Answer:
A. [OH⁻] = 2.188x10⁻³M
B. pKb = 6.02
Explanation:
When hydrazine is in equilbrium with water, its reaction is:
N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)
Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
A. From pH, you can find [OH⁻], thus:
pH = -log [H⁺]
11.34 = -log [H⁺]
4.57x10⁻¹² = [H⁺]
As 1x10⁻¹⁴ = [OH⁻] [H⁺]
1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]
[OH⁻] = 2.188x10⁻³MB. Concentrations in equilibrium of the species are:
[N₂H₄] = 5.0M - X
[HN₂H₄⁺] = X
[OH⁻] = X
Where X is reaction coordinate
As [OH⁻] = 2.188x10⁻³M
X = 2.188x10⁻³M
Replacing:
[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M
[HN₂H₄⁺] = 2.188x10⁻³M
[OH⁻] = 2.188x10⁻³M
Replacing in Kb expression:
Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]
Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]
Kb = 9.577x10⁻⁷
pKb is defined as -log Kb
pKb = -log 9.577x10⁻⁷
pKb = 6.02
Calculate the combustion of gaseous dimethyl ether CH 3 OCH 3 (g)+3O 2 (g) 2CO 2 (g)+3H 2 O(l) using standard molar enthalpies of formation Molecule AH H l ^ 0 (k)/mol) CH 3 OCH 3 (g) - 184.1
Answer:
[tex]\Delta _cH=-1328.3kJ/mol[/tex]
Explanation:
Helllo,
In this case, for the given chemical reaction in gaseous state:
[tex]CH_3OCH_3+3O_2\rightarrow 2CO_2+3H_2O[/tex]
We comoute the combustion enthalpy as the reaction enthalpy for one mole of fuel (dimethyl ether) considering the formation enthalpy of each given substance and whether they are reactants (subtracting) or products (adding), therefore we write:
[tex]\Delta _cH=2*\Delta _fH_{CO_2}+3*\Delta _fH_{H_2O}-\Delta _fH_{CH_3OCH_3}-3*\Delta _fH_{O_2}[/tex]
Whereas the formation enthalpies for carbon dioxide, water, dimethyl ether and oxygen are -393.5, -241.8, -184.1 and 0 kJ/mol respectively, thereby, the combustion enthalpy turns out:
[tex]\Delta _cH=2(-393.5)+3*(-241.8)-(-184.1)-3(0)\\\\\Delta _cH=-1328.3kJ/mol[/tex]
Notice that enthalpy of formation of oxygen is zero since forming an element has no chemical sense, it just exists as it has been early demonstrated.
Regards.
. Explain why, in the sample calculations, 0.1 g of the unknown produced a GREATER freezing point depression than~e same mass of naphthalene.
Answer
Naphthalene is a non electrolyte
If the unknown compound is an electrolyte it gives 2 or more ions in solution
( NaCl >> Na+ + Cl- => 2 ions
Ca(NO3)2 >> Ca2+ + 2 NO3- => 3 ions)
the f.p. lowering is directly proportional to the molal concentration of dissolved ions in the solution )
For naphthalene
delta T = 1.86 x m
for a salt that gives 2 ions
delta T = 1.86 x m x 2
hence the lowering in freezion point of unkown is greater then napthalene
Which of the following atoms would have the longest de Broglie wavelength, if all have the same velocity?
A) Li
B) Na
C) Fe
D) Pb
E) Not possible to tell with given information
Answer:
Li
Explanation:
The phenomenon of wave particle duality was well established by Louis deBroglie. The wavelength associated with matter waves was related to its mass and velocity as shown below;
λ= h/mv
Where;
λ= wavelength of matter waves
m= mass of the particle
v= velocity of the particle
This implies that if the velocities of all particles are the same, the wavelength of matter waves will now depend on the mass of the particle. Hence; the wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The longest wavelength will then be obtained from the smallest mass of matter. Hence lithium which has the smallest mass will exhibit the longest DeBroglie wavelength
The atom that have the longest de Broglie wavelength is ; ( A ) Li
Wave particle duality is a phenomenon by de Broglie. that shows that The wavelength associated with matter waves is related to its mass and velocity .
Wave particle duality is represented as ; λ = h / mv
λ= wavelength of matter waves
m= mass of the particle
v= velocity of the particle
Given that the elements have the same velocity the atom that would have the longest de Broglie wavelength is Li
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Which best describes the trends in electonegativity on the periodic table
Hey! :)
__________ ☆ ☆__________________________________
Answer:
The answer is Electronegativity increases up and to the right
Explanation:
When you move from left to right it increases ( in the periodic table )
But when you move down the table electronegativity decreases.
So “ Electronegativity increases up and to the right” describes the trends the best.
Hope this helps! :)
____________☆ ☆________________________________
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g The most common position for an double bond in an unsaturated fatty acid is delta _________(fill in the number).
Answer:
The most common position for an double bond in an unsaturated fatty acid is delta 9 (Δ⁹)
Explanation:
Unsaturated fatty acids are carboxylic acids which contains one or more double bonds. The chain length as well as the number of double bonds is written separated by a colon. The positions of the double bonds are specified starting from the carboxyl carbon, numbered as 1, by superscript numbers following a delta (Δ). For example, an 18-carbon fatty acid containing a single double bond between carbon number 9 and 10 is written as 18:1(Δ⁹).
In most monounsaturated fatty acids, the double bond is between C-9 and C-10 (Δ⁹), and the other double bonds of polyunsaturated fatty acids are generally Δ¹² and Δ¹⁵. This positioning is due to the nature of the biosynthesis of fatty acids. In the mammalian hepatocytes, double bonds are introduced easily into fatty acids at the Δ⁹ position, but cannot introduce additional double bonds between C-10 and the methyl-terminal end. However, plants are able to introduce these additional double bonds at the Δ¹² and Δ¹⁵ positions.
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent composition of fluorine
Answer:
The percent composition of fluorine is 65.67%
Explanation:
Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.
That is, the percentage composition is the percentage by mass of each of the elements present in a compound.
The calculation of the percentage composition of an element is made by:
[tex]percent composition element A=\frac{total mass of element A}{mass of compound} *100[/tex]
In this case, the percent composition of fluorine is:
[tex]percent composition of fluorine=\frac{39.6 g}{60.3 g} *100[/tex]
percent composition of fluorine= 65.67%
The percent composition of fluorine is 65.67%
Answer:
The percent composition of fluorine is 65.67%
Explanation:
Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.
That is, the percentage composition is the percentage by mass of each of the elements present in a compound.
The calculation of the percentage composition of an element is made by:
In this case, the percent composition of fluorine is:
percent composition of fluorine= 65.67%
The percent composition of fluorine is 65.67%
Name MgSo4.7H2O /Cs3PO4.H2O
Answer:
MgSo4.7H2O = Magnesium sulfate
Cs3PO4.H2O = Cesium Phosphate
Hope this helps!
A constant volume and mass of helium gas at 77°C is heated so that the pressure of the gas doubles. What is the new temperature of the gas in Celsius degrees?
Answer:
427°C .
Explanation:
Step 1:
Data obtained from the question. This include the following:
Initial temperature (T1) = 77°C
Initial pressure (P1) = P
Final pressure (P2) = 2P
Final temperature (T2) =?
Step 2:
Conversion of celsius temperature to Kelvin temperature.
This is illustrated below:
T(K) = T (°C) + 273
Initial temperature (T1) = 77°C
Initial temperature (T1) = 77°C+ 273 = 350K
Step 3:
Determination of the new temperature. The new temperature can be obtained as follow:
P1/T1 = P2/T2
P/350 = 2P/T2
Cross multiply
P x T2 = 350 x 2P
Divide both side by P
T2 = (350 x 2P ) / P
T2 = 700K
Step 4:
Conversion of Kelvin temperature to celsius temperature.
This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 700K
T(°C) = 700 – 273
T(°C) = 427°C
Therefore, the new temperature of the gas is 427°C
Which physical method can be used for obtaining a sample of salt from a small beaker of salt water?
boiling
freezing
chromatography
sorting
Answer:
a. boiling
Explanation:
Calculate the number of moles of C2H6 in 3.97×1023 molecules of C2H6.
3.97×1023 molecules C2H6 1 mol C2H6
------------------------------------------ x ------------------------------------ = 0.66 mol C2H6
6.022 x 1023 molec. C2H6
Identify the following substances:
(1) An acidic gas which gives dense white fumes with NH3
(11 An alkane which can also be called a green house gas.
(iii) A solid which when kept in the open, forms a solution after som
(iv) An alloy used in electrical fittings.
(v) A metal which gives hydrogen gas on reacting with both dilute a
Write equations for the following reactions:
Aluminium oxide and Sodium hydroxide.
(ii) Zine and dilute sulphuric acid.
(
11
Answer:
1) HCl (Hydrochloric acid reacts with NH3 and forms dense fumes)
2) Methane (It is from the group of alkanes and is a green house gas)
3) Deliquescent substance (It is a solid which when kept in open forms a solution after sometime)
4) Brass (It is a solid-in-solid solution used to make electrical fittings)
5) Aluminium
Question 2:
1) Al₂O₃ + 2NaOH ⇒ 2NaAlO₂ (Sodium Aluminate) + H₂O
2) Zn + H₂SO₄ (dilute) => ZnSO₄ (Zinc Sulphate) + H₂
Answer:
HCl is the gasmethanedeliquescent substances (they absorbs water out of the air until they dissolve themselves into liquid solutions. )brassequations
Aluminium oxide and Sodium hydroxide react to form water and sodium aluminate
Al2O3 + 2NaOH → 2NaAlO2 + H2OZinc reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas
Zn + H2SO4 → ZnSO4 +H2
Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page?
The question is incomplete; the complete question is: Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? Same (angles do not change) Different (angles change) Answer Bank | H2O | CO2, SO2, XeF2, BF3 CIF3, NH3, CH4, SF4, XeF4, BrF5, PCI5,SF6
Answer:
Compounds whose real bond angle are the same as ideal bond angle;
SF6, BF3, CH4, PCI5
Compounds whose real bond angles differ from ideal bond angles;
H2O, CO2, SO2, XeF2, CIF3, NH3, SF4, XeF4, BrF5
Explanation:
According to the valence shell electron pair repulsion theory (VSEPR), molecules adopt various shapes based on the number of electron pairs on the valence shell of the central atom of the molecule. The electron pairs usually orient themselves as far apart in space as possible leading to various observed bond angles.
The extent of repulsion of lone pairs is greater than that of bond pairs. Hence, the presence of lone pairs on the valence shell of the central atom in the molecule distorts the bond angles of molecules away from the ideal bond angles predicted on the basis of valence shell electron pair repulsion theory.
For instance, methane is a perfect tetrahedron having an ideal bond angle of 109°28'. Both methane and ammonia are based on a tetrahedron, however, the presence of a lone pair of electrons on nitrogen distorts the bond angle of ammonia to about 107°. The distortion of lone pairs in water is even more as the bond angles of water is about 104°.
When 75.5 grams of phosphorus pentachloride react with an excess of water, as shown in the unbalanced chemical equation below, how many moles of hydrochloric acid will be produced? Please show all your work for the calculations for full credit. PCl5 + H2O --> H3PO4 + HC
Answer:
Explanation: M(PCL5)= 31 + 5(35.5)
=208.5g/mol
M(H20)= 18g/mol
n(PCL5) = 75.5÷208.5
= 0.362mol
n(HCl)/n(PCL5)= 5/1
n(HCl)= 5×0.362
=1.81mol of HCl
Determine the radius of an Al atom (in pm) if the density of aluminum is 2.71 g/cm3 . Aluminum crystallizes in a face centered cubic structure with an edge leng
Answer:
143pm is the radius of an Al atom
Explanation:
In a face centered cubic structure, FCC, there are 4 atoms per unit cell.
First, you need to obtain the mass of an unit cell using molar mass of Aluminium and thus, obtain edge length and knowing Edge = √8R you can find the radius, R, of an Al atom.
Mass of an unit cell
As 1 mole of Al weighs 26.98g. 4 atoms of Al weigh:
4 atoms × (1mole / 6.022x10²³atoms) × (26.98g / mole) = 1.792x10⁻²²g
Edge length
As density of aluminium is 2.71g/cm³, the volume of an unit cell is:
1.792x10⁻²²g × (1cm³ / 2.71g) = 6.613x10⁻²³cm³
And the length of an edge of the cell is:
∛6.613x10⁻²³cm³ = 4.044x10⁻⁸cm = 4.044x10⁻¹⁰m
Radius:
As in FCC structure, Edge = √8 R, radius of an atom of Al is:
4.044x10⁻¹⁰m = √8 R
1.430x10⁻¹⁰m = R.
In pm:
1.430x10⁻¹⁰m ₓ (1x10¹²pm / 1m) =
143pm is the radius of an Al atomThe radius of the atom of Al in the FCC structure has been 143 pm.
The FCC lattice has been contributed with atoms at the edge of the cubic structure.
The FCC has consisted of 4 atoms in a lattice.
The mass of the unit cell of Al can be calculated as:[tex]\rm 6.023\;\times\;10^2^3[/tex] atoms = 1 mole
4 atoms = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles
The mass of 1 mole Al has been 26.98 g/mol.
The mass of [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles = [tex]\rm \dfrac{4}{6.023\;\times\;10^2^3}[/tex] moles × 26.98 g
The mass of 1 unit cell of Al has been = 1.792 [tex]\rm \bold{\times\;10^-^2^2}[/tex] g.
The volume of the Al cell can be calculated as:Density = [tex]\rm \dfrac{mass}{volume}[/tex]
Volume = Density × Mass
The volume of Al unit cell = 2.71 g/[tex]\rm cm^3[/tex] × 1.792 [tex]\rm \times\;10^-^2^2[/tex] g
The volume of Al cell = 6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex]
The volume of the cube has been given as:Volume = [tex]\rm edge\;length^3[/tex]
6.613 [tex]\rm \times\;10^-^2^3[/tex] [tex]\rm cm^3[/tex] = [tex]\rm edge\;length^3[/tex]
Edge length = [tex]\rm \sqrt[3]{6.613\;\times\;10^-^2^3}[/tex] cm
Edge length = 4.044 [tex]\rm \times\;10^-^8[/tex] cm
Edge length = 4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
In an FCC lattice structure, the radius of the atom can be given by:Edge length = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
4.044 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = [tex]\rm \sqrt{8\;\times\;radius}[/tex]
Radius = 1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m.
1 m = [tex]\rm 10^1^2[/tex] pm
1.430 [tex]\rm \bold{\times\;10^-^1^0}[/tex] m = 143 pm.
The radius of the atom of Al in the FCC structure has been 143 pm.
For more information about the FCC structure, refer to the link:
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What is the atomic mass of OsO4
Answer:
254.23 g/mol
Explanation:
Atomic mass for Osmium tetroxide would be 254.23 g/ml
Answer:254.2276
Explanation:
How does the carbon calculator estimate the amount of CO2 that each individual releases into the atmosphere?
Answer:
The calculator add the CO2 released from the use of electricity, released from driving and the CO2 from the waste that we disposed.
Explanation:
The carbon dioxide, CO2 is what the human body does not need, therefore, we breathe it out, hence taking in oxygen(respiration process). The plants need oxygen for the production of their own food.
The carbon calculator estimate the amount of CO2 that each individual releases into the atmosphere through the consideration of several factors such as the kind of food that we eat.
Therefore, if we are to use the carbon calculator to determine the amount of CO2 that each individual releases into the atmosphere we will have:
The amount of CO2 that each individual releases into the atmosphere =( CO2 released from the use of electricity) + (CO2 released from driving) + (the CO2 from the waste that we disposed).
If you combine 24.2 g of a solute that has a molar mass of 24.2 g/mol with 100.0 g of a solvent, what is the molality of the resulting solution
Answer: 10 moles/kg.
Explanation:
Given, Mass of solute = 24.2 g
Molar mass of solute = 24.2 g/mol
[tex]\text{Moles of solute =}\dfrac{\text{Mass of solute}}{\text{Molar mass of solute}}\\\\=\dfrac{24.2}{24.2}=1[/tex]
Mass of solvent = 100.0g = 0.1 kg [1 g=0.001 kg]
[tex]\text{Molality}=\dfrac{\text{Moles of solute}}{\text{kilograms of Solvent}}\\\\=\dfrac{1}{0.1}\\\\=10\ moles/kg[/tex]
Hence, the molality of the resulting solution is 10 moles/kg.
Identify the acid, base, conjugate acid and conjugate base in the following reactions:
a. NH_3(aq) + CH_3COOH(aq) NH_4^+ (aq) + CH_3COO^-(aq)
b. HClO_4(aq) + NH_4(aq) ClO_4^- (aq) + NH_4^+ (aq)
Answer:
a. NH₃ : base
CH₃COOH (acetic acid) : acid
NH₄⁺ : conjugate acid
CH₃COO⁻ : conjugate base
b. HClO₄ (perchloric acid) : acid
NH₃ : base
ClO₄⁻ : conjugate base
NH₄⁺ : conjugate acid
Hope this helps.
An electrode has a negative electrode potential. Which statement is correct regarding the potential energy of an electron at this electrode?
A. An electron at this electrode has the same potential energy as it has at a standard hydrogen electrode.
B. An electron at this electrode has a lower potential energy than it has at a standard hydrogen electrode.
C. An electron at this electrode has a higher potential energy than it has at a standard hydrogen electrode.
Answer:
C. An electron at this electrode has a higher potential energy than it has at a standard hydrogen electrode.
Explanation:
The standard hydrogen electrode (SHE) is used to measure the electrode potential of substances. The standard hydrogen electrode is arbitrarily assigned an electrode potential of zero. Recall that electrode potentials are always measured as reduction potentials in electrochemical systems.
For an electrode that has a negative electrode potential, electrons at this electrode have a higher potential energy compared to electrons at the standard hydrogen electrode. Electrons flow from this electrode to the hydrogen electrode.
On the other hand, a positive electrode potential implies that an electron at this electrode has a lower potential energy than it has at a standard hydrogen electrode. Hence electrons will flow from the standard hydrogen electrode to this electrode.
Which of the following is NOT a type of crystal structure? A. None of these B. Metallic C. Ionic D. Macromolecular (giant covalent)
Answer:
A. None of these.
Explanation:
A crystal structure is an arrangement of atoms or ions in a repeating three-dimensional array.
B. is wrong. Metal atoms, such as gold, arrange themselves into a crystal structure.
C. is wrong. Ionic solids, such as sodium chloride, arrange themselves into a crystal structure.
D. is wrong. Macromolecules (network solids), such as diamond, arrange themselves into a crystal structure.
The correct answer is None of these.
What is a crystal structure?A crystal structure is a three-dimensional collection of atoms or ions that repeats itself.Metal atoms(gold), Ionic solids (sodium chloride), and Macromolecules(network solids) arrange themselves into a crystal structure.Learn more about crystal structure here:-
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The thermochemical equation is for the reaction of hydrogen bromide gas to form hydrogen gas and bromine liquid. 2HBr(g) = H 2 (g)+ Br 2 (l) 72.6 kJ How many grams of HBr (g) would be made to react if 11.4 energy were provided?
Answer:
the mass of HBr that would react is 25.41 g of HBr
Explanation:
attached is the calculations.
At 25.0°C the Henry's Law constant for methane CH4 gas in water is ×1.410−3/Matm.
Calculate the mass in grams of CH4 gas that can be dissolved in 75.mL of water at 25.0°C and a CH4 partial pressure of 0.68atm. Round your answer to 2 significant digits.
Answer:
1.1 × 10⁻³ g
Explanation:
Step 1: Given data
Henry's Law constant for methane (k): 1.4 × 10⁻³ M/atm
Volume of water (=volume of solution): 75 mL
Partial pressure of methane (P): 0.68 atm
Step 2: Calculate the concentration of methane in water (C)
We will use Henry's law.
[tex]C = k \times P = 1.4 \times 10^{-3}M/atm \times 0.68atm = 9.5 \times 10^{-4}M[/tex]
Step 3: Calculate the moles of methane in 75 mL of water
[tex]\frac{9.5 \times 10^{-4}mol}{L} \times 0.075 L = 7.1 \times 10^{-5}mol[/tex]
Step 4: Calculate the mass corresponding to 7.1 × 10⁻⁵ mol of methane
The molar mass of methane is 16.04 g/mol.
[tex]7.1 \times 10^{-5}mol \times \frac{16.04g}{mol} = 1.1 \times 10^{-3} g[/tex]
The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles of salt in one tablespoon.
Finally, solve (remember significant figures):
Answer:
0.109 mol/tablespoon
Explanation:
6.37 g/ 58.5 mol = 0.10888888 mol (0.109 significantly)
Answer:
A: 0.109
Explanation:
Edge 2020
Draw a structural formula of an alkene or alkenes (if more than one) that undergo acid-catalyzed hydration and without re-arrangement give 2-butanol as the major product.
Answer:
See explanation
Explanation:
Hydration of alkenes is a common reaction in organic chemistry. Hydration is simply the addition of water to an alkene. This is an acid catalysed reaction as we can see from the mechanism attached.
Recall that our task is to carry out the synthesis of 2-butanol using an alkene starting material in which there will be no rearrangement of the intermediate carbocation. If we start with the compound shown in the image (but-2-ene), the first step is the formation of the secondary carbocation. This is followed by the addition of water. Subsequently, the added water is deprotonated by another water molecule to yield 2-butanol and the acid catalyst. All these steps have been clearly outlined in the image attached.
Silver crystallizes in a face-centered cubic structure. What is the edge length of the unit cell if the atomic radius of silver is 144 pm?
Answer:
Edge length of the unit cell is 4.07x10⁻¹⁰m
Explanation:
In a face-centered cubic structure, the edge, a, could be obtained using pythagoras theorem knowing the hypotenuse of the unit cell, b, is equal to 4r:
a² + a² = b² = (4r)²
2a² = 16r²
a = √8 r
That means edge lenght is = √8 r
adius
As radius of Silver is 144pm = 144x10⁻¹²m:
a = √8 r
a = √8 ₓ 144x10⁻¹²m
a = 4.07x10⁻¹⁰m
Edge length of the unit cell is 4.07x10⁻¹⁰mMany free radicals combine to form molecules that do not contain any unpaired electrons. The driving force for the radical–radical combination reaction is the formation of a new electron‑pair bond. Consider the formation of hydrogen peroxide. 2OH(g)⟶H2O2(g) Write Lewis formulas for the reactant and product species in the chemical equation. Include nonbonding electrons.
Answer:
In the attached image the Lewis equation is shown where it is shown how two oxygens react with two hydrogens to meet the octet of the electrons.
Explanation:
Hydrogen peroxide is one of the most named chemicals since it is not only sold as "hydrogen peroxide" in pharmacies but it is also one of the great weapons of immune defense cells to defend ourselves against anaerobic bacteria.
The disadvantage of this compound is that when dividing it forms free oxygen radicals that are considered toxic or aging for our body.
In the attached image below, you will see the Lewis equation is shown there. There, you will see how two oxygens react with two hydrogens to come about the octet of the electrons.
When two or more atoms bond with each other, they often form a molecule. When two hydrogens and an oxygen share electrons through covalent bonds, a water molecule is formed.
The octet rule is known as when most atoms want to gain stability in their outer most energy level by filling themselves that is the S and P orbitals of the highest energy level with eight electron.
HOOH is the compound that is form. It is called Hydrogen peroxide. This because it is has reactive oxygen species and the simplest peroxide.
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Which element has 4 valence electrons in the 3p sublevel?
The Periodic Table
A. Ga
B. Si
C. N
D. S
Answer:
D . Sulphur
Explanation:
the element with a 3p4 valence configuration, look in period 3 and group XVI, and that is ...
S, sulfur.
The compound barium nitrate is a strong electrolyte. Write the transformation that occurs when solid barium nitrate dissolves in water.
Answer:
Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻
Explanation:
A strong electrolyte is a salt (A compound that has an anion and a cation and are neutral) that, in water, dissociates completely in its ions.
In Barium nitrate, Ba(NO₃)₂, the cation is Ba²⁺ (Alkaline earth metal), and the anion is the nitrate ion, NO₃⁻.
Thus, when Ba(NO₃)₂ (s) is dissolved in water, its transformation is:
Ba(NO₃)₂(s) → Ba²⁺ + 2NO₃⁻When solid barium nitrate (Ba(NO₃)₂) dissolves in water, it undergoes a dissociation process where the compound breaks apart into its constituent ions.
Dissociation refers to the process in which a compound breaks apart into its constituent ions when dissolved in a solvent, typically water. In this process, the chemical bonds within the compound are disrupted, resulting in the separation of positive and negative ions.
The dissociation occurs due to the interaction between the solute particles and the solvent molecules, leading to the formation of hydrated ions.
The transformation can be represented as follows:
Ba(NO₃)₂(s) → Ba²⁺(aq) + 2NO₃⁻(aq)
In this process, the barium nitrate compound dissociates into barium ions (Ba²⁺) and nitrate ions (NO₃⁻) in the aqueous solution. The resulting ions are free to move and conduct electricity, indicating that barium nitrate is a strong electrolyte when dissolved in water.
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Calculate the mass of a body
Whose volume is
Is 2cm3 and
density is 520cm3
Answer:
The answer is
1040gExplanation:
Density = mass / volume
mass = density × volume
volume = 2cm³
density = 520g/cm³
mass = 2 × 520
= 1040g
Hope this helps you
How many valence electrons are in the electron dot structures for the elements in group 3A(13)?
Answer:
here, as we have known the elements of group 3A(13) such as aluminium , boron has three valance electron and in perodic table the elements are kept with similar proterties in same place so, their valance electron is 3.
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The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
What are Groups in the Periodic Table?The periodic table is organized into groups (vertical columns), periods (horizontal rows), and families (groups of elements that are similar). Elements in the same group have the same number of valence electrons.
Groups are the columns of the periodic table, and periods are the rows. There are 18 groups, and there are 7 periods plus the lanthanides and actinides.
There are two different numbering systems that are commonly used to designate groups, and you should be familiar with both.
The traditional system used in the United States involves the use of the letters A and B. The first two groups are 1A and 2A, while the last six groups are 3A through 8A. The middle groups use B in their titles.
Therefore, The number of valence electrons are in the electron dot structures for the elements in group 3A(13) is three.
Learn more about Groups in the periodic table, here:
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