Answer:
41.4 s
Explanation:
Given data
Rate constant (k): 8.42 × 10⁻² s⁻¹ at 800 °CInitial concentration of A ([A]₀): 5.00 MConcentration of A at a time t ([A]): 0.153 MLet's consider the following reaction of first order with respect to A.
2 A ⇒ B
We can find the time that it will take for A to decrease from 5.00 M to 0.153 M using the following expression.
[tex]ln([A]/[A]_0)=-k.t\\ln(0.153M/5.00M)=-8.42 \times 10^{-2}s^{-1} .t\\t = 41.4 s[/tex]
A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure surrounding the balloon remains constant?
Answer:
[tex]V_2 = 4.87 * 10^3[/tex]
Explanation:
This question is an illustration of ideal Gas Law;
The given parameters are as follows;
Initial Temperature = 25C
Initial Volume = 4.5 * 10³L
Required
Calculate the volume when temperature is 50C
NB: Pressure remains constant;
Ideal Gas Law states that;
[tex]PV = nRT[/tex]
The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n
Divide both sides by PT
[tex]\frac{PV}{PT} = \frac{nRT}{PT}[/tex]
[tex]\frac{V}{T} = \frac{nR}{P}[/tex]
Represent [tex]\frac{nR}{P}[/tex] with k
[tex]\frac{V}{T} = k[/tex]
[tex]k = \frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
At this point, we can solve for the required parameter using the following;
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;
From the given parameters;
V1 = 4.5 * 10³L
T1 = 25C
T2 = 50C
Convert temperatures to degree kelvin
V1 = 4.5 * 10³L
T1 = 25 +273 = 298K
T2 = 50 + 273 = 323K
Substitute values for V1, T1 and T2 in [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex]
[tex]\frac{4.5 * 10^3}{298} = \frac{V_2}{323}[/tex]
Multiply both sides by 323
[tex]323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323[/tex]
[tex]323 * \frac{4.5 * 10^3}{298} = V_2[/tex]
[tex]V_2 = 323 * \frac{4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{323 * 4.5 * 10^3}{298}[/tex]
[tex]V_2 = \frac{1453.5 * 10^3}{298}[/tex]
[tex]V_2 = 4.87 * 10^3[/tex]
Hence, the final volume at 50C is [tex]V_2 = 4.87 * 10^3[/tex]
Identify a reaction of sugars, starch, or cellulose to produce ethanol and carbon dioxide halogenation polymerization
Answer:
Fermentation
Explanation:
Fermentation is the general term used to describe the process by which sugars such as glucose, starch or cellulose are converted to ethanol and carbon (iv) oxide. It is anaerobic process meaning that it occurs in the absence of air or in very low oxygen concentrations.
Yeast and other microorganisms ferment glucose into ethanol and carbon (iv) oxide with the help of the enzyme zymase. Polysaccharides such as starch and cellulose are first broken down into glucose by enzymes such as diastases, maltase and cellulase, before it is then converted into ethanol and carbon (iv) oxide.
The equation for the conversion of glucose to ethanol and carbon (iv) oxide is as follows:
C₆H₁₂O₆(aq) -----> 2C₂H₅OH(aq) + 2CO₂(g)
Draw an aldohexose.
Answer:
See explanation
Explanation:
We have to remember that theory behind the carbohydrates. Carbohydrates are molecules with several hydroxyl groups in which the main functional group can be an aldehyde or a ketone.
If we have an aldehyde as a main functional group we will have an "aldose". If we have a ketone as a main functional group we will have a "ketose".
We can also, classify the carbohydrates using the number of carbons. So, for example, if we have 5 carbons and a ketone as the main functional group we will have a "keto-pentose". If we have for example 4 carbons and an aldehyde as the main functional group we will have a "tetra-aldose".
In this case, we have an aldohexose, so we will have 6 carbons and an aldehyde as main functional group. So, we can draw a structure with 6 carbons, in carbon 1 we have to put the aldehyde group and in the other carbons we have to put "OH" groups.
See figure 1
I hope it helps!
A naturally occurring oil co-distills with water to produce an oil/water distillate that is 20% oil by weight. If the molecular weight of the oil 100 g/mol, what was the partial pressure of the oil during distillation assuming atmospheric pressure is 760 mm Hg
Answer:
Explanation:
Partial pressure of oil = mole fraction of oil x total pressure
mole fraction of oil = mole of oil / mole of water + mole of oil
= mole of oil = mass of oil / molecular weight of oil
= 20 / 100 = .2
mole of water = 80 / 18
= 4.444
mole fraction of oil = .2 / .2 + 4.444
= .2 / 4.644
Partial pressure of oil = mole fraction of oil x total pressure
= (.2 / 4.644 ) x 760 mm
= 32.73 mm Hg .
What element forms an ion with an electronic configuration of 1s22s22p6 (or [Ne] ) and a −2 charge? Give the symbol for the element. g
Answer:
Mg²⁺
Explanation:
Electronic configuration = 1s22s22p6 (or [Ne] )
Charge = -2
This means the element has two extra electrons. So total electrons = 12.
The lement is Magnesium and the ion is Mg²⁺
Observing Gas Properties - Flammability
Chemicals Observations
HCl
Zn
HCl + Zn Reaction
Gas + Flame
Conclusion:
A. When performing this experiment, when is the tetradecanol in a condensed phase? Explain your answer.
B. The published melting point of H2O is 0°C, and the published boiling point is 100°C. Why would you have found slightly different values?
C. In this experiment, there were two possible identities for the gas produced: hydrogen or chlorine, which have quite different properties. Hydrogen is a colorless, odorless, flammable gas. Chlorine gas is greenish yellow with a pungent, bleach-like odor and is non-flammable. Use your observations to determine which gas was produced in this experiment.
D. Give some examples of how it might be useful to know the melting or boiling points of a substance. Think in terms of both scientific and commercial/industrial settings.
Answer:
The answer is C love.
Explanation:
Answer:
a) It is when a solid is going into the liquid stage. The molecules are really close together.
b) The water could have impurities, altitude could effect the boiling point or the temperature could of been measured wrong.
c) The gas was colorless and odorless therefore it is Hydrogen gas.
d) In a Scientific setting finding the boiling point of a new unknown substances can help identify and organize them. In a commercial setting chemicals need to be stored a certain way according to their boiling point. Stored incorrectly could cause the substance to evaporate.
what bonding is similar to ionic bonding, except there are no high-electronegativity atosms present to accept any electrons that the present atoms are willing to donate.
Answer:
Metallic bonding
Explanation:
Ionic bonding involves the transfer of electrons from a highly electropositive metal to a highly electronegative nonmetal.
The metallic bond is somewhat similar to the ionic bond since there are also charged positive metal ions. The only difference is that there isn't any electronegative element that accepts the electrons.
In a metallic bond, the positively charged metal ions are bound together by a sea of mobile electrons. The attractive force between the metal ions and the mobile electrons hold the metallic crystal lattice together.
need this asap , help please
Answer:
Path A-B-D involves a catalyst and is slower than A-C-D
Explanation:
The diagram above illustrates both the catalyzed path and the uncatalyzed path of a chemical reaction.
The catalysed path is the path expressed with broken lines and the uncatalyzed path is the path expressed with thick small line as shown in the diagram above.
The catalyzed path has a higher activation energy than the uncatalyzed path.
Therefore, the catalyzed path will be slower that the uncatalyzed path because, the catalyzed path will require a higher energy to overcome the activation energy in order for the reaction to proceed to product.
On the other hand, the uncatalyzed path has a lower activation energy and a lesser amount of energy is needed to overcome it in order for the reaction to proceed to product.
What concentration unit is measured in moles of solute per kilogram of
solvent?
Answer:
molality
Explanation:
The SI unit for molality is moles per kilogram of solvent. A solution with a molality of 3 mol/kg is often described as "3 molal", "3 m" or "3 m". hope this helps you :)
How many moles of carbon atoms are there in 0.500 mol of C2H6?
The number of moles of carbon atoms in 0.500 mol of ethane (C₂H₆) is equal to one mole.
What is a mole?A mole can be defined as a scientific unit that is utilized to calculate the quantities such as atoms, molecules, ions, or other particular particles. The mass of one mole of a given chemical element is atomic mass and that of 1 mole of a chemical compound is molar mass.
The number of entities found in one mole is equal to 6.023 × 10 ²³ which is known as Avogadro’s constant.
Given, the number of moles of C₂H₆ = 0.500 mole
One molecule of ethane has carbons = 2
One mole of ethane has moles of carbons = 2 moles
0.500 mol of ethane has moles of carbon atoms = 0.500×2 = 1 mol
Therefore, one mole of carbon atoms is present in 0.500 mol of ethane C₂H₆.
Learn more about the mole, here:
brainly.com/question/26416088
#SPJ5
A gas particle of mass 5.31 × 10^-23 kg has a velocity of 1.00 102 m/s. What is the kinetic energy of the molecule
Answer:
Kinetic energy = 1/2mv²
where m is the mass
v = velocity
m = 5.31 × 10^-23 kg
v = 1.00 × 10^2 m/s
Kinetic energy = 1/2 × 5.31 × 10^-23 × ( 1.00 × 10^2)²
= 2.655 × 10^-19 Joules
Hope this helps
How many formula units make up 36.0 g of magnesium chloride (MgCl2)?
Express the number of formula units numerically.
Answer: There are [tex]2.29\times 10^{23}[/tex] formula units
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{36.0g}{95g/mol}=0.38moles[/tex]
1 mole of [tex]MgCl_2[/tex] contains = [tex]6.023\times 10^{23}[/tex] formula units
Thus 0.38 moles of [tex]MgCl_2[/tex] contains = [tex]\frac{6.023\times 10^{23}}{1}\times 0.38=2.29\times 10^{23}[/tex] formula units
Thus there are [tex]2.29\times 10^{23}[/tex] formula units
After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Express the concentration to two significant figures and include the appropriate units.
Answer:
[tex]\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}[/tex]
Explanation:
Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.
1. Moles of Ni²⁺
[tex]n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}[/tex]
2. Moles of NH₃
[tex]n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}[/tex]
3. Initial concentrations after mixing
(a) Total volume
V = 135 mL + 190 mL = 325 mL
(b) [Ni²⁺]
[tex]c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}[/tex]
(c) [NH₃]
[tex]c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}[/tex]
3. Equilibrium concentration of Ni²⁺
The reaction will reach the same equilibrium whether it approaches from the right or left.
Assume the reaction goes to completion.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0
C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0
E/mol·L⁻¹: 0 0.1095 6.106×10⁻³
Then we approach equilibrium from the right.
Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺
I/mol·L⁻¹: 0 0.1095 6.106×10⁻³
C/mol·L⁻¹: +x +6x -x
E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x
[tex]K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}[/tex]
Kf is large, so x ≪ 6.106×10⁻³. Then
[tex]K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}[/tex]
In what ways are solid solid mixture categorised
Mixtures can be classified as homogeneous or heterogeneous . Mixtures are composed of substances that are not chemically combined.
Homogeneous mixtures are solutions. The components of a solution are evenly distributed throughout, so that every part of the solution is the same. The components that make up a solution include one or more solutes dissolved in a solvent. Solutes can be solids, liquids, or gases, and solvents can also be solids, liquids or gases.
Brass is an example of a solid/solid solution, saline solution is an example of a solid/liquid solution, diluted ethanol is an example of a liquid/liquid solution. There are many examples of solutions. The components of a solution can be separated by physical means, such as distillation, evaporation, and chromatography, among others.
Which of the following required Bohr's model of the atom to need modification ?
The question his incomplete, the complete question is;
Which of the following required Bohr's model of the atom to need modification
A. energies of electrons are quantized
b. electrons do not follow circular orbits around the nucleus
c. Quantized electrons energies are responsible for emission spectra lines
d. an electrons energy increases the farther it moves from the nucleus
Answer:
electrons do not follow circular orbits around the nucleus
Explanation:
Neils Bohr's model of the atom closely follows the planetary model of Rutherford, hence is it sometimes referred to as the Bohr-Rutherford's model of the atom. Its main points are that;
Electrons are found in circular orbits
Electronic transition within the atom leads to spectral lines
The energy of an orbit is related to its size.
One major refinement of the Bohr's model of the atom is the Bohr-sommerfield model. Sommerfeld added that electrons are found in elliptical rather than circular orbits. This Bohr-sommerfield model was able to explain atomic spectral effects, such as the Stark effect in spectral line splitting.
Hence, the fact that electrons are found in elliptical rather than circular orbits was a major failure of Bohr's original proposition.
Answer:
Electrons do not follow circular orbits around the nucleus
Explanation:
The percent yield (isolation yield) of guaifenesin isolated from a 650 mg tablet containing 400 mg dose of drug, can be expressed as: Group of answer choices
Answer:
61.54%
Explanation:
Hello,
To calculate the percent yield of a product, we express it as ratio between the actual yield to the theoretical yield multiplied by 100.
Percent yield = (actual yield / theoretical yield) × 100
Actual yield = 400mg
Theoretical yield = 650mg
Percent yield = (400 / 650) × 100
Percent yield = 0.6154 × 100
Percent yield = 61.54%
Percent yield of guaifenesin in the drug is 61.54%
Kinetic energy and gravitational potential energy are both forms of which type
of energy?
A. Internal energy
B. Mechanical energy
C. Potential energy
D. Thermal energy
Answer:
C. Potential energy
Explanation:
Kinetic energy and gravitational potential energy are both forms of potential energy. Potential energy is stored energy, when an object is not in motion it has stored energy. When an object is an motion it has kinetic energy. An object posses gravitational potential energy when it is above or below the zero height.
Identify the particle that must receive 2 electrons to acquire a charge of +1. a) K b) Fe2+ c) O2- d) Nee) Al3+ (URGENT) Needs to be done in 30 mins
Answer:
E) Al³⁺
Explanation:
A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.
In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;
A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.
K + 2e⁻ ----> K⁻²
B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.
Fe⁺ + 2e⁻ ----> Fe
C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.
O²⁻ + 2e⁻ ----> O⁴⁻
D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.
Ne + 2e⁻ ----> Ne²⁻
E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.
Al³⁺ + 2e⁻ ----> Al⁺
Which of the following are safety and health resources workers can use inside of their workplace
Answer:
Employer
coworkers
Safety data sheet (SDS)
Training material
BLogs
Explanation:
The resources which can be used inside of the workplaces is as follows
a. Employer: The employer is the person who is working in an organization plus for every employee the health and safety matters first
b. Coworkers: The coworkers are the person who is at similar level or for the same role plus for every coworker the health and safety matters first
c. Safety data sheet: In this sheet, it maintains the data regarding health and safety
d. Training material: This one is also used inside the workplace
e. Blogs: The blog is an article in which the health and related data could be written and provided
Therefore these all the resources are used inside the organization
Answer:
Employer
Co-workers
SDSs
Training materials
Explanation:
g Suppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read the base titrant volume as 1.94 mL. After running the titration and reaching the endpoint, you read the base titrant volume as 23.82 mL. What volume of base was required for the titration
Answer:
21.88mL is the volume of base required for the titration.
Explanation:
For an acid-base titration trying to find the concentration of an acid, you must add a known quantity of the acid and titrate it with an standarized base.
If you know the moles of base you add to the acid solution, these moles are equal to moles of acid.
In the buret of the titration, initial volume is 1.94mL and final volume is 23.82mL. The volume you are adding is the difference between initial and final volume, that is:
23.82mL - 1.94mL
21.88mL is the volume of base required for the titration.Vitamin c is known chemically by the name ascorbic acid determine the empirical formula of ascorbic acid if it is composed of 40.92% carbon, 4.58% hydrogen, and 54.50% oxygen.
Answer:
[tex]=C_3H_4O_3[/tex]
Explanation:
When percentage composition is given, and asked for the empirical formula, it is simplest to assume 100 g of material. Thus,
Mass C = 40.92 g. Moles C = 40.92 g x 1 mole/12 g = 3.41 moles C
Mass H = 4.58 g. Moles H = 4.58 g x 1 mole/1.0 g = 4.58 moles H
Mass O = 54.50 g. Moles O = 54.50 g x 1 mole/16 g = 3.41 moles O
Now, we want to get the moles into whole numbers, so we begin by dividing all by the smallest, i.e. divide all values by 3.41.
Moles C = 3.41/3.41 = 1
Moles H = 4.58/3.41 = 1.34
Moles O = 3.41/3.41 = 1
Now, in order to get 1.34 to be a whole number we multiply it (and all others) by 3
Moles C = 1x3 = 3
Moles H = 1.34x3 = 4
Moles O = 1x3 = 3
Empirical Formula [tex]=C_3H_4O_3[/tex]
Suppose you have a bucket of sand containing 5.5 billion grains of sand ( 5.5×109 grains). Most of the grains of sand are white, but there are some brown grains mixed evenly throughout.
If the concentration of brown sand is 6.0% , how many grains of brown sand are in the bucket?
brown grains
If the concentration of brown sand is 6.0 ppm, how many grains of brown sand are in the bucket?
brown grains
If the concentration of brown sand is 6.0 ppb, how many grains of brown sand are in the bucket?
brown grains
Explanation:
5.5 billion grains of sand ( 5.5×109 grains)
If the concentration of brown sand is 6.0% , how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
Brown grains = 0.06 * 5.5×10^9 = 0.33 x 10^9 = 3.3 x 10^8 grains
If the concentration of brown sand is 6.0 ppm, how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
6ppm = 6 / 1000000 = 0.000006
Brown grains = 0.000006 * 5.5×10^9 = 3.3 x 10^4 grains
If the concentration of brown sand is 6.0 ppb, how many grains of brown sand are in the bucket?
Number of grains = Concentration of brown side * Bucket of sand
6ppb = 6 / 1000000000 = 0.000000006
Brown grains = 0.000000006 * 5.5×10^9 = 3.3 x 10^1 = 33 grains
2 physical changes that happen in aquaponics
and endothermic reactions that happen in aquaponics
3 chemical changes that happend in aquaponics
A conclusion about matter and energy and explain how both can change ecosystems and earth surface with time.
Answer:
The physical and chemical change that occurs in the aquaponics are given below.
Explanation:
The plants and animals grow in size and decrease the mass of plant due to eating by the fishes is a physical changes which occurs in aquaponics. The sunlight has a heat energy which is absorb by the plants present in aquaponics which is a type of endothermic reaction. In aquaponics, the ammonia present in water is converted into nitrates which is used by the plants as a nutrients. When the mass is converted into energy, it increases the temperature of the ecosystem and also the earth surface. For example, if a wood is burn, it change into heat energy which increases the temperature and cause the global warming on the earth surface.
When 0.7521 g of benzoic acid was burned in a calorimeter containing 1000 g of water, a temperature increase of 3.60 ℃ was observed. What is the heat capacity of the bomb calorimeter, excluding water? The heat of combustion of benzoic acid is -26.42 kJ/g.
Answer:
The correct answer is 1335.4 J/ °C.
Explanation:
Based on the given information, the amount of benzoic acid burned is 0.7521 grams, the amount of water present within a calorimeter is 1000 grams and there is an elevation in temperature of 3.60 °C. The heat of combustion of benzoic acid given is -26.42 kJ per mole.
Now the amount of heat discharged in a reaction is:
= 0.7521 grams × -26.42 kJ / gram = -19.872 kJ or -19870 J
qreaction = - (qwater + qcal)
-19870 J = - (mwatersΔT + CcalΔT
= - [(1000 grams × 4.184 J/g degree C × 3.60 degree C) + (Ccal 3.60 °C)] (The specific heat of water is 4.184 J/gram °C)
(1000 gram × 4.184 J/g °C × 3.60 °C) + (Ccal 3.60 °C) = 19870 J
15062 J + Ccal 3.60 °C = 19870 J
Ccal 3.60 °C = 4807.6 J
Ccal = 4807.6 J / 3.60 °C = 1335.4 J/°C or 1.3 kJ/°C.
An unknown gas Q requires 2.67 times as long to effuse under the same conditions as the same amount of nitrogen gas. What is the molar mass (g/mol) of Q
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
1. Natural gas is used as a cooking fuel in many restaurants and homes. The primary chemical components of natural gas are hydrocarbons known as alkanes. Research and list the primary alkanes found in natural gas, and explain why these compounds excel as fuel sources.
Answer:
The main component of natural gas is methane (CH4) at 60 to 90% followed by various combination of ethane, propane, and butane whose percentage can vary from 0 to 20% each. For each unit mass of alkanes, the combustion energy (energy released when the fuel reacts with oxygen) released is very high about 13 to 15 kcal/g, which is higher than even those generated by petrol or diesel. So, for heating or other energy generation purpose for household purposes, this source of energy is used.
The equation for combustion of methane is shown below. Upon combustion, carbondioxide and water is produced with simultaneous generation of heat which is the source of energy used for consumption.
CH4 + 2O2 --> CO2+ 2H2O + heat [ For methane, the combustion energy is ~ 6kcal/g]
As the CH2 units are increased in the alkanes, the combustion energy increases, for e.g., ethane has combustion energy of 7 kcal/g and propane has about 12 kcal/g.
Explanation:
Calculate how many moles of NO2 form from 3.0 mol N2O5 2N2O5(g)→4NO2(g)+O2(g)Express your answer using two significant figures.
Answer:
6.0 moles NO2(g)
Explanation:
Based on the reaction every 2 moles N2O5(g) gives reaction with 4 moles NO2(g).Then when we have 3.0 mol N2O5(g),
2 moles N2O5(g) 4 moles NO2(g)
3 moles N2O5(g) ? moles NO2(g)
______________________________________
3 *4 / 2 = 6.00 moles NO2(g)
Answer:
There are 6 mol of NO2 with respect to 3 mol of N2O5
Explanation:
Approach 1 ( dimensional analysis ) :
3 Moles of N2O5 [tex]*[/tex] ( 4 moles of NO2 / 2 Moles of N2O5 ) - moles of N2O5 cancel out, leaving you with the moles of NO2 -
3 [tex]*[/tex] 4 / 2 = 12 / 2 = 6 moles of NO2
So as you can see in the formula there are 4 moles of NO2 present per 2 Moles of N2O5 - " 4NO2 and 2N2O5. " As we wanted the moles of N2O5 to cancel out, the 2 moles of N2O5 was kept as the denominator, and hence we received the fraction we needed.
Approach 2 :
There are 3 Moles of N2O5. The ratio of Moles of N2O5 to moles of NO2 is provided by the reaction -
Moles of N2O5 : Moles of NO2,
2 : 4,
1 : 2
Therefore the moles of NO2 will be two times as much as the given moles of N2O5, or 3 [tex]*[/tex] 2 = 6 moles of NO2
Name the following alkanes. Please need help last assignment. The questions in the picture.
Answer:
Hydrogen, carbon reaction
Considering the steps involved in dissolution, which of the following do you expect to speed up a dissolution process?
A. Sweeping all the solute particles into a pile within the solvent.
B. Stirring the solution vigorously.
C. Grinding the solute down into tiny particles.
D. Gently heating the solution.
Answer:
C. Grinding the solute down into tiny particles.
Explanation:
The dissolution of a solute has something to do with particle size. The size of solute particles usually determines how quickly a solute dissolves in a solvent. When large solute particles are introduced into the solvent, the large solute particles do not easily interact with solvent particles hence preventing easy dissolution in the solvent.
However, when the solute is ground into tiny particles, smaller solute particles interact more effectively with solvent particles hence dissolution is faster.
Therefore, tiny solute particles will dissolve faster in a solvent than a lump of solute. Summarily, small particle size enhances dissolution of a solute in the appropriate solvent.
Answer: stirring the solution vigorously
Grinding the solute down into tiny particles
gently heating the solution
Explanation:
A dissolution will proceed more readily when heated . Breaking up the solute as much as possible will aid in overcoming the solute-solute interaction, as will stirring the solution
Use your trendline equation to determine the gas pressure at 200 K and 400 K. (notice the temperature units) How many times greater is the pressure at 400 K in comparison to 200 K? Is this what you’d expect? Why?
Answer:
The pressure will be twice the initial pressure
Explanation:
Gay-Lussac's law states that the pressure of a gas is directely proportional to absolute temperature under constant volume. That is because vibrations of a gas increase when temperature increases, increasing the pressure of the gas.
That means if the temperature of a gas is doubled, the pressure will be twice the initial pressure.