Answer:
Presence of impurities in the product.
Explanation:
Laboratory experience has shown that the percent yield obtained from a reaction may be greater than 100% if the product obtained from the reaction contains impurities that causes its mass to be greater than it actually would be if the product was pure.
A typical impurity that may be present in a product is the solvent used. This can be removed by evaporation of the solvent or by washing the product thoroughly.
Reaction of solid products with gases may also increase the percentage yield above 100%.
Dynamic equilibrium between liquid and gas is established when __________ . a. the vapor pressure of the liquid equals the external pressureb. the liquid and vapor are at the same temperaturec. a liquid vaporizes and condenses at the same rated. the liquid is entirely evaporated
Answer:
The correct answer is option c. "a liquid vaporizes and condenses at the same rate"
Explanation:
In chemistry, dynamic equilibrium is defined as a state where the reaction rate of a forward reaction is equal to the reaction rate of the backward reaction. In this case, the reactions that participate in the system are the vaporization of a liquid and the condensation of a gas. Therefore, dynamic equilibrium is established when a liquid vaporizes and condenses at the same rate.
Are the bonds that chlorine forms with sodium (to form NaCl) and with carbon (to form CCl4) the same in both compounds?
Answer:
No, NaCl is formed by an ionic bond, and CCl4 would be formed by covalent bonds.
Explanation:
NaCl is an ionic bond because of their opposite charges while carbon and chlorine are sharing their electrons making it a covalent bond.
Taking into account the definition of ionic and covalent bond, the bonds that chlorine forms with sodium (to form NaCl) and with carbon (to form CCl₄) are different in both compounds.
Ionic bondAn ionic bond is a type of chemical bond that occurs when one atom gives up an electron to the other, in order for both to achieve electronic stability.
This union normally occurs between metal and nonmetal elements with different electronegativity, which means that the elements have different capacity to attract electrons.
In other words, an ionic bond is produced between metallic and non-metallic atoms, where electrons are completely transferred from one atom to another. During this process, one atom loses electrons and another one gains them, forming ions. Usually, the metal gives up its electrons forming a cation to the nonmetal element, which forms an anion.
Covalent bondThe covalent bond is the chemical bond between atoms where electrons are shared, forming a molecule. Covalent bonds are established between non-metallic elements, such as hydrogen H, oxygen O and chlorine Cl. These elements have many electrons in their outermost level (valence electrons) and have a tendency to gain electrons to acquire the stability of the electronic structure of noble gas.
Bonds of NaCl and CCl₄In the case of NaCl, Na is a metal while Cl is a nonmetal. So NaCl is an ionic bond.
On the other hand, in the case of CCl₄ both C and Cl are two non-metallic elements. So CCl₄ is a covalent bond.
In summary, the bonds that chlorine forms with sodium (to form NaCl) and with carbon (to form CCl₄) are different in both compounds.
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ionic bond:
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covalent bond:
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A water bottle has a mass of 14.0g. Given a density of 1.38g/cm^3, what is the volume of the plastic used to make the water bottle in g/cm^3?
Answer:
[tex] \boxed{\sf Volume \ of \ plastic \ used = 10.14 \ cm^3} [/tex]
Given:
Mass = 14.0 g
Density ([tex] \rho [/tex]) = 1.38 g/cm³
To Find:
Volume (V)of the plastic used to make water bottle
Explanation:
Formula:
[tex] \boxed{ \bold{Density \ (\rho) = \frac{Mass \ (m)}{Volume \ (V)}}}[/tex]
Substituting value of m & density in the formula:
[tex] \sf \implies 1.38 = \frac{14.0}{V} \\ \\ \sf \implies V = \frac{14.0}{1.38} \\ \\ \sf \implies V = 10.14 \ cm^3 [/tex]
[tex] \therefore [/tex]
Volume of the plastic used to make water bottle = 10.14 cm³
Calculate the molar free energy of combustion for liquid n-octane, C8H18 with the given data: AG'CH,,!) 6.4 kJ/mol AG®(00,,8) -394.4 kJ/mol AGʻ(H,0,8) -228.57 kJ/mol a. -5218.7 kJ/mol b. -5384.6 kJ/mol c. -629.4 kJ/mol d. +629.4 kJ/mol e. -2609.4 kJ/mol
Answer:
a. -5218.7 kJ/mol
Explanation:
Hello,
In this case, the combustion of liquid n-octane is represented by:
[tex]C_8H_{18}(l)+\frac{25}{2} O_2(g)\rightarrow 8CO_2(g)+9H_2O(g)[/tex]
Thus, the molar free energy of combustion (reaction) is:
[tex]\Delta _CH=8*\Delta _fH_{CO_2}+9*\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{18}}[/tex]
Hence, with the given data we obtain:
[tex]\Delta _CH=8*(-394.4kJ/mol)+9*\Delta _fH_{-228.57kJ/mol}-6.4kJ/mol\\\\\Delta _CH=-5218.73kJ/mol[/tex]
Therefore, answer is a. -52 kJ/mol18.7.
Regards.
The heat of combustion is -5218.7 kJ/mol.
The molar free energy of the reaction can be obtained as the sum of the free energy of the products minus the sum of the free energy of the reactants.
Now the reaction is;
C8H18(g) + 25/2O2(g) ----> 8CO2(g) + 9H2O(l)
Hence;
ΔGreaction = [8( -394.4) + 9(-228.57)] - [ 6.4 + 0]
=[(-3155.2) + (-2057.13)] - 6.4
= -5218.7 kJ/mol
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An unknown element X has the following isotopes: ⁵²X (79.50% abundant), ⁴⁹X (8.00% abundant), ⁵⁰X (12.50% abundant). What is the average atomic mass in amu of X?
Answer:
51.51 amu .
Explanation:
The atomic mass 52 amu has percentage of 79.5%
The atomic mass 49 amu has percentage of 8%
The atomic mass 50 amu has percentage of 12.5 %
Average atomic pass will be weighted average of their atomic weights
= 52 x .795 + 49 x .08 + 50 x .125
= 41.34 + 3.92 + 6.25
= 51.51 amu .
Which of the following is the basis for arranging the elements in the modern periodic table?
Answer:
The correct answer is - ascending order of atomic numbers in period and similar characters in groups.
Explanation:
The modern periodic table is arranged in periods and groups. In periods the arrangement of the elements are on the basis of the increasing order of atomic numbers.
In groups of the periodic table, elements are arranged on the basis of the similar characters of the elements and number of electrons in outer shell of the elements. The elements in the Modern Periodic Table are arranged in 7 Periods and 18 Groups.
Thus, the correct answer is - ascending order of atomic numbers in period and similar characters in groups.
In a coffee cup calorimeter, 100.0 mL of 1.0M NaOH and 100.0 mL of 1.0M HCl are mixed. Both solutions were originally at 24.6 degrees Celsius. After the reaction, the final temperature is 31.3 degrees Celsius. Assuming that all the solution have a density of 1.0g/cm3 and a specific heat capacity of 4.18 J/oCg, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.
Answer:
56.0kJ/mol
Explanation:
The reaction of NaOH with HCl is:
NaOH + HCl → H₂O + NaCl + ΔH
Where ΔH is the heat of reaction that is released per mole of reactants,
The moles that reacts are:
100mL = 0.1L * (1mol / L) = 0.1 moles reacts
To find the heat released in the coffee cup calorimeter, you use the equation:
Q = m×ΔT×C
Where Q is heat released,
m is mass of the solution
ΔT is change in temperature (Final temperature - Initial temperature)
C is specific heat of the solution (4.18J/g°C)
Mass of the solution is:
100mL + 100mL = 200mL
Density of the solution is 1.0g/mL. The mass is 200g
Change in temperature is 31.3°C - 24.6°C = 6.7°C
Replacing:
Q = m×ΔT×C
Q = 200g×6.7°C×4.18J/g°C
Q = 5601.2J
This is the heat released per 0.1mol. The heat released per mole (Enthalpy change for the neutralization of HCl by NaOH is:
5601.2J / 0.1 moles = 56012J / mol =
56.0kJ/mol
The stopper-puncturing needle should be completely pushed into the evacuated tube:__________. A. after the vein is entered. B. while the equipment is being assembled. C. just before the vein is entered. D. prior to anchoring the vein.
Answer:
The correct answer is:
after the vein is entered (A.)
Explanation:
The procedure in the question describes venopuncture.
Venopuncture is the method of aseptically collecting blood into the right sample containers. Using a stopper-punturing needle and a vacuum tube, the following procedure is followed:
assemble the needle and hub (needle holder)apply the tourniquet and examine the arm for palpable vein, palpate the veinrelease the tourniquet and swab the collection site with 70% alcoholallow the swab to dryreapply the tourniquetinsert the stopper-puncturing needle (attached to hub) into the appropriate veinafter the vein is entered, the stopper-puncturing needle should be completely pushed into the evacuated (vacuum) tubesblood will flow into the tube until vacuum is completely filledpull out the tube and invert gently several times if the tube contains anticoagulantsplace a gauze on the site of puncture (do not press down the gauze)remove the needle and press down the gauze immediatelydispose the needle into the sharps container or appropriatelylabel the tubes.An unknown acid solution has PH 3.4. 66% of the acid is ionized. Whats the pka?
Answer:
[tex]pKa=3.58[/tex]
Explanation:
Hello,
In this case, since the pH defines the concentration of hydrogen:
[tex]pH=-log([H^+])[/tex]
[tex][H^+]=10^{-pH}=10^{-3.4}=3.98x10^{-4}[/tex]
And the percent ionization is:
[tex]\% \ ionization=\frac{[H^+]}{[HA]}*100\%[/tex]
We compute the concentration of the acid, HA:
[tex][HA]=\frac{[H^+]}{\% \ ionization}*100\%=\frac{3.98x10^{-4}}{66\%} *100\%\\\\[/tex]
[tex][HA]=6.03x10^{-4}[/tex]
Thus, the Ka is:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}=\frac{3.98x10^{-4}*3.98x10^{-4}}{6.03x10^{-4}}\\ \\Ka=2.63x10^{-4}[/tex]
So the pKa is:
[tex]pKa=-log(Ka)=-log(2.63x10^{-4})\\\\pKa=3.58[/tex]
Regards.
Classify each of the following as a pure substance or a mixture. If a mixture, indicate whether it is homogeneous or heterogeneous:_______. (a) rice pudding, (b) seawater, (c) magnesium, (d) crushed ice.
Answer:
a) Heterogeneous mixture (b) Homogenous mixture (c) Pure substance (d) Pure substance
Explanation:
Homogenous mixtures contains mixture of substances with similar proportions while Heterogenous mixture contains substances with a varying proportion.
How can we be safe in a science laboratory
Answer:
We can be safe in a science laboratory by following all directions and wearing gloves, goggles to protect ourselves.
Explanation:
If we don't follow the rules then we might cuase a problem that we didn't want to happen and we need to wear protective gear because we need to stay safe
please answer this question
Answer:
1. False, water can't because its liquid
2. True, it dissolves more substances than any other liquid
3. False,stirring speeds up the rate of dissolving
Hope this helped
Predict the major product(s) for each of the following reactions. H3O (a) 4-Chlorobut-1-ene HBr (b) 1-Chlorobut-1-ene HBr (c) 4,4-Dimethylcyclopentene H2O, H (d) Propyne 2 HCl (e) Cyclopentylethene
Answer:
(a) 3-bromo-1-chlorobutane
(b) 2-bromo-1-chlorobutane
(c) 3,3-dimethylcyclopentan-1-ol
(d) 1,1,2,2-tetrachloropropane
(e) 1-ethylcyclopentan-1-ol
Explanation:
We can analyze each reaction:
(a) 4-Chlorobut-1-ene + HBr
In this reaction, we have an alkene. So, we will have an addition reaction, the nucleophile "[tex]Br^-[/tex]" will be added to the most substituted carbon. In this case carbon 3 to produce 3-bromo-1-chlorobutane.
(b) 1-Chlorobut-1-ene + HBr
In this reaction, we also have an alkene. So, we will have an addition reaction, the nucleophile "[tex]Br^-[/tex]" will be added to the most substituted carbon. In this case carbon 2 to produce 2-bromo-1-chlorobutane.
(c) 4,4-Dimethylcyclopentene + H2O, H+
In this reaction, we have an alkene. So, we will have an addition reaction, the nucleophile "[tex]H_2O[/tex]" will be added to the most substituted carbon. In this case carbon 3 to produce 3,3-dimethylcyclopentan-1-ol.
(d) Propyne + 2HCl
In this reaction, we have an alkyne. So, we will have an addition reaction, the nucleophile "[tex]Cl^-[/tex]" will be added to the most substituted carbon, but we have 2 moles of the nucleophile, so would be added 2 times and we will have as product 1,1,2,2-tetrachloropropane.
(e) Cyclopentylethene + H3O+
In this reaction, we have an alkene. So, we will have an addition reaction, the nucleophile "[tex]H_3O^+[/tex]" will be added to the most substituted carbon. But in this case, the carbon cation would be produced in carbon 1 of the ethene. So, we can have a hydride shift to produce a tertiary carbocation. With this in mind the nucleohile will be addde to this tertiary carbocation and we will have 1-ethylcyclopentan-1-ol.
See figure 1 to further explanations
How do humans affect sea turtles, seabirds and other marine animals?
Answer:
Fish, marine mammals and seabirds are being injured and killed by plastic corruption /pollution, and it is believed that 700 species could go deceased because of it. ... Losses are generally influenced by the ingestion of plastics, starvation, suffocation, virus, drowning and entrapment.
You can thank me and google
An unlit match contains approximately 1,000 J of chemical energy. When it burns, the match releases thermal energy and light energy. After it has burned, the remaining match contains 250 J of chemical energy. If the thermal energy was measured to be 400 J, how much light energy did the match emit?
Answer:
350J
Explanation:
Based on the first law of thermodynamics, which is the law of conservation of energy, energy can neither be created nor lost but can only be transformed. This implies that energy input of a system must equate the energy output as no energy can be lost.
In this case regarding an unlit match that contains 1000J of chemical energy. It burns to emit light energy and heat/thermal energy. If the amount of emitted thermal energy measured is 400J and the remaining/left over amount of chemical energy is 250J, this means that the missing amount of energy has been emitted as light energy.
To find the amount of light energy emitted, we say;
Amount of chemical energy - (thermal energy emitted + remaining chemical energy)
= 1000 - (400 + 250)
= 1000 - 650
= 350J
Hence, the amount of light energy emitted is 350J. This satisfies the law of conservation of energy because the output energy is equal to the total energy in the system.
Draw 1,2,3,4,5,6-hexachlorocyclohexane with :
a. all the chloro groups in axial positions.
b. all the chloro groups in equatorial positions.
Answer:
This is required answer.
Explanation:
Given that,
1,2,3,4,5,6-hexachlorocyclohexane
(a). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions
Using given data
We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in axial positions.
When we say that all the chloro groups in axial position that means axial bonds are vertical.
(b). We need to draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions
Using given data
We draw 1,2,3,4,5,6-hexachlorocyclohexane with all the chloro groups in equatorial positions.
When we say that all the chloro groups in equatorial position that means axial bonds are horizontal.
Hence, This is required answer.
Speed = Acceleration x Time
30. If an object accelerates at 5 m/s2 for
4 seconds, how fast will it be going?
a rectangle solid has a length of 3cm a height of 4 centimeters and a width of 5cm what is the solid’s volume?
Answer:
Explanation:
volume = 60cm^3
One liter of a 0.1M Tris buffer (pKa=8.3) is adjusted to a pH of 2.0. a. What are the concentrations of the conjugate base and weak acid at this pH? b. What is the pH when 1.5mL of 3.0M HCl is added to this buffer? Is Tris a good buffer at this pH? Why? c. What is the pH when 1.5mL of 3.0M NaOH is added to this buffer?
Answer:
a. [Weak acid] = 0.099999949M
[Conjugate base = 5.0119x10⁻⁸M
b. pH = 1.84. Is not a good buffer
c. pH = 6.97
Explanation:
a. To find pH of the buffer we must use Henderson-Hasselbalch formula:
pH = pKa + log [Conjugate base] / [Weak acid]
2.0 = 8.3 + log [Conjugate base] / [Weak acid]
5.01x10⁻⁷ = [Conjugate base] / [Weak acid] (1)
As concentration of the buffer is 0.1M = [Conjugate base] + [Weak acid] (2)
Replacing (2) in (1):
5.01x10⁻⁷ = 0.1M - [Weak acid] / [Weak acid]
5.01x10⁻⁷ [Weak acid] = 0.1M - [Weak acid]
[Weak acid] = 0.099999949M
[Conjugate base] = 5.0119x10⁻⁸M
b. The conjugate base reacts with the HCl. Moles of HCl are:
1.5x10⁻³L * (3.0mol / L) = 4.5x10⁻³ moles HCl
As the conjugate base contains just 5x10⁻⁸ moles. Almost all HCl doesn't react and pH =
[H⁺] = 4.5x10⁻³ moles + 1x10⁻² moles (Initial moles H⁺) / L = 0.0145M
pH = -log [H⁺]
pH = 1.84
As the concentration of the conjugate base is <<< than weak acid. In this pH, Tris is not a good buffer. As general rule a good buffer works pH between pKa ± 1.
c. Now, NaOH reacts with the weak acid producing conjugate base.
The new moles are:
[Weak acid] = [Weak acid] = 0.099999949M - 4.5x10⁻³ = 0.0955
[Conjugate base] = 5.0119x10⁻⁸M + 4.5x10⁻³ = 4.5x10⁻³
pH = pKa + log [Conjugate base] / [Weak acid]
pH = 8.3 + log [4.5x10⁻³] / [0.0955]
pH = 6.97
Following are the response to the given points:
For question a)
The concentration of Tris buffer [tex]\text{(Tris buffer)} = 0.1\ M[/tex]
Using formula:
[tex]\text{[ Tris buffer ] = [ Conjugate base ] + [ Acid ]}[/tex]
[tex]0.1 M = \text{ Conjugate base +Acid } \\\\ \text{Acid} = 0.1\ M - \text{ Conjugate base } .............(a)[/tex]
Using the Henderson-Hasselbalch equations
[tex]\to pH = pKa + \log \frac{\text{ Conjugate base }}{ Acid}\\\\\to 2.0 = 8.3 + \log \frac{\text{ Conjugate base }}{ Acid}\\\\\to \log \frac{\text{ Conjugate base }}{ Acid} = 2.0 - 8.3\\\\\to \log \frac{\text{ Conjugate base }}{ Acid} = - 6.3\\\\\to \frac{\text{ Conjugate base }}{ Acid}= 10-6.3\\\\\to \frac{\text{ Conjugate base }}{ Acid} = 5.01 \times 10^{-7}\\\\\to \text{Conjugate base} = 5.01 \times 10^{-7} \times \ Acid ..............(b)[/tex]
Putting the value of equation (b) in equation (a):
[tex]\text{Acid } = 0.1\ M - 5.01 \times 10^{-7}\\\\\text{Acid } = 0.1 M \ ( approx.)[/tex]
from equation (b)
[tex]\to \text{Conjugate base} = 5.01 \times 10^{-7} \times 0.1\ M= 5.01 \times 10^{-8}\ M\\\\[/tex]
Therefore,
[tex]\to \text{Acid} = 0.1\ M\\\\ \to \text{Conjugate base} = 5.01 \times 10^{-8}\ M[/tex]
For question b)
[tex]\to \text{Number of moles of HCl} = 3.0\ M \times 0.00152\ L= 0.0045 \ moles\\\\\to \text{Number of moles Conjugate base in 1 L of buffer} = 5.01 \times 10^{-8}\ M \times 1\ L = 5.01 \times 10^{-8}\ moles\\\\ \to \text{Number of moles of weak acid} = 0.1\ M \times 1\ L = 0.1\ moles\\\\[/tex]
Now we need all the conjugate bases for neutralizing the [tex]HCl[/tex]
therefore
Calculating the molarity of [tex]HCl[/tex] in the solution [tex]= \frac{0.0045\ moles}{1\ L} = 0.0045\ M[/tex]
[tex]\to H^+ \ in \ HCl = 0.0045 \ M\\\\p_H = 2.0\\\\\therefore \\\\ H^+ \ in \ buffer = 10^{-2}\ M = 0.01\ M \ ( \because \ p_H = - \log [ H^+]\ and\ [ H^+ ] = 10-p_H )[/tex]
Now Total concentration of [tex]H^+ \ i.e\ [H^+][/tex]
[tex]\to Total = 0.0045 + 0.01 = 0.0145\ M\\\\\to pH = - \log 0.0145\\\\\to pH = - ( - 1.84 )\\\\\to pH = 1.84\\\\[/tex]
For question c)
Number of moles of [tex]NaOH = 3.0 \ M \times 0.00152\ L= 0.0045\ moles\\\\[/tex]
[tex]\to \text{Number of moles Conjugate base ( say A-) in 1\ L of buffer} = 5.01 \times 10^{-8}\ M \times 1 \ L = 5.01 \times 10^{-8}\ moles \\\\ \to \text{Number of moles of weak acid ( say HA )} = 0.1\ M \times 1\ L = 0.1\ moles[/tex]
Now ICF table is :
[tex]HA + OH^- \longrightarrow A^- + H_2O[/tex]
[tex]\text{I 0.1 mol 0.0045 mol 5.01 x 10-8 mol}\\\\\text{C - 0.0045 mol - 0.0045 mol + 0.0045 mol}\\\\\text{F 0.0955 mol 0 mol 0.0045 mol ( approx. )}\\\\[/tex]
Using Henderson-Hasselbalch:
[tex]\to pH = 8.3 + \log \frac{0.0045}{ 0.0955} \\\\\to pH = 8.3 + \log 0.04712\\\\\to pH = 8.3 - 1.3268\\\\\to pH = 6.97[/tex]
Learn more:
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If u burn paper in the jar, is it the same mass as before u burn it or less or more.
Answer: the mass will be same
Answer:
the mass wouldnt be the same because of the ashes in the jar
Explanation:
EXPLAIN How did energy change form in the calorimeter lab?
As the food burned, thermal chemical nuclear energy was transformed into
thermal chemical | nuclear energy. Thus, a form of kinetic | potential energy was
converted to a form of kinetic potential energy.
Answer:
Chemical energy is converted into thermal and mechanical energy while kinetic energy into potential energy.
Explanation:
As the breakdown of food started, the chemical energy present in the food substances converted into thermal energy and mechanical energy. Thermal or heat energy is used to maintain the body's temperature while mechanical energy is used in other activities such as movement of muscles. Kinetic energy is also converted into potential energy when the object moves from top of a place such as hill.
As the food burned, chemical energy was transformed into thermal energy. Thus, a form of potential energy was converted to a form of kinetic energy.
Hope I helped! :)
What is the partial pressure of water vapor in an air sample when the total pressure is 1.00 atm, the partial pressure of nitrogen is 0.79 atm, the partial pressure of oxygen is 0.20 atm, and the partial pressure of all other gases (except the water) is 0.0044 atm?
Convert 1663.5 g to mg
Answer:
It will be 1663500mg
Explanation:
Ap chem, please help with number 31!!
Answer:
this old man looks so weird ..
Explanation:
but i m sorry .....u know the reason very well
do u wanna be my friend .
The compound F2O has two lone pairs on the central oxygen atom. What's the structure of an F2O molecule? Is this compound polar or non-polar?
Answer:
Bent or Anglar, Polar
Explanation:
Took Test
What is the mass, in grams, of a cube of copper that has a volume of 62.2 cm3?
) Copper density= 8.96 g/cm³ Copper volume= 51.3c.m.3 Copper density = M/V8.96= M/51. M= 459.68g.m.
Write the electron configuration for the following elements. (1) Neon, (2) Zinc, (3) Tungsten, (4) Magnesium
Explanation:
Hey there!!
We generally use s,p,d,f formula.
So, let's do with it.
1.Neon
Atomic no. = 10
So, Its electronic configuration is;
[tex] = 1 {s}^{2} .2 {s}^{2} 2 {p}^{6} [/tex]
2.Zinc.
Atomic no. = 30
So, it's electronic configuration is;
[tex] = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{10} [/tex]
3. Tungsten.
Atomic no. = 74.
So, it's electronic configuration is;
[tex] = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{10}4 {p}^{6} 5 {s}^{2} 4 {d}^{10} 5 {p}^{6} 6 {s}^{2} 4 {f}^{14} 5 {d}^{4} [/tex]
4. Magnesium.
Atomic no. = 12
So, the electronic configuration is ;
[tex] = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6}3 {s}^{2} [/tex]
Hope it helps...
Mr. Hall was conducting an experiment. He dissolved an unknown
ubstance in water. He performed multiple trials while varying the
emperature. What is the independent variable in his experiment?
a. The unknown substance, because it's the only thing he changed
b.The temperature, because it's the only thing he changed
Answer:
a
Explanation:
What happens to these physical properties as the strength of intermolecular forces increases?Increase or decrease?a) melting point - increasesb) vapor pressure-increasesc) boiling point- increasesd) viscosity - decreasese) surface tension - decreases
Boiling Point, Melting Point, Viscosity, Surface Tension. Decrease: Vapor Pressure.
Which of the following is true in regard to a proton?Choose one or more:A. A proton has a 1+ charge.B. A neutral atom must have an equal number of protons and neutrons.C. A proton has a mass of about 1.0 amu.D. A proton has a 1- charge.E. The number of protons determines the identity of an element.
Answer:
1. A PROTON HAS A +1 CHARGE
2. A PROTON HAS A MASS OF ABOUT 1.0 amu
3. THE NUMBER OF PROTONS DETERMINES THE IDENTITY OF AN ELEMENT
Explanation:
The proton is one of the three particles that make up the atom. Proton is a positively charge particle with a +1 charge. It neutralizes the negative charge of the electrons outside of the nucleus of the atom as the proton are located inside the nucleus of the atom. Proton has a mass of 1.0 amu which is about 1.67 * 10^-27 kg. The addition of the proton and the neutrons gives the molar mass of the atom.
Also, the identity of the cell that is the atomic number which gives the position of the element in the periodic table and also shows the individual characteristics of the atoms is determined by the number of protons in the nucleus of the atoms.
All the above options are true in regard to a proton.