Answer:
Caesium (Cs) > Barium (Ba) > Lead (Pb) > Silicon (Si) > Phosphorus (P) > Oxygen (O)
Explanation:
Before proceeding, we need to understand the trend of atomic sized.
In the periodic table, atomic size generally decreases form left to right in a period. It increases down the group.
Having known this, the order of the elements is thus given as;
Caesium (Cs) > Barium (Ba) > Lead (Pb) > Silicon (Si) > Phosphorus (P) > Oxygen (O)
g A molecule with 6 areas of electrons (6 electrons domains) consisting of 4 bonding domains and 2 non-bonding domains will have a
Answer:
a) octahedral electron domain geometry
b)square planar molecular geometry
c) a bond angle of 90°
Explanation:
According to the Valence Shell Electron Pair repulsion Theory, the shape of a molecule is dependent on the number of electron pairs on the valence shell of the central atom in the molecule. These electron pairs orient themselves as far apart in space as possible to minimize electron pair repulsion.
Electron pairs may be lone pairs or bond pairs. Lone pairs of electrons cause more repulsion than bond pairs. These lone pairs often cause the molecular geometry to depart from what is predicted on the basis of the electron domain geometry due to greater repulsion of lone pairs.
When a molecule has six electron domains consisting of four bond pairs and two lone pairs, the bonding pairs arrange themselves at the corners of a square at a bond angle of 90° with the lone pairs found above and below the plane of the bonding groups leading to a square planar molecular geometry
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.
Name the following alkanes. Please need help last assignment. The questions in the picture.
Answer:
Hydrogen, carbon reaction
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 :)
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]
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:
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.
the oceans ______ residents produce up to 70% of the oxygen found in the atmosphere. A bacterial B Photosynthetic C microscopic D macroscopic
The oceans photosynthetic residents produce up to 70% of the oxygen found in the atmosphere.
What are Photosynthetic organisms?These are organisms such as plants which are capable of producing food through the process known as photosynthesis.
By so doing, they oxygen is also produced which is used by other animals in the ecosystem for their cellular respiration and survival.
Read more about Photosynthetic organisms here https://brainly.com/question/1615639
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.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]
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.
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
What are the first and last terms of the following series ?
Answer:
B
Explanation:
In the reaction 2Cu2O + Cu2S --> 6Cu + SO2
When 10.5 grams of Copper I oxide is reacted with unlimited Copper I sulfide how many grams of
Copper will be produced?
14.0 grams
2.73 grams
28.0 grams
O 16.9 grams
35.2 grams
Answer:
14.0 g
Explanation:
Step 1: Write the balanced equation
2 Cu₂O + Cu₂S ⇒ 6 Cu + SO₂
Step 2: Calculate the moles corresponding to 10.5 g of copper (I) oxide
The molar mass of Cu₂O is 143.09 g/mol.
[tex]10.5g \times \frac{1mol}{143.09g} = 0.0734 mol[/tex]
Step 3: Calculate the moles of copper produced from 0.0734 moles of copper (I) oxide
The molar ratio of Cu₂O to Cu is 2:6. The moles of Cu produced are 6/2 × 0.0734 mol = 0.220 mol
Step 4: Calculate the mass corresponding to 0.220 mol of copper
The molar mass of Cu is 63.55 g/mol.
[tex]0.220mol \times \frac{63.55g}{mol} = 14.0 g[/tex]
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
Given that the molar mass of NAOH is 40.00G/MOL, what mass of NAOH is needed to make 2.500 L of a 2.000 M NaOH Solution?
Answer:
200g
Explanation:
n = CV
n = mass/molar mass
mass/molar mass = CV
mass/40 = 2 x 2.5
mass/40 = 5
mass = 5x 40
mass = 200g
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:
What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 100. mL of solution
Answer:
[tex]3.65~M[/tex]
Explanation:
We have to remember the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
So, we have to calculate "mol" and "L". The total volume is 100 mL. So, we can do the conversion:
[tex]100~mL\frac{1~L}{1000~mL}=~0.1~L[/tex]
Now we can calculate the moles. For this we have to calculate the molar mass:
O: 16 g/mol
H: 1 g/mol
C: 12 g/mol
[tex](16*1)+(1*4)+(12*1)=32~g/mol[/tex]
With the molar mass value we can calculate the number of moles:
[tex]1.7~g~of~CH_3OH\frac{1~mol~CH_3OH}{32~g~of~CH_3OH}=0.365~mol~CH_3OH[/tex]
Finally, we can calculate the molarity:
[tex]M=\frac{0.365~mol~CH_3OH}{0.1~L}=3.65~M[/tex]
I hope it helps!
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⁺
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:
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
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.
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!
Of the following, ________ should have the highest critical temperature. Of the following, ________ should have the highest critical temperature. CH4 H2 CCl4 CBr4 CF4
Answer:
H2
Explanation:
Critical temperature is the temperature above which gas cannot be liquefied, regardless of the pressure applied.
Critical temperature directly depends on the force of attraction between atoms, it means stronger the force of higher will be the critical temperature. So, from the given options H2 should have the highest critical temperature because of high attractive forces due to H bonding.
Hence, the correct option is H2.
Answer:
CBr4
Explanation:
Critical temperature is dependent on the strength of the intermolecular forces.
First consider the types of intermolecular forces and the order of their strengths.
Dispersion forces < dipole-dipole forces < hydrogen bonding < ionic bonding
Remember, dispersion forces are present in all cases
H2: only dispersion forces are present
CH4, CCl4, CBr4, CF4: only dispersion forces are present
In order to break the tie we must start considering molar mass because larger molar masses correspond to larger intermolecular forces. Calculating molar mass shows that CBr4 is the largest and will have the strongest intermolecular forces and therefore will have the highest critical temperature.
The answer is CBr4
Brown algae, such as giant kelp and sargassum, are an example of ? A. macroalgae B chemosynthetic algae C Flowering plants D Phytoplankton
Answer:
The answer is option A.
I hope this helps you.
Answer:
Macro Algae
Explanation:
probz
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
The molecular mass of propane-1,2-diol is 76.1 amuamu . Calculate the molecular mass of propane-1,3-diol, an isomer of propane-1,2-diol
Answer:
76.1 amu
Explanation:
Let us recall that isomers refer to two different compounds with the same molecular formula but different atom to atom connectivity and different chemical properties. When two compounds are isomers, we can essentially represent them with exactly the same molecular formula.
Now propane-1,2-diol and propane-1,3-diol are both represented by the molecular formula C3H8O2 since they are isomers of each other. When two compounds have the same molecular formula, they must essentially have the same molecular mass. Hence the molecular mass of propane-1,3-diol is also 76.1 amu.
Answer:
molecular mass of propane-1,3-diol, an isomer of propane-1,2-diol = 76.1 amu
Explanation:
An isomer of any compound will also have the same molecular mass of the comlund
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)
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