Answer:
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96rduttie
A gas mixture contains 3.50 moles of helium, 5.00 moles of krypton and 7.60 moles of neon. A) What is the mole fraction for each gas
Answer:
.217, .311, and .472, respectively.
Explanation:
The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).
X of helium=3.50/16.10 = .217
X of krypton=5.00/16.10 = .311
X of neon=7.60/16.10 = .472
Izopropanole doesn't form by Select one:
a. Reduction of propan-2-one
b. Hydration of 2-chloropropane
c. Hydration of propene
d. Reduction of propanal
Izopropanol doesn't form by Hydration of propene.
What is Reduction of propan-2-one?Acetone, propanone, or dimethyl ketone, exists as an organic compound with the formula (CH₃)₂CO. It stands for the simplest and smallest ketone. Reduction of ketones can be accomplished by hydrogenation (or) utilizing Grignard Reagent. Let us now decrease propanone to propan-2-ol by hydrogenation. The reduction of propane in the existence of catalyst platinum along with hydrogen provides the product propan-2-ol.
The correct answer is option A.
Izopropanol doesn't form by Hydration of propene.
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If we represent the equilibrium as:...N2O4(g) 2 NO2(g) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the temperature is increased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. 3. When the temperature is increased the equilibrium concentration of NO2: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question.
Answer:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases.
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Explanation:
Hello,
In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.
With the aforementioned, we can conclude that the chemical reaction:
[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]
Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:
[tex]G=H-TS[/tex]
As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:
[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Regards.
If a gas is kept in a container with a constant volume and the pressure is reduced, how will the temperature of the gas be affected?
Answer:
The pressure law states that for a constant volume of gas in a sealed container the temperature of the gas is directly proportional to its pressure. ... This means that they have more collisions with each other and the sides of the container and hence the pressure is increased.
The temperature of the gas is directly proportional to its pressure.
What is an ideal gas equation?The ideal gas equation can be given as:
PV = nRT
where P =pressure, V = volume, n = moles of gas, R = rydberg constant, and T = temperature.
The pressure law states that for a constant volume of gas in a container, the temperature of the gas is directly proportional to its pressure which means there are more collisions with each other and the sides of the container and hence the pressure is increased.
Hence, the temperature of the gas is directly proportional to its pressure.
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P.1 is a variant of SARS-CoV2. This is the so-called "Brazil variant". It has the amino acid the substitution N501Y in the spike protein on the virus surface. In cellular fluids at ~pH 7 the wild type spike protein has a net negative charge. This substitution will make the spike protein
more negative
less hydrophobic
able to absorb move UV light
less negative
There are various variants of Cov id virus. The Brazilian variant P also known as Gamma variant is the third variant of the original SARS-CoV2.
The correct answer is more negative
This variant has raised concerns since it has ability to spread more quickly then previous variants and this is more negative variant.
It is assumed that Cov id variant Gamma and Delta has ability to absorb move UV light but this is not proved yet and research is underway.
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When equal moles of an acid and a base are mixed, after reaction the two are compounds are said to be at the _______________. Select one: Indicator point Stoichiometric point Equilibrium point End point
Answer:
when equal moles of an acid and base are mixed,after reaction the two are compounds are said to be at the Equivalent point.
PLEASE HELP!! 40 POINTS
Answer:
1) 6.524779402×10^(-17)
2)521.1g
3)113
Explanation:
Answer: 1) 6.524779402×10^(-17)
2)521.1g
Explanation:
A certain reaction has an activation energy of 39.5 kJ/mol. As the temperature is increased from 25.0°C to a higher temperature, the rate constant increases by a factor of 5.90. Calculate the higher temperature (in °C).
Answer:
Explanation:
We shall apply Arrhenius equation which is given below .
[tex]ln\frac{k_2}{k_1} = \frac{E_a}{R} [\frac{1}{T_1} -\frac{1}{T_2} ][/tex]
K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .
Putting the given values
[tex]ln\frac{5.9}{1} = \frac{39500}{8.3} [\frac{1}{298} -\frac{1}{T_2} ][/tex]
[tex].000373= [\frac{1}{298} -\frac{1}{T_2} ][/tex]
T₂ = 335.27 K
= 62.27 °C
The higher temperature is 62.27°C.
Calculating the higher temperature:Given that the activation energy of the reaction is:
Eₐ = 39.5 kJ/mol
initial temperature T₁ = 25°C = 298K
Let the final temperature be T₂
The rate constant at temperature T₁ be K₁, and at a temperature T₂ be K₂.
According to the question: K₂/K₁ = 5.9
Now, applying the Arrhenius equation we get:
[tex]\ln\frac{K_2}{K1}=\frac{E_a}{R}[\frac{1}{T_1} -\frac{1}{T_2}]\\\\\ln(5.9)= \frac{39.5}{8.3}[\frac{1}{298} -\frac{1}{T_2}]\\\\0.000373=\frac{1}{298} -\frac{1}{T_2}[/tex]
T₂ = 335.27K
T₂ = 335.27 -273
T₂ = 62.27°C
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When balancing redox reactions under acidic conditions, hydrogen is balanced by adding: Select the correct answer below:
a. hydrogen gas
b. water molecules
c. hydrogen atoms
d. hydrogen ions
Answer:
water molecules
Explanation:
Redox reactions are carried out under acidic or basic conditions as the case may be.
If the reaction is carried out in an acid medium, then we must balance the hydrogen ions on the lefthand side of the reaction equation with water molecules on the righthand side of the reaction equation.
For instance, the equation for reduction of MnO4^- under acidic condition is shown below;
MnO4^-(aq) + 5e + 8H^+(aq) --------> Mn^2+(aq) + 4H2O(l)
Calculate the pH of a 0.20 M NH3/0.20 M NH4Cl buffer after the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Answer:
Explanation:
For pH of a buffer solution , the formula is
pH = pKa + log [ Base ] / [ conjugate acid ]
= pKa + log [ NH₃ ] / [ NH₄⁺ ]
Ka = Kw / Kb
Kb for NH₄OH = 1.8 x 10⁻⁵
Ka = 10⁻¹⁴ / 1.8 x 10⁻⁵
= 5.6 x 10⁻¹⁰
pH = - log ( 5.6 x 10⁻¹⁰ ) + log 0.2 / 0.2
= 10 - log 5.6
= 9.25
Effect of addition of HCl
H⁺ of HCl will react with NH₃ to produce NH₄⁺
25 mL of .1 HCl = 2.5 mM of HCl
25 mL of .1 NH₄⁺ = 2.5 mM of NH₄⁺
65 mL of .2 M NH₃ = 13 mM of NH₃
65 mL of .2 M NH₄⁺ = 13 mM of NH₄⁺
NH₃ + H⁺ = NH₄⁺
NH₄⁺ formed = 2.5 + 13 mM
15.5 mM of NH₄⁺
NH₃ = 13 mM
Concentration of NH₃ = 13 / 90
Concentration of NH₄⁺ = 15.5 / 90
pH of final buffer mixture
= 9.6 + log 13 / 15.5
= 9.25 - .076
= 9.174
The pH value is mathematically given as
pH= -6.332.
What is the pH of a 0.20 M NH3/0.20 M NH4Cl buffer?Question Parameters:
the pH of a 0.20 M NH3/0.20 M NH4Cl buffer
the addition of 25.0 mL of 0.10 M HCl to 65.0 mL of the buffer.
Generally, the equation for the Chemical Reaction is mathematically given as
HCl + NH3 --> NH4^+ + Cl^-
Therefore
pH= pka + log(13/14).
pH= -6.3 + log 0.93.
pH= -6.3+ (-0.032).
pH= -6.332.
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I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water in °C? The density of water is approximately 1g/mL.
Answer:
THE FINAL TEMPERATURE OF WATER IS -4.117 °C
Explanation:
Mass of the aluminium = 50 g
c = 0.88 J/g C
Initial temperature of aluminium = 225 °C
Volume of water = 100 ml
Density of water = 1 g/ml
Mass of water = density * volume of water
Mass of water = 1 * 100 = 100 g of water
Initial temperature of water = 20 C
It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,
So therefore,
heat lost by aluminium = heat gained by water
H = mass * specific heat capacity * temeprature change
So:
m c ( T2- T1) = m c (T2-T1)
50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)
44 ( T2 - 225 ) = 418 ( T2 - 20)
44 T2 - 9900 = 418 T2 - 8360
-9900 + 8360 = 418 T2 - 44 T2
-1540 = 374 T2
T2 = - 4.117
So therefore the final temperature of water is -4.117 °C
Briefly workout the relationship between these constants:
[tex]{ \bf{K _{sp} \: and \: K _{c} }}[/tex]
In consideration of the decopmposition of hydrogen iodide.
[tex]{ \sf{2HI _{(g)} →H _{2(g)} +I _{2(g)} }}[/tex]
[tex]{ \tt{any \: help \: is \: appreciated}}[/tex]
Kc require (aqueous/gaseous) products to be on the numerator and (aqueous/gaseous) reactants to be in the denominator, whereas Ksp will require (aqueous) products to be on the numerator and (aqueous) reactants to be in the denominator. Both require products on top and reactants in the bottom.
K = [products] / [reactants]
Kc is used when a reaction reaches dynamic equilibrium, whereas Ksp is used when an insoluble ionic solid dissolved by a tiny amount in a solution, as well as in determining whether or not a precipitate will form.
Kc can be used to measure equilibrium concentration for all reactions, whereas Ksp is limited to only ionic compounds' solubility.
The decomposition of HI (g) will required the use of Kc since the species are all gaseous, and gases cannot be ionic.
What is the final volume V2 in milliliters when 0.551 L of a 50.0 % (m/v) solution is diluted to 23.5 % (m/v)?
Answer:
[tex]V_2=1.17L[/tex]
Explanation:
Hello,
In this case, for dilution processes, we must remember that the amount of solute remains the same, therefore, we can write:
[tex]V_1C_1=V_2C_2[/tex]
Whereas V accounts for volume and C for concentration that in this case is %(m/v). In such a way, the final volume V2 turns out:
[tex]V_2=\frac{V_1C_1}{C_2}= \frac{0.551L*50.0\%}{23.5\%}\\ \\V_2=1.17L[/tex]
Best regards.
In the reaction 2 AgI + HgI 2 → Ag 2HgI 4, 2.00 g of AgI and 3.50 g of HgI 2 were used. What is the limiting reactant?
Answer:
AgI I the limiting reactant.
Explanation:
The balanced equation for the reaction is given below:
2AgI + HgI2 → Ag2HgI4
Next, we shall determine masses AgI and HgI2 that reacted from the balanced equation.
This is illustrated below:
Molar mass of Agl = 108 + 127 = 235 g/mol
Mass of AgI from the balanced equation = 2 x 235 = 470 g
Molar mass of HgI2 = 201 + (2x127) = 455 g/mol
Mass of HgI2 from the balanced equation = 1 x 455 = 455 g
From the balanced equation above,
470 g of AgI reacted with 455 g of HgI2.
Finally, we shall determine the limiting reactant as follow:
From the balanced equation above,
470 g of AgI reacted with 455 g of HgI2.
Therefore, 2 g of AgI will react with
= (2 x 455)/470 = 1.94 g of HgI2.
From the calculations made above, we can see that only 1.94 g out of 3.50 g of HgI2 given is needed to react completely with 2 g of AgI.
Therefore, AgI I the limiting reactant.
The decomposition of ethylene oxide(CH₂)₂O(g) → CH₄(g) + CO(g)is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 629 K.
Answer:
Half-life at 629K = 252.4min
Explanation:
Using Arrhenius equation:
[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]
And as Half-life in a first order reaction is:
[tex]t_{1/2}=\frac{ln2}{K}[/tex]
We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:
[tex]58.0min=\frac{ln2}{K}[/tex]
K = 0.01195min⁻¹ = K₁
[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]
[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]
[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]
K₂ = 2.75x10⁻³ min⁻¹
And, replacing again in Half-life expression:
[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]
Half-life at 629K = 252.4minThe half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.
The activation energy of a reaction is related to its rate constant as follows:
[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex] (1)
Where:
k: is the rate constant A: is the pre-exponential factor[tex]E_{a}[/tex]: is the activation energy of the reaction = 218 kJ/mol R: is the gas constant = 8.314 J/(K*mol)T: is the temperature
We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:
[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex] (2)
Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min
Hence, the rate constant at 652 K is:
[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]
Now, from equation (1) we can find the pre-exponential factor (A):
[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]
With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):
[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]
Finally, the half-life at 629 K is (eq 2):
[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]
Therefore, the half-life at 629 K is 251.1 min.
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Rubric #2
Forensic Science
1. Define Nucleus.
2. Define Cytoplasm.
3. Define Cell Membrane.
4. Define DNA.
5. Define Plant.
6. Define Chlorophyll.
7. Define Photosynthesis.
8. How do Plant cells and Animal cells differ?
9. Define Cell Wall.
10. Define Vacuole.
11. Why do cells differentiate in multicellular organisms?
12. Define Multicellular.
13. Complete the Eukaryotic cells and Cell Differentiation assessment.
https://clever.discoveryeducation.com/learn/techbook/units/95c20a43-6d3d-40d3-
848d-89929101140d/concepts/co0fef01-33e7-4116-8819.
143e289e15ba/tabs/6e1551ab-57b8-42d4-8e5b-25549791c760/pages/de4182af-aa 60-
454f-ae5e-28df6f4eb3ac
Explanation:
1. Nucleus is a memberane bound organelle that contains cell,s chromosomes.
Check 0/1 ptRetries 5 Element R has three isotopes. The isotopes are present in 0.0825, 0.2671, and 0.6504 relative abundance. If their masses are 97.62, 109.3, and 138.3 respectively, calculate the atomic mass of element R.
Answer:
Atomic mass = 127.198 amu
Explanation:
The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.
Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)
Atomic mass = 8.05365 + 29.19403 + 89.95032
Atomic mass = 127.198 amu
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat. If the radiation wavelength is 12.5 cm, how many photons of this radiation would be required to heat a container with 0.250 L of water from a temperature of 20.0oC to a temperature of 99oC
Answer:
The total photons required for this radiation = 5.1938 × 10²⁸ photons
Explanation:
Given that:
A microwave oven heats by radiating food with microwave radiation, which is absorbed by the food and converted to heat.
If the radiation wavelength is 12.5 cm,
density of water = 1g/cm³
volume of the container = 0.250 L = 250 cm³
density = mass/volume
mass of the water = density × volume
mass of the water = 1g/cm³ × 250 cm³
mass of the water = 250 g
specific heat capacity of water = 4.182 J/g°C
The change in temperature was from 20.0° C to 99° C
ΔT =( 99 -20.0)° C
ΔT = 79.0° C
The heat absorbed in the process is calculated by using the formula,
q = mcΔT
q = 250 g × 4.182 J/g°C × 79.0° C
q = 82594.5 Joules
Recall that the radiation wavelength λ = 12.5 cm = 0.125 m
The amount of energy of one photon of the radiation wavelength is determined by using the formula:
E = hv
since v = c/λ
E = hc/λ
where;
h = Planck's constant = 6.626 × 10⁻³⁴ J.s
c = velocity of light = 3.0 × 10⁸ m/s
∴
E = (6.626 × 10⁻³⁴ J.s × 3.0 × 10⁸ m/s)/ 0.125 m
E = 1.59024⁻²⁴ Joules
The total photons required for this radiation = total heat energy/energy of radiation
The total photons required for this radiation = 82594.5 Joules/1.59024⁻²⁴ Joules
The total photons required for this radiation = 5.1938 × 10²⁸ photons
Atomic mass is calculated by _____. subtracting protons from neutrons averaging the mass of isotopes adding protons and neutrons subtracting neutrons from protons
Answer:
Atomic mass is calculated by adding protons and neutrons.
Explanation:
Atomic mass is the sum of protons and neutrons in an atomic nucleus. For example, the element Oxygen has 8 protons (derived from the atomic number) and 8 neutrons (derived from subtracting the amount of protons from the atomic mass).
We can craft an equation to show the relationship between these variables.
M - N = P, where M = Mass, N = Neutrons, and P = Protons
This equation can be rearranged to show the relationship between the neutrons and protons leading to the atomic mass. Simply add N to both sides of the equation.
M = N + P
This shows that atomic mass is equivalent to the sum of protons and neutrons in an atom's nucleus.
How many kg/hr of steam are produced by a 50HP boiler?
Answer:
Explanation:
50 HP = 50 x 746 watt
= 37300 watt
= 37300 J /s
heat produced in one hour = 60 x 60 x 37300 J
= 134280 x 10³ J
latent heat of vaporization = 2260 x 10³ J / kg .
for evaporation of water of mass 1 kg , 2260 x 10³ J of heat is required .
kg of water being evaporated by boiler per hour
= 134280 x 10³ / 2260 x 10³
= 59.41 kg
rate of production of steam
= 59.41 kg / hr .
Please Help! Use Hess’s Law to determine the ΔHrxn for: Ca (s) + ½ O2 (g) → CaO (s) Given: Ca (s) + 2 H+ (aq) → Ca2+ (aq) + H2 (g) ΔH = 1925.9 kJ/mol 2 H2 (g) + O2 (g) → 2 H2O (l) ΔH = −571.68 kJ/mole CaO (s) + 2 H+ (aq) → Ca2+ (aq) + H2O (l) ΔH = 2275.2 kJ/mole ΔHrxn =
Answer:
ΔHrxn = -635.14kJ/mol
Explanation:
We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:
(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol
(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole
(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole
Reaction (1) - (3) produce:
Ca(s) + H2O(l) → H2(g) + CaO(s)
ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol
Now this reaction + 1/2(2):
Ca(s) + ½ O2(g) → CaO(s)
ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)
ΔHrxn = -635.14kJ/molAre acids harmful to work with.
Answer:
yes it is
Explanation:
because there are some acid which really harm skin.
I need to know what is the median of the data
Answer:
The median is also the number that is halfway into the set. To find the median, the data should be arranged in order from least to greatest. If there is an even number of items in the data set, then the median is found by taking the mean (average) of the two middlemost numbers.
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Click on ALL of the following that use sound waves to communicate with their surroundings and find their way! bats cars dolphins whales birds submarines buses school
Which of the following provides a characteristic of
MgO(s) with a correct explanation?
Choose 1 answer:
А
It is hard because its ions are held together by strong
electrostatic attractions.
B
It is malleable because its atoms can easily move past
one another without disrupting the bonding.
It is a poor conductor of electricity because its
electrons are tightly held within covalent bonds and
lone pairs.
It has a high melting point because its molecules
interact through strong intermolecular forces.
Answer:
А It is hard because its ions are held together by strong electrostatic attractions.
B It is malleable because its atoms can easily move past one another without disrupting the bonding.
Explanation:
These are correct explanations of the properties of magnesium.
C is wrong. Mg is a good conductor of electricity and it has metallic bonds.
D is wrong. Mg has no molecules. It has no intermolecular forces.
Match each compound with its appropriate pKa value.
(a) 4-Nitrobenzoic acid, benzoic acid, 4-chlorobenzoic acid pKa=4.19,3.98, and 3.41pKa =4.19,3.98, and 3.41
(b) Benzoic acid, cyclohexanol, phenol pKa=18.0,9.95, and 4.19pK a =18.0,9.95, and 4.19
(c) 4-Nitrobenzoic acid, 4-nitrophenol, 4-nitrophenylacetic acid pKa=7.15,3.85, and 3.41pK a =7.15,3.85, and 3.41
Answer:
Explanation:
a) 4-nitrobenzoic acid pKa= 3.41
benzoic acid pKa= 4.19
4-chlorobenzoic acid pKa= 3.98
b) benzoic acid pKa= 4.19
cyclohexanol pKa= 18.0
phenol pKa= 9.95
c) 4-Nitrobenzoic acid pKa= 3.41
4-nitrophenol pKa= 7.15
4-nitrophenylacetic acid pKa= 3.85
There are eight consitutional isomers with the molecular formula C4H11N.
name and draw a structural formulas for each amine.
Answer:
See figure 1
Explanation:
We have to remember that in the isomer structures we have to change the structure but we have to maintain the same formula, in this case [tex]C_4H_1_1N[/tex].
In the formula, we have 1 nitrogen atom. Therefore we will have as a main functional group the amine group.
In the amines, we have different types of amines. Depending on the number of carbons bonded to the "N" atom. In the primary amines, we have only 1 C-H. In the secondary amines, we have two C-N bonds and in the tertiary amines, we have three C-N bonds.
With this in mind, we can have:
-) Primary amines:
1) n-butyl amine
2) sec-butyl amine including 2 optical isomers
3) isobutyl amine
4) tert-butyl amine
-) Secondary amines:
5) N-methyl n-propyl amine
6) N-methyl isopropyl amine
7) N, N-diethyl amine
-) Tertiary amines:
8) N-ethyl N, N-dimethyl amine
See figure 1
I hope it helps!
Nombre para la siguiente estructura de compuesto orgánico
Explanation:
Introducción:
La designación de los compuestos orgánicos puede hacerse utulizando alguno de los
siguientes sistemas:
a) Mediante nombres triviales o comunes, que expresen alguna propiedad característica
(sabor, color, acción fisiológica, etc.) o hagan referencia a la materia de la cual se extrajo el
compuesto.
b) Mediante nombres racionales que proporcionen una idea de su constitucion química y
destaquen sus analogías estructurales.
La necesidad de una nomenclatura sistemática, que expresara en forma clara, conforme
a normas precisas, el nombre y la estructura de los compuestos orgánicos, ha sido motivo de
preocupación permanente y observada a través de los numerosos congresos internacionales
que, al efecto, se han realizado en diversas oportunidades.
Las bases del actual sistema de nomenclatura fueron establecidas por una comisión que
se reunió en Ginebra en 1892. Posteriormente, fue perfeccionado y ampliado por el Comité de
Nomenclatura de la Unión Internacional de Química Pura y Aplicada, por lo que se conoce
como sistema I.U.P.A.C. (International Union of Pure and Applied Chemistry).
En las reglas aprobadas se ha tratado de introducir los menores cambios posibles a la
terminología universalmente adoptada. El sistema tiene la necesaria flexibilidad como para
adaptar la forma precisa de las palabras, de las terminaciones, etc. a las características de
distintos idiomas.
El nombre de los hidrocarburos consta de tres partes: a) la raíz, que indica el esqueleto
carbonado; b) la terminación o sufijo, que indica el grado de saturación, y c) el prefijo que
diferencia las distintas estructuras isoméricas (distintas estructuras construidas con exactamente
los mismos átomos).
Ej.: CH3-CH2-CH2-CH2-CH3 pentano (tambien llamado n-pentano)
penta: raíz que señala el número de átomos de carbono que componen la cadena principal
del compuesto.
-ano: sufijo que indica que el hidrocarburo es saturado
What happens when an atom of sulfur combines with two atoms of chlorine to produce SCl2?
Answer:
Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom
What are some geographic features that could be found in the hydrosphere?
Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.
hydro = water
Answer:
Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!
Explanation:
The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)