given that,
Draw the Lewis structure of acetaldehyde (CH₃CHO).
We know that,
The Lewis structure shows the number of electrons around an atom.
According to structure,
We need to find the molecular geometries of the two central atoms
Using molecular geometries
For first central atom,
Number of bond pair = 2
Here, double bond to O count as single bond
The number of lone pair is zero.
The geometry is Trigonal planar.
For second central atom,
Number of bond pair = 4
The number of lone pair is zero.
The geometry is tetrahedral
Hence, The molecular geometries of the two central atoms are trigonal planar and tetrahedral.
(d) is correct option.
The central carbon atoms in acetaldehyde have a tetrahedral geometry and a trigonal planar geometry respectively.
Acetaldehyde has two central carbon atoms. The Lewis structure of acetaldehyde shows the arrangement of electrons around the atoms in the compound. The lone pairs are shown as dots while the bond pairs are represented using a single dash.
The first central carbon atom in acetaldehyde has a tetrahedral geometry while the second central carbon atom in acetaldehyde has a trigonal planar geometry.
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A four carbon chain; the second carbon is also single bonded to CH3. Spell out the full name of the compound
Answer:
This description shows a methyl group.
Explanation: