The electron dot structure for CI is
A. Cl (w/ a dot above the C)
B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1)
C. Cl (w/ 2 dots above and to the left of the C, 1 underneath, and 2 dots to the right of the 1)
D. Cl (w/ 2 dots above, to the left of, and under the C, as well as 2 dots to the right of the 1)
The electron dot structure for Cl is ... (P.S. The l in Cl is a lowercase "L", for Chlorine, in case anyone else may have been confused.)
The electron dot structure for Cl is B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1).
Explanation:The electron dot structure for Cl is B. Cl (w/ a dot above, to the left of, and under the C, as well as 2 dots to the right of the 1). The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons.
Ydrogen and fluorine combine according to the equation h2(g) + f2(g) → 2 hf(g) if 5.00 g of hydrogen gas are combined with 38.0 g of fluorine gas, the maximum mass of hydrogen fluoride that could be produced is
Methanol (ch3oh) can be made by the reaction of co with h2: co(g)+2h2(g)⇌ch3oh(g) to maximize the equilibrium yield of methanol, would you use a high or low temperature?
Calculate the molarity of each solution. 28.33 g c6h12o6 in 1.28 l of solution
Explanation:
Molarity is the number of moles present in a liter of solution.
Mathematically, Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
And, no. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
Molar mass of [tex]C_{6}H_{12}O_{6}[/tex] is 180.15 g/mol. Therefore, number of moles present will be as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex] = [tex]\frac{28.33 g}{180.15 g/mol}[/tex]
= 0.157 mol
Hence, calculate the molarity as follows.
Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.157 mol}{1.28 L}[/tex]
= 0.122 M
Thus, we can conclude that molarity of the solution is 0.122 M.
What is the final step in the scientific method
A gas sample contains 4.00g of CH4 and 2.00g of He. What is the volume of the sample at STP?
To calculate the volume of the gas sample at standard temperature and pressure (STP), we find the moles of each gas (CH4 and He) using their molar masses and then multiply by the molar volume (22.4 L/mol) to get their individual volumes at STP. Adding these volumes together gives us the total volume of the gas sample, which is 16.8 L.
Explanation:To calculate the volume of a gas sample at standard temperature and pressure (STP), we can use the molar volume concept where one mole of any gas occupies 22.4 liters at STP. Methane (CH4) has a molar mass of 16.00 g/mol, and helium (He) has a molar mass of 4.00 g/mol.
To find the moles of CH4, we divide 4.00 g by its molar mass, 16.00 g/mol, which gives us 0.250 moles. Similarly, for He, we divide 2.00 g by its molar mass, 4.00 g/mol, which gives us 0.500 moles. To find the total volume, we sum the volumes of CH4 and He after multiplying their mole quantities by the molar volume (22.4 L/mol).
Total volume of CH4 = 0.250 moles × 22.4 L/mol = 5.60 L
Total volume of He = 0.500 moles × 22.4 L/mol = 11.2 L
Therefore, the total volume of the gas sample at STP is 5.60 L + 11.2 L = 16.8 L.
The voltage generated by the zinc concentration cell described by the line notation zn(s) ∣∣ zn2+(aq,0.100 m) ∥∥ zn2+(aq,? m) ∣∣ zn(s) is 14.0 mv at 25 °c. calculate the concentration of the zn2+(aq) ion at the cathode.
Answer: The concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M
Explanation:
The half reactions for the cell is:
Oxidation half reaction (anode): [tex]Zn(s)\rightarrow Zn^{2+}(0.100M,aq.)+2e^-[/tex]
Reduction half reaction (cathode): [tex]Zn^{2+}(?M,aq.)+2e^-\rightarrow Zn(s)[/tex]
In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]_{anode}}{[Zn^{2+}]_{cathode}}[/tex]
where,
n = number of electrons in oxidation-reduction reaction = 2
[tex]E_{cell}[/tex] = 14.0 mV = 0.014 V (Conversion factor: 1 V = 1000 mV)
[tex][Zn^{2+}]_{anode}[/tex] = 0.100 M
[tex][Zn^{2+}]_{cathode}[/tex] = ? M
Putting values in above equation, we get:
[tex]0.014=0-\frac{0.0592}{2}\log \frac{0.100M}{[Zn^{2+}]_{cathode}}[/tex]
[tex][Zn^{2+}]_{cathode}=0.295M[/tex]
Hence, the concentration of [tex]Zn^{2+}[/tex] ion at cathode is 0.295 M
The concentration of Zn²⁺ ion at cathode is 0.295 M
Redox half equationsThe redox half equations are those of oxidation and reduction
Oxidation half reaction: Zn(s) ---> Zn²⁺ + 2 e⁻ [Zn⁺] = 0.100 M
reduction half reaction: Zn²⁺ (aq) + 2 e⁻ ----> Zn (s) [Zn⁺] = y
The Ecell = 14.0mV = 0.014 V
Using the Nernst equation:
Ecell = E°cell - 0.0592/n * log([Zn⁺]anode/[Zn⁺]cathode)since it is a concentration cell, E°cell = 0where n is number moles; n = 2
0.014 = 0 - 0.0592/2 * log (0.1/y)
y = 0.295 M
Therefore, the concentration of Zn²⁺ ion at cathode is 0.295 M
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The theoretical yield for the reaction above was 2.78 grams silver (Ag). The experimental yield was 2.55 grams. Calculate the percent yield
To calculate the percent yield of silver, divide the experimental yield (2.55 g) by the theoretical yield (2.78 g) and multiply by 100, resulting in a percent yield of 91.73%.
The concept of percent yield is a key aspect of laboratory work in chemistry that compares what was actually obtained from a reaction to what could be obtained based on stoichiometry. The percent yield is calculated using a simple formula: (actual yield/theoretical yield) imes 100. In the student's experiment where the theoretical yield was 2.78 grams of silver and the experimental yield was 2.55 grams, we calculate the percent yield by dividing 2.55 by 2.78 and then multiplying by 100.
Therefore, the calculation is:
(2.55 g / 2.78 g) imes 100 = 91.73%
So, the percent yield for the reaction is 91.73%.
Classify these solids as molecular, ionic, or atomic. co2 c cacl2 c6h12o6 kbr pbs
Explanation:
Molecular compounds : Molecular compounds are formed by the sharing of electrons with each elements.
Ionic compound : When compounds are formed by the transfer of electrons, they are called Ionic compounds.
Atomic species : Species comprised of only atom.
Co2 -- Molecular
C -- Atomic
CaCl2 -- Ionic
C6H12O6 --- Molecular
KBr --- Ionic
PbS ---- Ionic
The above compounds can be classified as;
CO₂- Molecular compound
C - Atomic compound
CaCl₂ - ionic compound
C₆H₁₂O₆ - Molecular compound
KBr - Ionic compound
PbS - Ionic compound
Further Explanation:A compound A compound is a substances that contains two or more different atoms that are bonded together.When the atoms are similar the substance is known as a molecule, therefore not all molecules are compounds.Types of compounds Ionic compounds Ionic compounds are compounds that contain ions. They are as a result of ionic bonding between a metal atom and a non-metal atom.During ionic bonding formation the metallic atom looses electrons while the non-metallic atom gains electrons.Ionic compounds contains both negatively charged ion(anion) and positively charged ion (cation).Examples of ionic compounds, NaCl, LiF, KCl, Cs2S, KBr, PbS etc.
Molecular compounds
Molecular compounds are covalent compounds that are formed when atoms of elements share electrons in a covalent bond to form molecules.Molecular covalent compounds are formed between non-metal atoms as a result of covalent bond between non-metal atoms.These compounds are electrically neutral.Examples of molecular compounds include; CO₂, H₂O, NO, SF₆, etc.Atomic covalent compounds
Atomic compounds are also an example of covalent compounds that result from sharing of electrons between non-metal atoms in a covalent bond.Usually atoms of the same element such as carbon are joined together by a covalent bond.Examples of such compounds are, diamond and graphite.Keywords: Compound, types of compounds, ionic compounds, molecular compounds, atomic compounds
Learn more about:
A compound: https://brainly.com/question/2272966Types of compounds: https://brainly.com/question/861309Ionic compound: https://brainly.com/question/2450458Examples of ionic compounds: https://brainly.com/question/2450458Molecular compound; https://brainly.com/question/8129823Level : High school
Subject: Chemistry
Topic: Structure and bonding
Sub-topic: Types of structures and compounds
According to the valence bond theory the triple bond in ethyne consists of
When water reaches its boiling point and turns into water vapor what happens to the molecules structure
What is the emf of a cell consisting of a pb2+ / pb half-cell and a pt / h+ / h2 half-cell if [pb2+] = 0.83 m, [h+] = 0.064 m and ph2 = 1.0 atm ?
The emf of the electrochemical cell has been calculated to be 0.0532 V.
The emf has been the potential of the cell in the reaction with the change in the electrons in the reaction. The emf of the cell has been given by the Nernst equation as;
[tex]emf=E^\circ _{cell}-\dfrac{0.059}{n}\;log\;\dfrac{1}{\rm concentration} [/tex]
Computation for the emf of the cellThe given cell has the number of electrons transfer, [tex]n=2[/tex]
The concentration of [tex]\rm Pb^2^+=0.83\;M[/tex]
The concentration of [tex]\rm H^+=0.064\;M[/tex]
The cell potential of the reaction has been, [tex]E^\circ =-0.126\;\text V[/tex]
Substituting the values for the emf of the cell:
[tex]emf=-0.126\;-\dfrac{0.059}{2}\;\times\;log\;\dfrac{1}{[0.83]\;\times\;[0.064]^2}\\ emf=0.0532\;\text V [/tex]
The emf of the electrochemical cell has been calculated to be 0.0532 V.
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Carbon tetrachloride will have which shape?
A) bent
B) linear
C) tetrahedral
D) triangular
What is the ph of a 0.0055 m ha (weak acid) solution that is 8.2% ionized?
Identify the lowest energy lewis structure for nitrogen oxide
What is the effect of removing some so3 from a system initially at equilibrium?
Answer:
Right
Left
Left
Right
Explanation:
For the equilibrium system described by this equation, what will happen if SO3 is removed?
The equilibrium shifts to the
✔ right
What will happen if NO is added?
The equilibrium shifts to the
✔ left
.For the equilibrium system described by this equation, what will happen if SO2 is removed?
The equilibrium shifts to the
✔ left
.
What will happen if NO2 is added?
The equilibrium shifts to the
✔ right
Question 1 a sample of 0.255 mole of gas has a volume of 748 ml at 28°c. calculate the pressure of this gas. (r= 0.0821 l ∙ atm / mol ∙ k) 0.784 atm 8.42 atm 0.00842 atm 7.84 × 10-4 atm none of the above
Answer : The pressure of the gas is, 8.42 atm
Explanation :
Using ideal gas equation,
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = ?
V = volume of the gas = 748 ml = 0.748 L
conversion used : (1 L = 1000 ml)
T = temperature of the gas = [tex]28^oC+273+28=301K[/tex]
n = number of moles of the gas = 0.255 mole
R = gas constant = 0.0821 L.atm/mole.K
Now put all the given values in the ideal gas equation, we get the pressure of the gas.
[tex]P\times (0.748L)=0.255mole\times (0.0821L.atm/mole.K)\times (301K)[/tex]
[tex]P=8.42atm[/tex]
Therefore, the pressure of the gas is, 8.42 atm
Determine the amount of heat(in Joules) needed to boil 5.25 grams of ice. (Assume standard conditions - the ice exists at zero degrees Celsius, melts at zero degrees Celsius, and boils at 100 degrees Celsius. Remember that you need to take into account three changes: melting ice, heating water, and vaporizing the water.)
The amount of heat needed to boil 5.25 grams of ice when we take into account three changes as melting, heating and vaporizing the water is 15797.93 J.
How do we calculate total heat?Total heat for the given condition will be calculated by the addition of the heat of fusion, heat of vaporization and specific heat of water.
In the question it is given that,
mass of ice = 5.25 grams
Change in temperature = 100 - 0 = 100 degree celsius
For the melting of ice:We know that heat of fusion of water = 334 J/g
Required heat for the melting of ice = 334 × 5.25 = 1753.5 J
For heating water:Amount of heat will be used by using the formula as,
Q = mcΔT, where
c = specific heat of water = 4.18 J/gK
Required heat for heating of water = 4.18 × 5.25 × 100 = 2195.18 J
For vaporization of water:We know that heat of vaporization of water = 2257 J/g
Required heat for vaporization of ice = 2257 × 5.25 = 11849.25 J
Total amount of heat involved = 1753.5 + 2195.18 + 11849.25 = 15797.93 J
Hence total amount of heat is 15797.93 J.
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Calculate the molarity of a solution that contains 0.175 mol of zncl2 n exactly 150 ml of solution.
Explanation:
Molarity is the number of moles present in liter of a solution.
Mathematically, Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
It is given that no. of moles present into the solution are 0.175 mol and volume is 150 ml.
As 1 ml = 0.001 L. So, 150 ml will be equal to 0.15 L.
Hence, calculate the molarity as follows.
Molarity = [tex]\frac{\text{no. of moles}}{\text{volume in liter}}[/tex]
= [tex]\frac{0.175 mol}{0.15 L}[/tex]
= 1.16 M
Thus, we can conclude that molarity of the given solution is 1.16 M.
Give the oxidation state of the metal species in each complex. ru(cn)(co)4 -
The oxidation state of the Ru metal in the complex [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex] is [tex]\boxed{{\text{zero}}}[/tex] .
Further explanation:
Oxidation number:
The oxidation number is used to represent the formal charge on an atom. It also shows that gain or loss of electrons by the atom. Oxidation number can be a positive or negative number but cannot be fractional.
The rules to identify the oxidation state:
(1) The oxidation state of an atom in the elemental form is zero.
(2) The total charge on the species is equal to the sum of the oxidation state of individual atoms.
(3) The oxidation state of halides is -1. For example, fluorine, chlorine, bromine, iodine have -1 oxidation state.
(4)Hydrogen has a +1 oxidation state.
(5) Oxygen has an oxidation state -2.
(6) In a coordination compound, neutral ligands have zero oxidation state and negative ligands such as CN have -1 oxidation.
The given compound is [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex]
Here, CN is a negative ligand thus oxidation state is -1 and CO is a neutral ligand thus it has 0 oxidation state. Also, the complex has -1 negative charge.
The expression to calculate the oxidation state in [tex]{\left[{{\text{Ru}}\left({{\text{CN}}}\right){{\left({{\text{CO}}}\right)}_4}}\right]^-}[/tex] is,
[tex]\left[{\left({{\text{oxidation state of Ru}}}\right)+\left({{\text{oxidation state of CN}}}\right)+4\left({{\text{oxidation state of CO}}}\right)}\right]=-1[/tex]
…… (1)
Rearrange equation (1) for the oxidation state of Ru.
[tex]{\text{Oxidation state of Ru}}=\left[{-\left({{\text{oxidation state of CN}}}\right)-4\left({{\text{oxidation state of CO}}}\right)-1}\right][/tex]
…… (2)
Substitute -1for the oxidation [tex]{\text{state}}[/tex] of CN and 0 for the oxidation state of CO in equation (2).
[tex]\begin{aligned}{\text{Oxidation state of Ru}}&=\left[{-\left({-{\text{1}}}\right)-4\left({\text{0}}\right)-1}\right]\\&=0\\\end{aligned}[/tex]
The oxidation state of Ru is zero.
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1. General statement that is not applied on metals: https://brainly.com/question/2474874
2. The neutral element represented by the excited state electronic configuration: https://brainly.com/question/9616334
Answer details:
Grade: Senior school
Subject: Chemistry
Chapter: Coordination complex
Keywords: Oxidation state, metal complex, ru(cn)(co)4-, formal charge, cynide, carbonyl, zero, hydrogen, oxygen, 0 and -1.
How many structural and geometrical isomers are there of chloropropene?
What is the mass occupied by 44.8 l of nitrogen (n2) at standard conditions?
A weak acid is a dilute acid that is not very powerful
a. True
b. False
If 4.27 g sucrose (c12h22o11) are dissolved in 15.2 g water, what is the boiling point of the resulting solution? kb for water = 0.512c/m.
Answer : The boiling point of a solution is, [tex]100.42^oC[/tex]
Explanation :
Formula used for Elevation in boiling point :
[tex]\Delta T_b=k_b\times m[/tex]
or,
[tex]T_b-T^o_b=\frac{1000\times k_b\times w_2}{w_1\times M_2}[/tex]
where,
[tex]T_b[/tex] = boiling point of solution = ?
[tex]T^o_b[/tex] = boiling point of pure water = [tex]100^oC[/tex]
[tex]k_b[/tex] = boiling point constant for water = [tex]0.512^oC/m[/tex]
m = molality
[tex]w_2[/tex] = mass of solute (sucrose) = 4.27 g
[tex]w_1[/tex] = mass of solvent (water) = 15.2 g
[tex]M_2[/tex] = molar mass of solute (sucrose) = 342.3 g/mole
Now put all the given values in the above formula, we get the boiling point of a solution.
[tex]T_b-100^oC=\frac{1000\times 0.512^oC/m\times 4.27g}{15.2g\times 342.3g/mole}[/tex]
[tex]T_b=100.42^oC[/tex]
Therefore, the boiling point of a solution is, [tex]100.42^oC[/tex]
The temperature of the solution is 100.42°C.
Given that;
ΔT = K m i
Where;
ΔT = boiling point elevation
K = boiling point constant for water
m = molality of the solution
i = Van't Hoff factor
Number of moles of solute = 4.27 g /342 g/mol = 0.0125 moles
Molality of the solution = 0.0125 moles/15.2 × 10^-3 Kg = 0.822 m
Since the boiling point of pure water = 100°C
Let the boiling point of pure water be Ta
Let the boiling point of the solution be Tb
ΔT = Tb - Ta = Tb - 100
Substituting values;
Tb - 100 = 0.512c/m × 0.822 m × 1
Note that the Van't Hoff factor (i) = 1 because the solute is molecular
Tb = [0.512°C/m × 0.822 m × 1] + 100°C
Tb = 100.42°C
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Vanillin, c8h8o3 (m = 152 g/mol), is the molecule responsible for the vanilla flavor in food. how many oxygen atoms are present in a 45.0 mg sample of vanillin?
In a 45.0 mg sample of vanillin, there are approximately 5.36 x 10^20 oxygen atoms. The calculation involves converting the mass of the sample to grams, calculating the number of moles, and using Avogadro's number to determine the number of oxygen atoms.
Explanation:The question is asking how many oxygen atoms are present in a 45.0 mg sample of vanillin, which is a molecule responsible for the vanilla flavor in food. The molecular formula of vanillin is C8H8O3 and its molar mass is 152 g/mol.
First, we have to convert the mass of the sample from milligrams (mg) to grams (g) because the molar mass is in grams. Therefore, 45.0 mg equals 0.045 g.
Next, we calculate the number of moles of vanillin in the sample using the equation:
Number of moles = Mass / Molar mass
Substituting the given values:
Number of moles = 0.045 g / 152 g/mol = 2.96 x 10^-4 moles.
Each molecule of vanillin contains 3 oxygen atoms. Therefore, in one mole of vanillin, there are 3 moles of oxygen atoms. From the concept of Avogadro's number, we know that one mole of any substance contains 6.02 x 10^23 entities (atoms, molecules, ions etc.).
Therefore, in 2.96 x 10^-4 moles of vanillin, we have (3 moles O atoms / mole of vanillin) x (2.96 x 10^-4 moles of vanillin) x (6.02 x 10^23 O atoms / mole of O atoms) = 5.36 x 10^20 oxygen atoms.
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The presence of which magnetic feature best explains why a magnet can act at a distance on other magnets or on objects containing certain metals?
Answer:
Magnetic fields.
Explanation:
Hello,
The presence of a magnetic fields best explains why a permanent magnet can act on another magnetic objects when are separated by a certain distance. There exist magnetic field lines which are emanated from the north pole to the south pole whose path is curved. Now, take into account that the longer the distance between the objects, the lower the magnetic field or force.
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AlCl3 + Na NaCl + Al Did Cl change oxidation number?
Answer : There is no change takes place in the oxidation number of Cl.
Explanation :
The given chemical reaction is,
[tex]AlCl_3+Na\rightarrow NaCl+Al[/tex]
This reaction is an unbalanced reaction because the chlorine atoms are not balanced.
In order to balance the chemical reaction, the coefficient 3 is put before the Na and NaCl.
The balanced chemical reaction will be,
[tex]AlCl_3+3Na\rightarrow 3NaCl+Al[/tex]
Now we have to calculate the oxidation number of all the elements.
In [tex]AlCl_3[/tex], the oxidation number of Al and Cl are, (+3) and (-1) respectively.
The oxidation number Na and Al are, zero (0)
In [tex]NaCl[/tex], the oxidation number of Na and Cl are, (+1) and (-1) respectively.
From this we conclude that the oxidation number of Cl changes from (-1) to (-1) that means remains same.
Therefore, there is no change takes place in the oxidation number of Cl.
Magnesium hydroxide (Mg(OH)2): g/mol Iron(III) oxide (Fe2O3): g/mol
Answer:
Magnesium hydroxide (Mg(OH)2): 58.33 g/mol
Iron(III) oxide (Fe2O3): 159.70 g/mol
Explanation:
To find the molar masses of both magnesium hydroxide and iron(III) oxide, add each molar mass in each compound.
Solving:
[tex]\section*{Molar Mass of Magnesium Hydroxide (Mg(OH)_2):}\textbf{Identify Atomic Masses}\begin{itemize} \item Magnesium (Mg): 24.31 g/mol \item Oxygen (O): 16.00 g/mol (2 oxygen atoms) \item Hydrogen (H): 1.01 g/mol (2 hydrogen atoms)\end{itemize}[/tex]
[tex]\textbf{Calculate Molar Mass:}\[\text{Molar mass of Mg(OH)}_2 = \text{Mg} + 2 \times (\text{O} + \text{H})\]\[= 24.31 \, \text{g/mol} + 2 \times (16.00 \, \text{g/mol} + 1.01 \, \text{g/mol})\]\\\[= 24.31 \, \text{g/mol} + 2 \times 17.01 \, \text{g/mol}\]\[= 24.31 \, \text{g/mol} + 34.02 \, \text{g/mol} = \boxed{58.33 \, \text{g/mol}}\][/tex]
[tex]\hrulefill[/tex]
[tex]\section*{Molar Mass of Iron(III) Oxide (Fe_2\text{O}_3):}\textbf{Identify Atomic Masses}\begin{itemize} \item Iron (Fe): 55.85 g/mol (2 iron atoms) \item Oxygen (O): 16.00 g/mol (3 oxygen atoms)\end{itemize}[/tex]
[tex]\textbf{Calculate Molar Mass:}\[\text{Molar mass of Fe}_2\text{O}_3 = 2 \times \text{Fe} + 3 \times \text{O}\]\\\[= 2 \times 55.85 \, \text{g/mol} + 3 \times 16.00 \, \text{g/mol}\]\\\[= 111.70 \, \text{g/mol} + 48.00 \, \text{g/mol} = \boxed{159.70 \, \text{g/mol}}\][/tex]
[tex]\hrulefill[/tex]
In a lead-acid storage battery, pbo2 is reduced to
Click to select the correct answers. Click again to unselect answers. Leave the incorrect answers unselected.
Nuclear fission and fusion both affect the nucleus of an atom. Choose all of the items below that are correct.
The final products of fission and fusion are elements that are different than the original
Fusion involves the splitting apart of large atoms into smaller ones
Fission occurs mostly with elements heavier than lead on the periodic table
Two hydrogen atoms can go through fission to form helium
Fusion and fission involve a loss or gain of electrons
The answer options about a nuclear reaction which are correct include the following:
A. Nuclear fission and fusion both affect the nucleus of an atom.
B. The final products of fission and fusion are elements that are different than the original.
D. Fission occurs mostly with elements heavier than lead on the periodic table.
A nuclear reaction refers to a reaction in which the nucleus of an atom is transformed or transmuted by either joining (fusion) or splitting (fission) the nucleus of another atom of a radioactive element. Also, a nuclear reaction is always accompanied by a release of energy.
Generally, the two (2) main types of nuclear reaction are:
Nuclear fusion: it involves the joining of two smaller nuclei of atoms to form a single massive or heavier nucleus with the release of energy. Nuclear fission: it involves the collision of a heavy atomic nucleus with a neutron, thereby causing a split and release of energy.From the above, we can deduce the following about a nuclear reaction:
I. The nucleus of an atom is affected by both nuclear fission and fusion.
II. The final products of both nuclear fission and fusion are chemical elements that are different than the original radioactive chemical elements.
III. Nuclear fission occurs mostly with chemical elements heavier than lead (Pb) on the periodic table because the nuclear force holding their atoms together is lesser than the electromagnetic force pushing the nucleus apart.
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Fill in the missing blank 2C4H6 + _______ → 8CO2 + 6H2O