Answer:
1. 17.5 g of CO₂
2. The limiting reactant is carbon (graphite), and its formula is C(graphite)
3. 3.7 g of O₂
Explanation:
First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:
Reactants: carbon(graphite) ⇒ C(graphite) ; oxygen gas ⇒ O₂(g)Products: carbon dioxide ⇒ CO₂(g)Thus, we write the chemical equation:
C(graphite) + O₂(g) → CO₂(g)
The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).
Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:
Mw(C) = 12 g/mol
moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol
Mw(O₂) = 16 g/mol x 2 = 32 g/mol
moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol
Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):
stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂
actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂
We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).
The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).
Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:
moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂
Now, we convert the moles of CO₂ to mass by using the Mw:
Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g
Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.
Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:
remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂
Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :
mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g
Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.
What is the concentration of Htions at a pH = 11?
mol/L
What is the concentration of Htions at a pH = 6?
mol/L
How many fewer Htions are there in a solution at a
pH = 11 than in a solution at a pH = 6?
1) Recall the two written definitions of an oxidation-reduction reaction provided in our lessons. Which of these definitions is
most inclusive of redox reactions? Explain your answer:
A redox reaction is where the oxidation and reduction reaction takes place at the same time, the oxidation half
reaction involves losing electrons and in the reduction half reaction involves gaining electrons. So in a redox
reaction an electron is lost by the reducing agent.
Explain how the reaction below meets these definitions. Which substance is being oxidized and which is
being reduced?
4Ag(s) + 2H2S(g) + O2(g)
2Ag2S(s) + 2H20(9)
Answer:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
Silver atoms are oxidized while oxygen atoms are reduced by a loss of electrons and a gain of electrons respectively.
Explanation:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
In a redox reaction,the two reactions occurring simultaneously can be divided into two half reactions; an oxidation half-reaction and a reduction half-reaction.
The oxidation half-reaction involves losing electrons and thus an increase in oxidation number of the species being oxidized. Whereas, the reduction half reaction involves gaining electrons and thus, a reduction innthe oxidation number of the species being reduced.
The species which oxidizes another species is known as an oxidizing agent and isnitself reduced due to its accepting electrons from the species being oxidized. Th reducing agent reduces another species and is itself oxidized as it loses electrons to the oxidized agent.
In the given reaction as shown below:
4 Ag (s) + 2 H₂S (g) + O₂ (g) ---> 2 Ag₂S (s) + 2 H₂0 (g)
The reaction is a redox reaction as a change innthe oxidation number of the reacting species; both oxidation and reduction occurs simultaneously and to the same extent.
The metallic silver atoms, have an oxidation number of zero initially. However, each of the four moles of atoms give up one mole of a electrons each to become oxidized to silver (i) ions, Ag+.
On the other hand, molecular oxygen gas also having oxidation number of zero becomes reduced to oxygen ion, O²-. Each of the two moles of atom in the oxygen gas molecule accept two electrons each donated by the metallic silver atoms to become reduced to oxygen ion, O²-.
The oxidation numbers of hydrogen ion and sulfide ion do not change.
9. Consider a magnesium atom with charge +2. How many overall electrons are on this particle?
Hint: Magnesium's atomic number is 12.
10
12
14
using the balanced equation below how many grams of lead(||) sulfate would be produced from the complete reaction of 23.6 g lead (|V) oxide
Answer:
59.8 g of PbSO₄.
Explanation:
The balanced equation for the reaction is given below:
Pb + PbO₂ + 2H₂SO₄ —> 2PbSO₄ + 2H₂O
Next, we shall determine the mass of PbO₂ that reacted and the mass of PbSO₄ produced from the balanced equation. This can be obtained as follow:
Molar mass of PbO₂ = 207 + (16×2)
= 207 + 32
= 239 g/mol
Mass of PbO₂ from the balanced equation = 1 × 239 = 239 g
Molar mass of PbSO₄ = 207 + 32 + (16×4)
= 207 + 32 + 64
= 303 g/mol
Mass of PbSO₄ from the balanced equation = 2 × 303 = 606 g
SUMMARY:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Finally, we shall determine the mass of PbSO₄ that will be produced by the reaction of 23.6 g of PbO₂. This can be obtained as follow:
From the balanced equation above,
239 g of PbO₂ reacted to produce 606 g of PbSO₄.
Therefore, 23.6 g of PbO₂ will react to produce = (23.6 × 606) / 239 = 59.8 g of PbSO₄.
Thus, 59.8 g of PbSO₄ were obtained from the reaction.
Analyze the data and determine the actual concentration of calcium chloride in the solution. Show all calculations and report in % wt/v concentration.
Known; Mass of CaCl2 present in original solution, based on actual yield= 1.77g moles
CaCl2 present in original solution, based on actual yield= 1.77g/molar mass of CaCl2=1.77g/110.98g/mol=0.016 moles
Total Volume of solution =V, which is 80ml
Answer:
2.21% wt/v
Explanation:
The mass/volume percentage, %wt/v, is an unit of concentration used in chemistry defined as 100 times the ratio of the mass of solute in g (In this case, CaCl2 = 1.77g) and the volume of solution in mL = 80mL
The %wt/v of this solution is:
%wt /v = 1.77g / 80mL * 100
%wt/v = 2.21% wt/v
A student was asked to determine the percent mass of sodium nitrate in a mixture of sodium nitrate (NaNO3) and calcium carbonate (CaCO3). The mass of the mixture used was 3.2 g. The student extracted NaNO3 from the mixture with water and separated the insoluble CaCO3 from the solution by filtration. After evaporating the filtrate, the student recovered and dried the NaNO3, and found that it weighted 0.45 g. The student dried the insoluble residue of CaCO3 and found that it weighted 2.23 g. Calculate the percent mass of NaNO3 in the mixture and round your final answer to the correct number of significant figures.
Answer:
自分の仕事をする translate to english
Explanation:
Sofia orders a spare part for her custom-built bike from Oregon Technologies Inc. The company makes use of a computer-aided design model to produce the spare part at its location closest to Sofia's home. In this case, which of the following technologies is used to produce the spare part?
a. Molding
b. Additive manufacturing
c. Lenticular printing
d. Tampography
Answer:
b. Additive manufacturing
Explanation:
Additive manufacturing is defined as that manufacturing process where light parts and components are being developed or manufactured in 3D form by adding materials to it.
It is a process of adding materials to produce the final product. It is also known as 3D printing.
In the context, Oregon Technologies Inc. uses computer-aided design model in order to manufacture a spare part required by Sofia for her custom made bike by using a process called additive manufacturing.
Thus the correct option is (b).
The reaction A + B <-------> C + D has been studied at five widely different temperature and the equilibrium tabulated.
Equilibrium constant K (at varies temperatures)
K at T1 1 x 10^-2
K at T2 2.25
K at T3 1.0
K at T4 81
K at T5 4 x 10^1
Which temperature is the products favored?
If K is greater than 1, then products are favored
All of the following are TRUE for activities and activity coefficients, except: A) activity for a chemical species is the product of concentration and activity coefficient. B) the activity coefficient corrects for non-ideal behavior due to ionic strength. C) as ionic strength increases, the value of the activity coefficient increases. D) for ions, the activity coefficient approaches unity as the ionic strength approaches 0. E) the activity coefficient for neutral molecules is approximately unity.
Answer:
C) as ionic strength increases, the value of the activity coefficient increases.
Explanation:
The effective concentration of ions available for reactions is known as the activity of the ion.
The activity coefficient important in chemistry because it accounts for the deviation of a solution from ideal behaviour.
The activity of a chemical species is defines as the product of concentration and activity coefficient.
Following the Debye–Hückel limiting law; log γ = −0.509z2I1/2. The ionic strength of a solution tends to increase as the activity coefficient (γ) of the ion decreases.
A laboratory utilizes a mixture of 10% dimethyl sulfoxide (DMSO) in the freezing and long-term storage of embryonic stem cells. If DMSO has a specific gravity of 1.1004, calculate the specific gravity, to four decimal places, of the mixture (assume water to be the 90% portion).
Answer:
The correct answer is "1.0100".
Explanation:
Let the volume of mixture be 100 ml.
then,
The volume of DMSO will be 10 mL as well as that of water will be 90 mL.
DMSO will be:
= [tex]10\times 1.1004[/tex]
= [tex]11.004 \ g[/tex]
The total mass of mixture will be:
= [tex]90+11.004[/tex]
= [tex]101.004 \ g[/tex]
Density of mixture will be:
= [tex]\frac{Mass}{Volume}[/tex]
= [tex]\frac{101.004}{100}[/tex]
= [tex]1.01004 \ g/mL[/tex]
hence,
Specific gravity of mixture will be:
= [tex]\frac{Density \ of \ mixture}{Density \ of \ water}[/tex]
= [tex]\frac{1.01004}{1}[/tex]
= [tex]1.0100[/tex]
Using the molarity of vinegar, calculate the mass percent of acetic acid in the original sample. Assume the density of vinegar is 1.00 g/mL. (The formula for acetic acid is C2H4O2).
Answer:
5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/L
Explanation:
Assuming the molarity of vinegar is 0.8935mol/L:
Mass percent is defined as 100 times the ratio between mass of solute (In this case, acetic acid), and the mass of the solution
To solve this question we need to find the mass of acetic acid from the moles using the molar mass and the mass of the solution from the volume in liters using the density:
Mass Acetic acid -Molar mass: 60.052g/mol-
0.8935mol * (60.052g / mol) = 53.656g Acetic Acid
Mass Solution:
1L = 1000mL * (1.00g/mL) = 1000g Solution
Mass Percent:
53.656g Acetic Acid / 1000g Solution * 100 =
5.37% w/w is the mass percent of vinegar assuming a molarity of 0.8935mol/LThe mass percent of acetic acid in the original sample of vinegar of molarity 0.8935mol/L is 5.37% w/w.
How do we calculate the mass percent?Mass percent of any solute present in any solution will be calculated as the:
Mass % of solute = (mass of solute / mass of solution) × 100
Let the molarity of vinegar = 0.8935mol/L
Means 0.8935 moles of vinegar present in the 1 liter of the solution.
Now we calculate mass from moles as:
n = W/M, where
W = required mass
M = molar mass = 60.052g /mol
W = (0.8935mol)(60.052g/mol) = 53.656g
Mass of solution = 1L = 1000mL×(1.00g/mL) = 1000g Solution
Then the mass % of acetic acid:
Mass % = (53.656g / 1000g) × 100 = 5.37% w/w
Hence the required % mass is 5.37% w/w.
To know more about mass percent, visit the below link:
https://brainly.com/question/26150306
A substance that donates a pair of electrons to form coordinate covalent bond is called
Lewis base: any species that can donate a pair of electrons and form a coordinate covalent bond. ligand: molecule or ion that surrounds a transition metal and forms a complex ion; ligands act as Lewis bases
Standard hydrogen electrode acts as both anode and cathode.Explain.
Answer:
A Standard Hydrogen Electrode is an electrode that scientists use for reference on all half-cell potential reactions. The value of the standard electrode potential is zero, which forms the basis one needs to calculate cell potentials using different electrodes or different concentrations. It is important to have this common reference electrode, just as it is important for the International Bureau of Weights and Measures to keep a sealed piece of metal that is used to reference the S.I. Kilogram.
Answer:
The role of an electrode as cathode or anode depends on the nature and electrode potential of the other electrode with which it forms the complete electrochemical cell.
When a cell is to be made with zinc electrode and hydrogen electrode, the hydrogen electrode will behave as a cathode and the zinc electrode will behave as anode because zinc is present above hydrogen in the activity series. That is zinc is more electropositive than hydrogen.
If the cell is made with a copper electrode and hydrogen electrode, the hydrogen electrode will behave as anode and the copper electrode as a cathode. This is due to the fact that Cooper is present below hydrogen in the activity series. Copper is less electropositive than hydrogen.
Explanation:
Which of the following reasons correctly explains the color changes that take place when ethylenediamine (C2N2H8) is added to the solution of cobalt(II) chloride?
a. Addition of the liquid ethylenediamine dilutes the concentration of cobalt(II) chloride in the solution resulting in a color change.
b. The ethylenediamine is oxidized and the resulting solution is deeply colored.
c. The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.
Answer:
The water ligands surrounding the cobalt metal center are being replaced by ethylenediamine and chloride ligands which results in a different crystal field splitting. Thus, the energy associated with electron transitions between the do-orbitals will differ for the two compounds showing a color change.
Explanation:
The five d-orbitals are usually degenerate. Upon approach of a ligand, the d-orbitals split into two sets of orbitals depending in the nature of the crystal field.
The magnitude of crystal field splitting is affected by the nature of the ligand. Ligands having filled p-π orbitals such as ethylenediamine lead to greater crystal field splitting.
The change in the colour that takes place when ethylenediamine is added to the solution of cobalt(II) chloride occurs due to a different crystal field splitting pattern. Thus, the energy associated with electron transitions between the d-orbitals now differ for the two compounds showing a color change.
Nucleophilic aromatic substitution involves the formation of a resonance-stabilized carbanion intermediate called a Meisenheimer complex as the nucleophile attacks the ring carbon carrying the eventual leaving group.
a. True
b. False
Answer:
True
Explanation:
Aromatic rings undergo nucleophillic substitution reactions in the presence of a electron withdrawing group which stabilizes the Meisenheimer complex.
When the nucleophile attacks the ring carbon atom carrying the eventual leaving group. A resonance-stabilized carbanion intermediate called a Meisenheimer complex is formed.
Subsequent loss of the leaving group from the intermediate complex yields the product of the reaction.
For which of the following reactions is the enthalpy change equal to the second ionization energy of nitrogen?
Answer:
"[tex]N^+(g) \rightarrow N^{2+}(g) + e^-[/tex]" is the appropriate answer.
Explanation:
Whenever one electron or particle must be removed from some kind of gas atom or molecule, it requires that the very first amount of energy necessary.Two electrons must be removed from such a mono-positive exhaust gases structure or position of ion before they may become a dipositive gaseous ion.Thus the above is the correct answer.
(in the image)
What Is the noble gas configuration for strontium (atomic number = 38)? Given: He (atomic number = 2) Ne (atomic number = 10) Ar (atomic number = 18) Kr (atomic number = 36) B В A [Kr 45Kr 5
Answer:
[Kr] 5s²
Explanation:
From the question given above, the following data were obtained:
Atomic number of strontium (Sr) = 38
Electronic configuration =?
Next, we shall determine the electronic configuration of the noble gas element before strontium (Sr).
The noble gas element before strontium (Sr) is krypton (Kr). Thus, the electronic configuration of krypton (Kr) is given below:
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Finally, we shall determine the electronic configuration of strontium (Sr). This can be obtained as follow:
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Sr (38) =>?
Sr (38) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶ 5s²
But
Kr (36) => 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s²4p⁶
Therefore,
Sr (38) => [Kr] 5s²
How many moles of HCl are contained in 0.600 L of 0.120 M HCl?
Please explain and show work.
We know
[tex]\boxed{\Large{\sf Molarity=\dfrac{No\:of\:moles\:of\:solute}{Volume\:of\:solution\:in\;\ell}}}[/tex]
[tex]\\ \Large\sf\longmapsto No\:of\:moles\:of\:HCl=0.6\times 0.12[/tex]
[tex]\\ \Large\sf\longmapsto No\:of\:moles\:of\:HCl=0.072mol[/tex]
Answer:
0.0.72
Explanation:
moles = V*CM=0.6*0.12=0.0.72
A chlorine (CI) atom has 7 valence electrons. Which of the following would be the most likely way for a chlorine atom to become stable?
A. Lose 5 electrons
B. Gain 2 electrons
C. Gain 1 electron
D. Lose 7 electrons
Answer:
Option C. Gain 1 electron
Explanation:
Valence electron(s) are the electron(s) located on the outermost shell of an atom. Valency is simply defined as the combining power of an atom.
Chlorine (Cl) atom has 7 valence electron. This implies that Cl needs just one electron to complete it's octet configuration. It will be difficult for Cl to lose any of it's valence electron(s). Cl can either gain or share 1 electron to become stable.
Thus, considering the options given in the question above, option C gives the correct answer to the question.
Consider the following chemical reaction:
2SO2 (g) + O2 (g) -----------> 2SO3 (g)
1.50 L. of sulfur trioxide at the pressure of 1.20 atm. and temperature of 25 oC is mixed with excess of oxygen.
Calclate volume of the product in L. at STP.
A. 11.2 L.
B. 1.65 L.
C. 16.5 L.
D. 0.129 L.
Answer:
B. 1.65 L
Explanation:
Step 1: Write the balanced equation
2 SO₂(g) + O₂(g) ⇒ 2 SO₃(g)
Step 2: Calculate the moles of SO₂
The pressure of the gas is 1.20 atm and the temperature 25 °C (298 K). We can calculate the moles using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.20 atm × 1.50 L / (0.0821 atm.L/mol.K) × 298 K = 0.0736 mol
Step 3: Calculate the moles of SO₃ produced
0.0736 mol SO₂ × 2 mol SO₃/2 mol SO₂ = 0.0736 mol SO₃
Step 4: Calculate the volume occupied by 0.0736 moles of SO₃ at STP
At STP, 1 mole of an ideal gas occupies 22.4 L.
0.0736 mol × 22.4 L/1 mol = 1.65 L
We know that water is purified before it is supplied to our houses. Then why do we have filters installed in our houses? What do they serve?
Answer:
Water filters remove elements that cause drinking water to have an unpleasant taste and smell, such as lead, chlorine and bacteria. Home water filtration system will improve the overall purity, taste and smell of your drinking water. It also lowers the pH level of the water that you drink.
What is the hydrogen atoms in 39.6g of ammonium sulphate,NH4 2SO4
Propane gas reacts with oxygen according to this balanced equation: C subscript 3 H subscript 8 space (g )space plus space 5 space O subscript 2 space (g )space rightwards arrow 3 space C O subscript 2 space (g )space plus space 4 space H subscript 2 O space (g )How many liters of carbon dioxide are produced at STP when 44 g of C3H8 completely reacts with oxygen
Explanation:
The balanced chemical equation of the reaction is:
[tex]C_3H_8(g)+ 5O_2 (g)->3CO_2(g)+4H_2O(g)[/tex]
From the balanced chemical equation,
1 mole of propane forms ------ 3 mol. of [tex]CO_2[/tex] gas.
The molar mass of propane is 44.1 g/mol.
One mole of any gas at STP occupies --- 22.4 L.
Hence, 44 g of propane forms (3x22.4 L=) 67.2 L of CO2 gas at STP.
Answer:
Thus, 67.2 L of CO2 is formed at STP.
What is ethane?
A. A polymer
B. An alkyne
C. An alkane
D. An alkene
Answer:
D. An alkene
Explanation:
because Ethane is C2H4
Answer:
It's a alkANE. C.
Explanation:
The easiest way to memorize this is to look at the endings. Substances that end in -ANE are alkANEs. Substances that end in -ENE are alkENEs. Substances that end in -YNE are alkYNEs.
PLEASE HELP ASAP
A total of 132.33g C3H8 is burned in 384.00 g O2. Use the following questions to determine the amounts of products formed.
• How many grams of CO2 and H2O will be produced? (2 points)
b. If the furnace is not properly adjusted, the products of combustion can include other gases, such as CO and unburned hydrocarbons. If only 269.34 g of CO2 were formed in the above reaction, what would the percent yield be? (2 points)
In an analysis of interhalogen reactivity, 0.350 mol ICl was placed in a 5.00 L flask and allowed to decompose at a high temperature.
2 ICl(g) I2(g) + Cl2(g)
Calculate the equilibrium concentrations of I2, Cl2, and ICl. (Kc = 0.110 at this temperature.)
I2 M
Cl2 M
ICl M
Answer:
[ICl] = 0.0420 M
[I₂] = [Cl₂] = 0.0140 M
Explanation:
Step 1: Calculate the initial concentration of ICl
[ICl] = 0.350 mol / 5.00 L = 0.0700 M
Step 2: Make an ICE chart
2 ICl(g) ⇄ I₂(g) + Cl₂(g)
I 0.0700 0 0
C -2x +x +x
E 0.0700-2x x x
The concentration equilibrium constant (Kc) is:
Kc = 0.110 = [I₂] [Cl₂] / [ICl]² = x² / (0.0700-2x)² = (x/0.0700-2x)²
0.332 = x/0.0700-2x
x = 0.0140
The concentrations at equilbrium are:
[ICl] = 0.0700-2x = 0.0700-0.0280 = 0.0420 M
[I₂] = [Cl₂] = x = 0.0140 M
What is Bose Einstein state of matter
Which of the following ions is the less likely to be formed?
A) Li+3
B) Na+
C) I-
D) Sr2+
Ε) Η+
Answer:
Li^3+
Explanation:
The electronic configuration of lithium is ; 1s2 2s1. This means that lithium has one electron in its outermost shell and two core electrons.
We know that it is difficult to remove these core electrons during ionization. Lithium belongs to group 1 hence Li^+ is formed more easily.
It is very difficult to form Li^3+ because it involves loss of core electrons which requires a lot of energy.
To draw a Lewis structure for a polyatomic ion, begin by calculating A, the available electrons, and N, the needed electrons. What is N for CIO3-, the chlorate ion?
A = 26
N = ?
Answer:
16
Explanation:
Because the sum of all electron in that compound should be 41 and as it has one electron extra ,total no. of electrons are 42 .
So if we add 26 +16 we get 42
Hence it's correct answer
Write an essay on sensipar (cinacalcet)
Explanation:
The menstrual cycle is a reproductive cycle that takes place in the females of the group of primates. The menstrual cycle is divided into four phases:
(i) Menstrual phase: It extends from 1
st
to 4
th
day of the cycle. It occurs in the absence of fertilisation. During this phase, bleeding occurs as the endometrium of the uterus is sloughed off. The menstrual flow consists of secretion of endometrial glands, cell debris, unfertilized ovum. After 4
th
day, once again the FSH secretion from the pituitary is resumed and the new follicle starts developing.
(ii) Follicular phase: When the ovary is in this phase, the uterus enters in the proliferative phase. This takes place from 5
th
to 13
th
day of the cycle. During this phase new primordial follicle in the ovary develops due to the action of FSH from the pituitary. It gradually changes into the Graafian follicle and the production of estrogen starts. Only one follicle develops in one cycle. Corresponding to the changes in the ovary, the uterus also undergoes proliferation. Endometrial glands, stimulated by estrogen do repair process of the uterus.
(iii) Ovulatory phase: During this phase, ovulation takes place. It usually occurs on 14
th
day. Mature Graafian follicle ruptures due to LH secreted by the pituitary. Graafian follicle bursts and releases the ovum. This ovum along with the follicular fluid is picked up by the fimbriae of the infundibulum of the fallopian tube. It passes through the fallopian tube, where, if it happens to meet a sperm, it is fertilised. If not fertilised, the ovum degenerates.
(iv) Luteal phase: It corresponds with the secretory phase in the uterus. It takes place between 15
th
to 28
th
day of the cycle.
Ovarian changes: In the ovary, corpus luteum is formed from an empty Graafian follicle. Progesterone is secreted now. If the ovum is fertilised, corpus luteum is retained. LH and LTH from pituitary help in the maintenance of corpus luteum. If the ovum is not fertilised, corpus luteum degenerates and forms corpus albicans.
Uterine changes: Under the influence of progesterone, there is an increase in the thickness of the endometrium. Endometrial glands grow and become secretory. Progesterone is responsible for the maintenance of pregnancy. When fertilised ovum reaches the uterus, it is implanted and the placenta is formed. Till placenta becomes functional corpus luteum keeps on producing progesterone. But when progesterone source is cut off, endometrium sloughs off and menstruation begins
