Answer: A molecule contains hydrogen bonding if it contains hydrogen covalently bonded to (fluorine, oxygen, or nitrogen)
Explanation:
A hydrogen bond is a strong dipole-dipole attraction which occurs between
--> the hydrogen atom attached to a strongly electronegative atom, and
--> another strongly electronegative atom with a lone pair of electrons.
When an electronegative atom such as fluorine, oxygen or Nitrogen is bonded to hydrogen, a dipole develops causing the hydrogen to be partially negative. The electrostatic attraction between the partially positive hydrogen atom in one molecule and the partially negative atom of the more electronegative element in another molecule gives rise to the strong dipole-dipole attraction called hydrogen bonding.
Hydrogen fluoride, water and ammonia contain the three most electronegative elements, fluorine, oxygen and nitrogen respectively, linked directly to hydrogen. In addition, lone pairs of electrons are present in the fluorine, oxygen and nitrogen atoms of the three hydrides, making hydrogen bonds to form easily between them. These compounds which exhibits hydrogen bonding always have higher melting and boiling points.
Using the boh model of a He ion, what transition is most likelu to result in the emission of radiation with a wavelength of approximately 274 nm
Answer:
[tex]n=6\ to\ n=3[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=274 *10^{-9}m[/tex]
Bohr's constant [tex]R = 1.097 × 10^7 / m (or m−1)[/tex]
Helium atom [tex]z=2[/tex]
Generally the equation for Wavelength is mathematically given by
[tex]\frac{1}{\lambda}=Rz^2(\frac{1}{nf^2}-\frac{1}{nf^2})[/tex]
[tex]0.083=(\frac{1}{nf^2}-\frac{1}{nf^2})[/tex]
Therefore
The Range of n fall at
[tex]n=6\ to\ n=3[/tex]
In order for a substitution reaction to take place, at a minimum, __________ and ___________ are needed. carbocation rearrangement proton transfer loss of a leaving group nucleophilic attack
Answer: In order for a substitution reaction to take place, at a minimum, loss of a leaving group and nucleophilic attack are needed.
Explanation:
When one group in a molecule is replaced by another one then it is called a substitution reaction.
For a substitution reaction reaction to occur at a minimum, it is necessary to have a leaving group that helps in creating a carbocation and hence a nucleophile can easily attack.
For example, [tex]CH_{4} + Cl_{2} + h\nu \rightarrow CH_{3}Cl + HCl[/tex]
Thus, we can conclude that in order for a substitution reaction to take place, at a minimum, loss of a leaving group and nucleophilic attack are needed.
What is the
energy
2) The energy transition from n = 1 →n= 3 in hydrogen is 12.09 eV (1.6022 X 10J= 1 eV)-
of light emitted from n = 3 →n=1?
E e fiz
Answer:
energy is the ability to do work
What effect does hybridization have on chemical bond
Write the balanced half-equations for silver + oxygen= silver oxide:
Answer: The balanced half-equations for silver + oxygen= silver oxide are:
Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]
Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]
Explanation:
The word equation is as follows.
silver + oxygen = silver oxide
In terms of chemical formulas this equation can be written as follows.
[tex]Ag + O_{2} \rightarrow Ag_{2}O[/tex]
The removal on electron(s) from an atom, ion or molecule in a chemical reaction is called oxidation.
The gain of electron(s) by an atom, ion or molecule in a chemical reaction is called reduction.
Hence, half-reaction equations for the given reaction is as follows.
Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]
Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]
As the number of atoms participating in the reaction are equal. Hence, the half-equations are balanced.
Thus, we can conclude that the balanced half-equations for silver + oxygen = silver oxide are:
Oxidation-half reaction: [tex]Ag \rightarrow 2Ag^{+} + 2e^{-}[/tex]
Reduction-half reaction: [tex]O_{2} + 2e^{-} \rightarrow 2O^{-}[/tex]
༒How much does the earth weigh?☆☆☆☆☆☆☆☆☆☆☆☆
Answer:
Earth has a mass of 5.9736×1024 kg
5.972×10²⁴kghope it is helpful to you
4.106
Calculate the moles and the mass of solute in each of the following solutions.
(a) 150.0 mL of 0.245 M CaCl2
molarity = moles of solute / volume of solution
moles of solute = molarity × volume of solution
moles of solute = 0.245 mol/L × 0.1500 L
moles of solute = 0.03675 mol
moles of solute = 0.0368 mol
-----------------------------------------------------------
Solution: (mass of solute)Step 1: Calculate the molar mass of solute.
molar mass of solute = (40.08 g/mol × 1) + (35.45 g/mol × 2)
molar mass of solute = 110.98 g/mol
Step 2: Calculate the mass of solute.
mass of solute = moles of solute × molar mass of solute
mass of solute = 0.03675 mol × 110.98 g/mol
mass of solute = 4.08 g
Note: The volume of solution must be expressed in liters (L).
Answer:
[tex]\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}[/tex]
[tex]\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}[/tex]
Explanation:
1. Moles of SoluteMolarity is a measure of concentration in moles per liter.
[tex]molarity= \frac {moles \ of \ solute}{liters \ of \ solution}[/tex]
In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.
First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.
[tex]{150 \ mL * \frac{1 \ L}{1000 \ mL}= \frac{150}{1000} \ L = 0.150 \ L[/tex]Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.
[tex]0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}[/tex]
We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.
[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L[/tex]
[tex]0.150 \ L *0.245 \ mol \ CaCl_2 /L=x[/tex]
The units of liters cancel.
[tex]0.150 *0.245 \ mol \ CaCl_2 =x[/tex]
[tex]0.03675 \ mol \ CaCl_2[/tex]
The original measurements have 3 significant figures, so our answer must have the same.
We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.
[tex]\bold {0.0368 \ mol \ CaCl_2}[/tex]
2. Mass of the SoluteWe can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.
The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.
Ca: 40.08 g/mol Cl: 35.45 g/molNotice that chlorine has a subscript of 2. We must multiply the molar mass by 2.
Cl₂: 35.45 *2= 70.9 g/molAdd calcium's molar mass.
CaCl₂: 40.08 + 70.9 =110.98 g/molUse the molar mass as a ratio.
[tex]\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]
Multiply the moles of calcium chloride we calculated above.
[tex]0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}[/tex]
The units of moles of calcium chloride cancel.
[tex]0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }[/tex]
[tex]4.084064 \ g\ CaCl_2[/tex]
Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.
[tex]\bold {4.08 \ g \ CaCl_2}[/tex]
ELECTROLYSIS OF MOLTEN NaCl IS DONE IN A DOWNS CELL OPERATING AT 7.0 VOLTS AND 4.0X10^4A. HOW MUCH Na(s) AND Cl2(g) CAN BE PRODUCED IN 8 HOURS IN SUCH A CELL?
Answer:
Mass of sodium produced = 2.74 × 10⁵ g of Na
Mass of chlorine produce = 4.24 × 10⁵ g of Cl₂
Explanation:
In the electrolysis of molten NaCl as described above, the quantity of charge used is given by the formula, Q = I × t
Where I isnthe current passed in amperes and t is time in seconds.
Q = 4.0 × 10⁴ A × (8 × 60 × 60) s = 1.152 × 10⁹ C
Equation for the discharge of sodium is; Na+ + e- ---> Na (s)
One mole of electrons is required to discharge one mole of Na
One mole of electron = 1 faraday = 96500 C
One mole of Na has a mass of 23 g
96500 C produces 23 g of Na
1.152 × 10⁹ C will produce 23 g × 1.152 × 10⁹ C / 96500 C = 2.74 × 10⁵ g of Na
Equation for the discharge of chlorine gas is; 2 Cl- ---> Cl₂(g) + 2e-
Two mole of electrons are required to discharge one mole of chlorine gas
Two moles of electron = 2 faraday = 2 × 96500 C = 193000
One mole of Cl₂ has a mass of 71 g
193000 C produces 71 g of Cl₂
1.152 × 10⁹ C will produce 71 g × 1.152 × 10⁹ C / 193000 C = 4.24 × 10⁵ g of Cl₂
The amount of Na produced is 274551 g and the amount of Cl₂ produced is 423763.5 g.
Current passed through the cell = [tex]4.0\times10^4 A[/tex]
Time = 8 Hours
We have to calculate the amount of Na and [tex]Cl_2[/tex] produced in 8 hours in the downs cell.
What is a down cell?The Downs process is an electrochemical method for the commercial preparation of metallic sodium, in which molten NaCl is electrolyzed in a special apparatus called the Downs cell.
The total charge passed through the cell is calculated by the given formula as
Charge(Q) = Current(I) × time(t)
Q = [tex]4.0\times10^4 A \times t[/tex]
[tex]t = 8 \times 60\times 60[/tex] sec
t = 28800 sec
Q = [tex]4.0\times10^4 A \times 28800[/tex] sec
Q = [tex]115200\times 10^4 A\ sec[/tex]
We know that, Ampere = Coulombs per sec
Q = [tex]115200\times 10^4\ C[/tex]
1 mol of electrons 96500 C charge
Therefore, the number of mols of electrons carries this [tex]115200\times 10^4\ C[/tex] charge = [tex]\frac{115200\times10^4}{96500}[/tex] = 1.1937 × 10⁴ = 11937 mol electrons
In the Down's cell
Half cell reactions are:
Reduction half-reaction: [tex]2Na^+(aq)+2e^- \to 2Na(s)[/tex]
Oxidation half-reaction: [tex]2Cl^-(aq)\to Cl_2(g) + 2e^-[/tex]
We know that no. of moles = [tex]\frac{given \ mass}{molar \ mass}[/tex]
Molar mass of Na = 23 g/mol
The mass of Na formed = 11937 mol × 23 g/mol = 274551 grams
The molar mass of Cl₂ = 71 g/mol
The mass of Cl₂ = [tex]\frac{11937}{2}\times 71 = 423763.5 \ grams[/tex]
Hence, the amount of Na produced is 274551 g and the amount of Cl₂ produced is 423763.5 g.
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https://brainly.com/question/13891856
What does Gibbs free energy predict?
A. It predicts what the rate of the reaction will be.
B. It predicts how high the activation energy is.
C. It predicts if entropy will increase or decrease.
D. It predicts whether or not a reaction will be spontaneous.
Answer:
It predicts whether or not a reaction will be spontaneous.
Explanation:
The equation;
∆G= ∆H - T∆S enables us to obtain the Gibbs free energy of a chemical process.
The Gibbs free energy value tells us whether a chemical process will be spontaneous or not.
When;
∆G>0 the reaction is not spontaneous
∆G=0 the reaction is at equilibrium
∆G<0 the reaction is spontaneous
Excluding any secondary chemical reactions,
which would be more effective as an antifreeze; a
solution containing
Select one:
25m CH3OH
the combination of 25m CH3OH and 25m
KCI.
O 25m KCI.
O
none of these
Answer:
Excluding any secondary chemical reactions,
which would be more effective as an antifreeze; a
solution containing
Select one:
25m CH3OH
the combination of 25m CH3OH and 25m
KCI.
25m KCI.
Explanation:
Antifreeze is the one that reduces the freezing point of a solvent further and will not allow the solvent to freeze.
Among the given options, the correct option is:
25 m CH3OH and 25m KCl.
Since, KOH is a strong electrolyte and dissociates into two ions.
So, the freezing point of solvent decreases further.
43.0 mL of 1.49 M perchloric acid is added to 14.0 mL of calcium hydroxide, and the resulting solution is found to be acidic.
29.1 mL of 0.498 M barium hydroxide is required to reach neutrality.
What is the molarity of the original calcium hydroxide solution?
Answer:
2.29 M
Explanation:
Equation of the reaction;
Ca(OH)2(aq) + 2HClO4(aq) → 2H2O(l) + Ca(ClO4)2(aq)
Concentration of acid CA = 1.49 M
Concentration of base CB= ????
Volume of acid VA= 43.0 ml
Volume of base VB= 14.0 ml
Number of moles of acid NA = 2 moles
Number of moles of base NB = 1 mole
CAVA/CBVB = NA/NB
CAVANB =CBVBNA
CB= CAVANB/VBNA
CB= 1.49 × 43.0 × 1/14.0 × 2
CB= 2.29 M
Which of the following is an alkaline earth metal?
A Carbon
B aluminum
C magnesium
D silicon
Answer:
Magnesium.
Explanation:
Because it is in group II
A scientist is conducting a Sanger's sequencing experiment to determine the number of polypeptides present in an oligomeric protein. The molecular weight of the protein is 18000 g/mol . After the reaction of 520 mg of the protein with 1‑fluoro‑2,4‑dinitrobenzene, the peptide bonds were hydrolyzed with an acid. As a result, the scientist obtained 39 mg of 2,4‑dinitrophenyl serine. What is the number of the polypeptide chains present in the oligomer?
Answer:
Depends on molecule.
Explanation:
The number of the polypeptide chains present in the oligomer depends on the molecule. Some molecules have more polypeptide chains whereas some of them have less polypeptide chains. For example, Hemoglobin is a oligomer that consists of four Polypeptide Chains, two of these Polypeptide Chains are α-globin molecules, each comprise of 141 amino acids, and the other two are (β, γ, δ, or ε) globins, each consist of 146 amino acids.
How many molecules are present in 1296 g of dinitrogen pentoxide (N2O5)
Answer:
The molar mass of N2O5 is 108 g/mol. 1296g of N2O5 has 1296 / 108 = 12 moles. 1 mole contains 6.022 x 10^23 molecules
? Question
In an ozone molecule, the three atoms must be connected, so there must at least be a single bond between them. Place
dots in pairs around the oxygen atoms until each oxygen atom has eight valence electrons, starting with the atoms on the
outside and doing the central atom last if there are enough. Do not exceed the total number of valence electrons
identified in part A. Remember that the dashes between the oxygen atoms, which represent single bonds, each indicate
the presence of two valence electrons.
A is the answer
In an ozone molecule, the three atoms must be connected, so there must at least be a single bond between them. Place
dots in pairs around the oxygen atoms until each oxygen atom has eight valence electrons, starting with the atoms on the
outside and doing the central atom last if there are enough. Do not exceed the total number of valence electrons
identified in part A. Remember that the dashes between the oxygen atoms, which represent single bonds, each indicate
the presence of two valence electrons
Answer:
Explanation: i did it
Describe how you would prepare your assigned ester from a carboxylic acid and an alcohol. You do not need to include a detailed procedure, but you should include any necessary reagents or catalyst (solvents are not needed).
Answer:
The general preparation of esters( for example ethyl ethanoate) is through a process known as ESTERIFICATION.
Explanation:
The formation of an ester by the reaction between an alkanol and an acid is known as esterification. This reaction is extremely slow and reversible at room temperature, and is catalyzed by a high concentration of hydrogen ions.
In the preparation of one of the simpler esters known as ETHYL ETHANOATE the reactants include ethanol(an alcohol) and glacial ethanoic acid(a carboxylic acid) in the presence of concentrated tetraoxosulphate VI acid as a CATALYST. Note that, a catalyst is any substance that is able to increase the rate of a chemical reaction.
The mixture is warmed in a water bath( hot but not boiling) for about 25 minutes. The mixture is poured into a beaker partially filled with a sodium or calcium chloride to remove interacted ethanol. The ethyl ETHANOATE floats on the mixture as oily globules.
Na2CO3 reacts with dil.HCl to produce NaCl, H2O and CO2. If 21.2 g of pure Na2CO3 are added in a solution containing 21.9g HCl , a. Find the limiting reagent. (2) b. Calculate the number of moles of excess reagent left over.(2) c. Calculate the number of molecules of H2O formed.(1) d. Calculate volume of CO2 gas produced at 270C and 760mm Hg pressure.(2) e. Write significance of limiting reagent
Answer:
See explanation
Explanation:
Equation of the reaction;
Na2CO3(aq) + 2HCl(aq) -------> 2NaCl(aq) + H2O(l) + CO2(g)
Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3
Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl
1 mole of Na2CO3 reacts with 2 moles of HCl
0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl
Hence Na2CO3 is the limiting reactant
Since there is 0.6 moles of HCl present, the number of moles of excess reagent=
0.6 moles - 0.4 moles = 0.2 moles of HCl
1 mole of Na2CO3 forms 1 mole of water
0.2 moles of Na2CO3 forms 0.2 moles of water
Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water
1 mole of Na2CO3 yields 1 mole of CO2
0.2 moles of Na2CO3 yields 0.2 moles of CO2
1 mole of CO2 occupies 22.4 L
0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP
Hence;
V1=4.48 L
T1 = 273 K
P1= 760 mmHg
T2 = 27°C + 273 = 300 K
P2 = 760 mmHg
V2 =
P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
V2 = P1V1T2/P2T1
V2 = 760 × 4.48 × 300/760 × 273
V2= 4.9 L
The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.
Which substance has nonpolar covalent bonds?
CO
NO2
H2
NaBr
Answer:
H2
Explanation:
Answer:
[tex]{ \bf{H _{2} }} \\ { \tt{hydrogen \: gas}}[/tex]
In a pure metal, the electrons can be thought of as [ Select ] throughout the metal. Using molecular orbital theory, there [ Select ] an energy gap between the filled molecular orbitals and empty molecular orbitals. The [ Select ] orbitals are typically higher in energy and are mostly [ Select ] .
Answer:
Explanation:
In a pure metal, the electrons can be thought of as [concentrated] around atoms throughout the metal. Using molecular orbital theory, there [is ] an energy gap between the filled molecular orbitals and empty molecular orbitals. The [antibonding] orbitals are typically higher in energy and are mostly (filled]
How many moles are contained in .984 molecules F2?
Answer:
1.6 x 10⁻²⁴ moles
Explanation:
We have 0.984 molecules of F₂. We know that 1 mol is equal to 6.022 x 10²³ molecules. Thus, we have a conversion factor: 1 mol/6.022 x 10²³ molecules
So, we multiply the molecules of F₂ by the conversion factor to calculate the moles:
0.984 molecules x 1 mol/6.022 x 10²³ molecules = 1.6 x 10⁻²⁴ moles
Given the following list of densities, which materials would float in a molten vat of lead provided that they do not themselves melt? Densities (g/mL): lead = 11.4, glass = 2.6, gold = 19.3, charcoal = 0.57, platinum = 21.4.
a. gold and platinum
b. glass and charcoal
c. gold, platinum, glass and coal
d. gold and charcoal
e. None of these
Answer:
b. glass and charcoal
Explanation:
Step 1: Given data
Density of Pb: 11.4 g/mLDensity of Glass: 2.6 g/mLDensity of Au: 19.3 g/mLDensity of charcoal: 0.57 g/mLDensity of platinum: 21.4 g/mLStep 2: Determine which material will float in molten lead
Density is an intrinsic property of matter. Less dense materials float in more dense materials. The materials whose density is lower than that of lead and will therefore float on it are glass and charcoal.
What is the molecular formula of the structure below?
Picture is attached pls help I’ll mark as brainliest for the right answer
Answer:
C₆H₆
Explanation:
Each border of the figure represents 1 atom of carbon. We have 6 borders = 6 atoms of carbon.
Each atom of carbon form 4 bonds. All the carbons are doing a double bond and a single bond with other carbons. That means are bonded 3 times. The other bond (That is not represented in the figure. See the image) comes from hydrogens. As we have 6 carbons that are bonded each 1 with one hydrogen. There are six hydrogens and the molecular formula is:
C₆H₆This structure is: Benzene
How many alkenes are formed by E2 elimination of HBr from 3-bromo -3,4-dimethylhexane using a strong base such as sodium methoxide (NOTE: draw 3,4-dimethyl as anti configuration)
Answer:
2
Explanation:
2 alkenes are formed by E2 elimination of HBr from 3-bromo -3, 4 - dimethylhexane using a strong base such as sodium methoxide
If I have 21 liters of gas held at a pressure of 78 atm and a temperature of 900. K, what will be the volume of the gas if I decrease the pressure to 52 atm and decrease the temperature to 750 K?
SHOW YOUR WORK
Answer:
txyxyc8uviciycuyc
Explanation:
gghvj ucy7cyvyfy
According to the following pKa values listed for a set of acids, which would lead to the strongest conjugate base?
a. -2.
b. 1.
c. 7.
d. 25.
e. 50.
Complete and balance the equations for the given single displacement reactions. Write the reaction in molecular form. Phases are optional. If you need to clear your work and reset the equation, click the CLR button.Li(s) + H2O---------- Ca(s) + H2O ---------------
Answer:
Li(s) + H2O(l) -----> LiOH(aq) + H2(g)
Ca(s) + 2H2O(l) ---------------> Ca(OH)2(aq) + H2(g)
Explanation:
Metals react with water to yield the metal hydroxide and hydrogen gas as shown in the answer above.
The reaction equations were balanced, the number of atoms of each element on both side of the reaction equation is exactly the same.
This is the way to write a balanced reaction equation for the species shown in the question.
A compound was analyzed and found to contain 76.57% carbon, 6.43% hydrogen, and 17.00% oxygen by mass. Calculate the empirical formula of the compound. If the molar mass of the compound is 94.11 g/mol, what is the molecular formula of the compound?
A second compound is composed of 53.30% Carbon 11.19% Hydrogen and 35.51% Oxygen by mass.Please Calculate the empirical formula of the compound of the molar mass of the compound is 90.12g/mol, what is the molecular formula for that compound?
Answer:
See explanation
Explanation:
First we divide the percentage by mass of each element by it's relative atomic mass then we divide the quotients obtained by the lowest ratio obtained in the first step.
C- 76.57/12, H= 6.43/1, O = 17.00/16
C- 6.38/1.06, H= 6.43/1.06, O= 1.06/1.06
C- 6, H- 6, O- 1
Empirical formula: C6H6O
[(12 ×6) + (6 × 1) + (16 × 1)]n=94.11
[72 + 6 +16]n = 94.11
n = 94.11/94
n= 1
Molecular formula = C6H6O
2)
C- 53.30/12, H- 11.19/1, O- 35.51/16
C- 4.44/2.22, H- 11.19/2.22, O- 2.22/2.22
C- 2, H- 5, O- 1
Empirical formula: C2H5O
[(2×12) + (5× 1) + (1×16)]n = 90.12
[24 + 5 + 16] n = 90.12
n= 90.12/45
n= 2
Molecular formula = C4H10O2
According to the Arrhenius equation, changing which factors will affect the
rate constant?
A. Temperature and the ideal gas constant
B. The activation energy and the constant A
C. The constant A and the temperature
D. Temperature and activation energy
Answer:
e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature
Answer:
D. Temperature and activation energy is the correct answer
Explanation:
^_^
Excited sodium atoms may emit radiation having a wavelength of 589nm. a) What is the wavelength in meters
Answer: When excited sodium atoms may emit radiation having a wavelength of 589nm. It's wavelength in meters is [tex]589 \times 10^{-9} m[/tex].
Explanation:
Given: Wavelength = 589 nm
It is known that,
[tex]1 nm = 10^{-9} m[/tex]
Hence, 589 nm is converted into meters as follows.
[tex]589 nm = 589 nm \times \frac{10^{-9}m}{1 nm}\\= 589 \times 10^{-9} m[/tex]
Thus, we can conclude that when excited sodium atoms may emit radiation having a wavelength of 589nm. It's wavelength in meters is [tex]589 \times 10^{-9} m[/tex].
show in chemistry the fermentation of carbohydrate to form alcohol
hope it helps to everyone
