The document discusses the mole concept in chemistry. Some key points:
- A mole is equal to 6.022x10^23 particles and can refer to atoms, molecules, etc.
- The molar mass of an element is its atomic mass in grams and the molar mass of a compound is the sum of the atomic masses of its elements.
- One mole of an ideal gas occupies 22.4 liters at STP.
- Questions at the end calculate things like moles, mass, volume using molar mass and mole ratios in chemical equations.
Easy way to understand moles. 1 mole has the same number of particles present in 12 grams of Carbon-12. Carbon-12 is an isotope of carbon which is taken as standard. Mass of 1 mole of atoms = Atomic mass, Mass of 1 mole of molecules = molecular mass.
Easy way to understand moles. 1 mole has the same number of particles present in 12 grams of Carbon-12. Carbon-12 is an isotope of carbon which is taken as standard. Mass of 1 mole of atoms = Atomic mass, Mass of 1 mole of molecules = molecular mass.
Presentation is for the first chapter of class 11th Chemistry CBSE board. Presentation is having detailed description for some of the basic concepts like mole concept, matter in our surrounding etc.
Presentation is for the first chapter of class 11th Chemistry CBSE board. Presentation is having detailed description for some of the basic concepts like mole concept, matter in our surrounding etc.
valencies, criss-cross method to find chemical formulae, formula unit mass, gram atomic mass, gram molecular mass, gram formula unit mass , mole concept, formulae
THE IMPORTANCE OF MARTIAN ATMOSPHERE SAMPLE RETURN.Sérgio Sacani
The return of a sample of near-surface atmosphere from Mars would facilitate answers to several first-order science questions surrounding the formation and evolution of the planet. One of the important aspects of terrestrial planet formation in general is the role that primary atmospheres played in influencing the chemistry and structure of the planets and their antecedents. Studies of the martian atmosphere can be used to investigate the role of a primary atmosphere in its history. Atmosphere samples would also inform our understanding of the near-surface chemistry of the planet, and ultimately the prospects for life. High-precision isotopic analyses of constituent gases are needed to address these questions, requiring that the analyses are made on returned samples rather than in situ.
This pdf is about the Schizophrenia.
For more details visit on YouTube; @SELF-EXPLANATORY;
https://www.youtube.com/channel/UCAiarMZDNhe1A3Rnpr_WkzA/videos
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Observation of Io’s Resurfacing via Plume Deposition Using Ground-based Adapt...Sérgio Sacani
Since volcanic activity was first discovered on Io from Voyager images in 1979, changes
on Io’s surface have been monitored from both spacecraft and ground-based telescopes.
Here, we present the highest spatial resolution images of Io ever obtained from a groundbased telescope. These images, acquired by the SHARK-VIS instrument on the Large
Binocular Telescope, show evidence of a major resurfacing event on Io’s trailing hemisphere. When compared to the most recent spacecraft images, the SHARK-VIS images
show that a plume deposit from a powerful eruption at Pillan Patera has covered part
of the long-lived Pele plume deposit. Although this type of resurfacing event may be common on Io, few have been detected due to the rarity of spacecraft visits and the previously low spatial resolution available from Earth-based telescopes. The SHARK-VIS instrument ushers in a new era of high resolution imaging of Io’s surface using adaptive
optics at visible wavelengths.
Introduction:
RNA interference (RNAi) or Post-Transcriptional Gene Silencing (PTGS) is an important biological process for modulating eukaryotic gene expression.
It is highly conserved process of posttranscriptional gene silencing by which double stranded RNA (dsRNA) causes sequence-specific degradation of mRNA sequences.
dsRNA-induced gene silencing (RNAi) is reported in a wide range of eukaryotes ranging from worms, insects, mammals and plants.
This process mediates resistance to both endogenous parasitic and exogenous pathogenic nucleic acids, and regulates the expression of protein-coding genes.
What are small ncRNAs?
micro RNA (miRNA)
short interfering RNA (siRNA)
Properties of small non-coding RNA:
Involved in silencing mRNA transcripts.
Called “small” because they are usually only about 21-24 nucleotides long.
Synthesized by first cutting up longer precursor sequences (like the 61nt one that Lee discovered).
Silence an mRNA by base pairing with some sequence on the mRNA.
Discovery of siRNA?
The first small RNA:
In 1993 Rosalind Lee (Victor Ambros lab) was studying a non- coding gene in C. elegans, lin-4, that was involved in silencing of another gene, lin-14, at the appropriate time in the
development of the worm C. elegans.
Two small transcripts of lin-4 (22nt and 61nt) were found to be complementary to a sequence in the 3' UTR of lin-14.
Because lin-4 encoded no protein, she deduced that it must be these transcripts that are causing the silencing by RNA-RNA interactions.
Types of RNAi ( non coding RNA)
MiRNA
Length (23-25 nt)
Trans acting
Binds with target MRNA in mismatch
Translation inhibition
Si RNA
Length 21 nt.
Cis acting
Bind with target Mrna in perfect complementary sequence
Piwi-RNA
Length ; 25 to 36 nt.
Expressed in Germ Cells
Regulates trnasposomes activity
MECHANISM OF RNAI:
First the double-stranded RNA teams up with a protein complex named Dicer, which cuts the long RNA into short pieces.
Then another protein complex called RISC (RNA-induced silencing complex) discards one of the two RNA strands.
The RISC-docked, single-stranded RNA then pairs with the homologous mRNA and destroys it.
THE RISC COMPLEX:
RISC is large(>500kD) RNA multi- protein Binding complex which triggers MRNA degradation in response to MRNA
Unwinding of double stranded Si RNA by ATP independent Helicase
Active component of RISC is Ago proteins( ENDONUCLEASE) which cleave target MRNA.
DICER: endonuclease (RNase Family III)
Argonaute: Central Component of the RNA-Induced Silencing Complex (RISC)
One strand of the dsRNA produced by Dicer is retained in the RISC complex in association with Argonaute
ARGONAUTE PROTEIN :
1.PAZ(PIWI/Argonaute/ Zwille)- Recognition of target MRNA
2.PIWI (p-element induced wimpy Testis)- breaks Phosphodiester bond of mRNA.)RNAse H activity.
MiRNA:
The Double-stranded RNAs are naturally produced in eukaryotic cells during development, and they have a key role in regulating gene expression .
The increased availability of biomedical data, particularly in the public domain, offers the opportunity to better understand human health and to develop effective therapeutics for a wide range of unmet medical needs. However, data scientists remain stymied by the fact that data remain hard to find and to productively reuse because data and their metadata i) are wholly inaccessible, ii) are in non-standard or incompatible representations, iii) do not conform to community standards, and iv) have unclear or highly restricted terms and conditions that preclude legitimate reuse. These limitations require a rethink on data can be made machine and AI-ready - the key motivation behind the FAIR Guiding Principles. Concurrently, while recent efforts have explored the use of deep learning to fuse disparate data into predictive models for a wide range of biomedical applications, these models often fail even when the correct answer is already known, and fail to explain individual predictions in terms that data scientists can appreciate. These limitations suggest that new methods to produce practical artificial intelligence are still needed.
In this talk, I will discuss our work in (1) building an integrative knowledge infrastructure to prepare FAIR and "AI-ready" data and services along with (2) neurosymbolic AI methods to improve the quality of predictions and to generate plausible explanations. Attention is given to standards, platforms, and methods to wrangle knowledge into simple, but effective semantic and latent representations, and to make these available into standards-compliant and discoverable interfaces that can be used in model building, validation, and explanation. Our work, and those of others in the field, creates a baseline for building trustworthy and easy to deploy AI models in biomedicine.
Bio
Dr. Michel Dumontier is the Distinguished Professor of Data Science at Maastricht University, founder and executive director of the Institute of Data Science, and co-founder of the FAIR (Findable, Accessible, Interoperable and Reusable) data principles. His research explores socio-technological approaches for responsible discovery science, which includes collaborative multi-modal knowledge graphs, privacy-preserving distributed data mining, and AI methods for drug discovery and personalized medicine. His work is supported through the Dutch National Research Agenda, the Netherlands Organisation for Scientific Research, Horizon Europe, the European Open Science Cloud, the US National Institutes of Health, and a Marie-Curie Innovative Training Network. He is the editor-in-chief for the journal Data Science and is internationally recognized for his contributions in bioinformatics, biomedical informatics, and semantic technologies including ontologies and linked data.
Richard's aventures in two entangled wonderlandsRichard Gill
Since the loophole-free Bell experiments of 2020 and the Nobel prizes in physics of 2022, critics of Bell's work have retreated to the fortress of super-determinism. Now, super-determinism is a derogatory word - it just means "determinism". Palmer, Hance and Hossenfelder argue that quantum mechanics and determinism are not incompatible, using a sophisticated mathematical construction based on a subtle thinning of allowed states and measurements in quantum mechanics, such that what is left appears to make Bell's argument fail, without altering the empirical predictions of quantum mechanics. I think however that it is a smoke screen, and the slogan "lost in math" comes to my mind. I will discuss some other recent disproofs of Bell's theorem using the language of causality based on causal graphs. Causal thinking is also central to law and justice. I will mention surprising connections to my work on serial killer nurse cases, in particular the Dutch case of Lucia de Berk and the current UK case of Lucy Letby.
A brief information about the SCOP protein database used in bioinformatics.
The Structural Classification of Proteins (SCOP) database is a comprehensive and authoritative resource for the structural and evolutionary relationships of proteins. It provides a detailed and curated classification of protein structures, grouping them into families, superfamilies, and folds based on their structural and sequence similarities.
(May 29th, 2024) Advancements in Intravital Microscopy- Insights for Preclini...Scintica Instrumentation
Intravital microscopy (IVM) is a powerful tool utilized to study cellular behavior over time and space in vivo. Much of our understanding of cell biology has been accomplished using various in vitro and ex vivo methods; however, these studies do not necessarily reflect the natural dynamics of biological processes. Unlike traditional cell culture or fixed tissue imaging, IVM allows for the ultra-fast high-resolution imaging of cellular processes over time and space and were studied in its natural environment. Real-time visualization of biological processes in the context of an intact organism helps maintain physiological relevance and provide insights into the progression of disease, response to treatments or developmental processes.
In this webinar we give an overview of advanced applications of the IVM system in preclinical research. IVIM technology is a provider of all-in-one intravital microscopy systems and solutions optimized for in vivo imaging of live animal models at sub-micron resolution. The system’s unique features and user-friendly software enables researchers to probe fast dynamic biological processes such as immune cell tracking, cell-cell interaction as well as vascularization and tumor metastasis with exceptional detail. This webinar will also give an overview of IVM being utilized in drug development, offering a view into the intricate interaction between drugs/nanoparticles and tissues in vivo and allows for the evaluation of therapeutic intervention in a variety of tissues and organs. This interdisciplinary collaboration continues to drive the advancements of novel therapeutic strategies.
Seminar of U.V. Spectroscopy by SAMIR PANDASAMIR PANDA
Spectroscopy is a branch of science dealing the study of interaction of electromagnetic radiation with matter.
Ultraviolet-visible spectroscopy refers to absorption spectroscopy or reflect spectroscopy in the UV-VIS spectral region.
Ultraviolet-visible spectroscopy is an analytical method that can measure the amount of light received by the analyte.
2. What is a Mole?
• Mole is a number just like 1 dozen = 12, 1 Score =
20, 1 Gross= 144
• 1 Mole = 6.022*10^23 = Avogadro Number (NA)
– 1 Mole Apples = 6.022*10^23 Apples
– 2 Mole Apples = 2*6.022*10^23 Apples
– 1 Mole Carbon Atoms= 6.022*10^23 Atoms of Carbon
– 5 Mole Carbon Atoms = 5*6.022*10^23 Atoms of
Carbon
Bharti Study Circle 2
3. Atomic Mass of C atom
• 1 Carbon (C) atom contains 6 Protons, 6 Neutrons and
6 electrons
• Atomic Mass of C = Mass of 1 atom of C = Mass of 6
Protons + Mass of 6 Neutrons + Mass of 6 electrons
• Mass of Proton = Mass of Neutron = 1.67*10^-24 gm =
1 Atomic Mass Unit (amu) = 1 u
• Mass of Electron = 9.1*10^-28 gm = 1/1837*Mass of
Proton/Neutron. Hence, mass of electron is considered
negligible as compared to mass of proton/neutron.
• Hence, Atomic Mass of C = Mass of 6 Protons + Mass of
6 Neutrons = 12 amu. Or 1 amu = 1/12th of mass of 1 C
atom = 1.67*10^-24 gm
Bharti Study Circle 3
4. Molar Mass of Atom
• Since Atomic Mass of C = Mass of 1 atom of C =
Mass of 6 Protons + Mass of 6 Neutrons = 12 amu
= 12 *1.67*10^-24 gm
• Hence Mass of 1 mole of C atom = Mass of
6.022*10 ^23 atoms of C = (6.022*10 ^23)*(12
*1.67*10^-24 gm) = 12 gm = Gram Atomic Mass
(GAM) of C = Molar Mass of C atom = Atomic
Mass of C in grams
• (6.022*10 ^23 = NA)*(1.67*10^-24 gm = 1 amu) =
1 gm
Bharti Study Circle 4
NA * 1 amu = 1 gm
5. Questions
Q1: Find the molar mass or GAM of Oxygen (O) atom.
A1: 1 atom of O contains 8 Protons, 8 Neutrons and 8 electrons.
Atomic Mass of O = Mass of 8 Protons + Mass of 8 Neutrons (As mass of electron is negligible
compared to mass of proton/neutron)
Atomic Mass of O = 8 amu + 8 amu = 16 amu
Molar Mass = GAM of O atom = Mass of 1 mole of O atoms = (6.022*10 ^23)*(16 *1.67*10^-24
gm) = 16 gm [As (6.022*10 ^23)*(1.67*10^-24 = 1]
Or Simply Molar Mass of O atom = Atomic Mass of O in gm = 16 gm
Q2: Find number of moles of O atoms present in 64 gm of O atom.
A2: Molar Mass of O atom = Mass of 1 mole of O atoms = 16 gm
16 gm of O atoms contains 1 mole of O atom
1 gm of O atom contains 1/16 moles of O atom
64 gm of O atom contains 1/16*64= 4 moles of O atom
By using above formula, Number of Moles of O atom = 64gm/16gm= 4
Bharti Study Circle 5
Number of Moles = Mass Given/Molar Mass
6. Questions
Q3: Find number of moles of atoms present in
i. 28 gm of N atoms
ii. 32 gm of S atoms
iii. 14 gm of H atoms
A3: Number of Moles = Mass Given/Molar Mass
i. Molar Mass of N = 14 gm. Hence, Number of moles
of N in 28 gm of N atoms = 28 gm/14 gm = 2
ii. Molar Mass of S = 32 gm. Hence, Number of moles of
S in 32 gm of S atoms = 32 gm/32 gm = 1
iii. Molar Mass of H = 1 gm. Hence, Number of moles of
H in 14 gm of H atoms = 14 gm/1 gm = 14
Bharti Study Circle 6
7. Molecular Mass
• Molecular Mass = Sum Total of Mass of Individual
Atoms in the Molecule. For example:
Molecular Mass of O2 molecule = Mass of 2 atoms of O
= 2* 16 amu = 32 amu
Molecular Mass of NH3 molecule = Mass of 1 atom of
N + Mass of 3 atoms of H = 1* 14 amu + 3 * 1 amu =
17 amu
Molecular Mass of H2SO4molecule = Mass of 2 atoms
of H + Mass of 1 atom of S + Mass of 4 atoms of O =
2* 1 amu + 1 * 32 amu + 4 * 16 amu = 98 amu
Bharti Study Circle 7
8. Molar Mass of Molecule
• Molar Mass of a Molecule = Mass of 1 mole of molecules =
Mass of 6.022*10 ^23 molecules = Gram Molecular Mass
(GMM) = Molecular Mass expressed in gm.
• For example:
Molar Mass of O2 molecule = Mass of 6.022*10 ^23 molecules
of O2 = (6.022*10 ^23) * (Molecular Mass of O2 molecule) =
(6.022*10 ^23) * 32 amu = 32 gm [As (6.022*10 ^23) * 1 amu =
1 gm]
Molar Mass of NH3 molecule = Mass of 6.022*10 ^23 molecules
of NH3 = (6.022*10 ^23) * (Molecular Mass of NH3 molecule) =
(6.022*10 ^23) * 17 amu = 17 gm
Molar Mass of H2SO4molecule = Mass of 6.022*10 ^23
molecules of H2SO4 = (6.022*10 ^23) * (Molecular Mass of
NH2SO4 molecule) = (6.022*10 ^23) * 98 amu = 98 gm
Bharti Study Circle 8
9. Questions
Q4: Find molar mass of
i. N atom
ii. N2 molecule
iii. S atom
iv. S8 molecule
A4: Molar Mass = Atomic Mass/Molecular Mass
expressed in gm. Hence, molar mass
i. N atom = 14 gm
ii. N2 molecule = 2 * 14 gm = 28 gm
iii. S atom = 32 gm
iv. S8 molecule = 8 * 32 gm = 256 gm
Bharti Study Circle 9
10. Questions
Q5: Find number of moles of O2 molecule in 64 gm
of O2 gas. Also, find the total number of O atoms.
A5: Number of moles of O2 molecule = Mass Given/
Molar Mass of O2 molecule = 64 gm/32 gm = 2
Since 1 molecule of O2 molecule contains 2 atoms of
O.
Hence, 1 mole of O2 molecule contains 6.022*10 ^23
molecules of O2 =2 * 6.022*10 ^23 atoms of O.
Hence, 2 moles of O2 molecule contains 2 *
6.022*10 ^23 molecules of O2 =2 * 2 * 6.022*10
^23 atoms of O = 4 * 6.022*10 ^23 atoms of O.
Bharti Study Circle 10
11. Questions
Q6: Find in 98 gm of H2SO4
i. Moles of H2SO4 molecule
ii. Moles of S atoms
iii. Number of S atoms
iv. Moles of H atoms
v. Gram atom of H
vi. Mass of S atom
vii. Moles of O atoms
A6:
i. Moles of H2SO4 molecule = Mass Given/ Molar Mass = 98 gm/98 gm = 1
ii. Since 1 molecule of H2SO4 contains 1 atom of S. Similarly, 1 mole of H2SO4 shall contain 1
mole of S atoms
iii. 1 mole of S atoms contains 6.022*10 ^23 atoms of S
iv. Since 1 molecule of H2SO4 contains 2 atom of H. Similarly, 1 mole of H2SO4 shall contain 2
moles of H atoms
v. Gram atom of H = Number of moles of H = 2
vi. Mass of S atom = Number of moles of S atom * Molar Mass of S atom = 1 mole * 32
gm/mole = 32 gm
vii. Since 1 molecule of H2SO4 contains 4 atoms of O. Similarly, 1 mole of H2SO4 shall contain 4
moles of O atoms
Bharti Study Circle 11
12. Molar Volume of Ideal Gas
• 1 mole of any ideal gas at STP occupies 22.4
litres (L) of Volume
Note: Standard Temperature and Pressure implies
Temperature = 273K and
Pressure = 1 atm = 760 mm of Hg
• 1 L = 1 dm ^3 = 1000 ml = 1000 cm ^3
Bharti Study Circle 12
Number of moles = Volume Given/ Molar Volume = Volume Given/ 22.4 L
13. Questions
Q7: Find volume occupied by 3 moles of CO2 at STP.
A7: Number of moles = 3
Hence, Volume = Number of moles * 22.4 L = 3 *
22.4 L = 67.2 L
Q8: Find number of moles of O3 gas present in 5.6
dm ^3 of O3 gas at STP.
A7: Number of moles = Volume Given/ 22.4 dm ^3
= 5.6 dm ^3 /22.4 dm ^3 = 0.25
Bharti Study Circle 13
14. Questions
Q8: Find number of atoms of H present in 11.2 L of
H2 gas at STP.
A8: Number of moles of H2 = Volume Given/ 22.4 L
= 11.2 L/22.4 L = 0.5 moles
1 molecule of H2 gas contains 2 atoms of H
1 mole of H2 contains 2 moles of H atoms
0.5 moles of H2 contains 2 * 0.5 moles of H atoms =
1 mole of H atom = 6.022*10 ^23 atoms of H
Bharti Study Circle 14
15. Questions
Q9: Find volume occupied by 8 gm of O2 gas at STP.
A9: Number of moles of O2 gas = Mass Given/ Molar Mass
= 8 gm/32 gm = 0.25 moles
Also, Number of Moles = Volume / 22.4 L
=> Volume = 0.25 * 22.4 L = 5.6 L
Q10: Find volume occupied by 128 gm of SO2 gas at STP.
A10: Number of moles of SO2 gas = Mass Given/ Molar
Mass = 128 gm/(32 gm + 2 *16 gm) = 2 moles
Also, Number of Moles = Volume / 22.4 L
=> Volume = 2 * 22.4 L = 44.8 L
Bharti Study Circle 15
16. Questions
Q11: Find mass of H2 required and NH3 produced from 14 gm of N2 gas in
below Haber’s reaction
N2 (g) + H2 (g) -> NH3 (g)
A11: First step is to balance the chemical equation:
N2 (g) + 3 H2 (g) -> 2 NH3 (g)
Number of moles of N2 gas = Mass Given/ Molar Mass of N2 gas = 14 gm/28
gm = 0.5 moles
As per above balanced equation:
1 mole of N2 requires 3 moles of H2 and produces 2 moles of NH3
=> 0.5 moles of N2 requires 1.5 moles of H2 and produces 1 mole of NH3
Mass of H2 = Number of moles * Molar Mass of H2 = 1.5 * 2 gm = 3 gm
Mass of NH3 = Number of moles * Molar Mass of NH3 = 2 * 17 gm = 34 gm
Bharti Study Circle 16
17. Questions
Q12: If 16 gm of CH4 is burnt, find mass and volume of CO2
produced.
CH4 (g) + O2 (g) -> CO2 (g) + H2O (l)
A12: First step is to balance the chemical equation:
CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (l)
Number of moles of CH4 gas = Mass Given/ Molar Mass of CH4 gas = 16
gm/( 12 gm + 4 * 1 gm = 16 gm) = 1 mole
As per above balanced equation:
1 mole of CH4 produces 1 mole of CO2. Hence,
Mass of 1 mole of CO2 = 12 gm + 2 * 16 gm = 44 gm
Volume of 1 mole of CO2 = 1 * 22.4 L = 22.4 L
Bharti Study Circle 17
18. Questions
Q13: Find mass of O2 required /CO2 produced when 5.8 gm of butane (C4H10) is burnt.
A13: First step is to balance the chemical equation:
2 C4H10 (g) + 13 O2 (g) -> 8 CO2 (g) + 10 H2O (l)
Number of moles of C4H10 gas = Mass Given/ Molar Mass of C4H10 gas = 5.8 gm/( 4 *12
gm + 10 * 1 gm = 58 gm) = 0.1 mole of C4H10 gas
As per above balanced equation:
2 moles of C4H10 requires 13 moles of O2 gas to produce 8 moles of CO2 => 1 mole of
C4H10 requires 13/2 (=6.5) moles of O2 gas to produce 8/2 (=4) moles of CO2
Moles of O2 required by 0.1 mole of C4H10 = 0.1 * 6.5 moles = 0.65 moles of O2
Mass of O2 produced = Number of moles * Molar Mass of O2 = 0.65 * 32 gm =
20.8 gm of O2
Similarly, Moles of CO2 produced by 0.1 mole of C4H10 = 0.1 * 4 moles = 0.4 moles of
CO2
Mass of CO2 produced = Number of moles * Molar Mass of CO2 = 0.4 * (12 gm + 2
* 16 gm = 44 gm) = 17.6 gm of CO2
Bharti Study Circle 18
