What is the oxidation state of fluorine (F)?
The Periodic Table
A.-2
B. +1
C. -1
D. +2
Answers
Answer:
It is C. -1
Explanation:
[tex] { \underline{ \sf{ \blue{christ \: † \: alone}}}}[/tex]
Related Questions
I need help!!!! please
Answers
6.94 is much more closet to 7.016, so Li 7 is more abundant
A piece of sodium metal can be described as?
Answers
The reactivity of elements can be predicted based on the position of an element in the periodic table. Based on the periodic table, which of the following elements would be expected to be the most reactive?
Iodine (I)
Iodine (I)
Chlorine (Cl)
Chlorine (Cl)
Fluorine (F)
Fluorine (F)
Bromine (Br)
Answers
Answer:
Chlorine (Cl) is most reactive out the the listed elements.
Answer:
iodine
Explanation:
It is the most reactive on the periodic table.
Liquids which do not wer the glass surface form a convex meniscus.True or false
Answers
Answer:
true is your answer
Explanation:
hope it helps
hii pls help me to balance the equation and state the symbols thanksss
Answers
Ca(s)+ H2O (l)----->H2 +Ca(oH)2
Ca(s)+2H2O(l)----->H2(g)+Ca(oH)2
I hope this helps
sorry if it's wrong
A chemist determines by measurements that 0.0550 moles of nitrogen gas participate in a chemical reaction. Calculate the mass of nitrogen gas that participates.
Be sure your answer has the correct number of significant digits.
Answers
The mass of nitrogen gas that participated in the chemical reaction is 1.54g
HOW TO CALCULATE MASS OF AN ELEMENT:
Mass of a substance can be calculated by multiplying the number of moles in mol of the substance by its molecular mass in g/mol. That is;mass (M) = molar mass (MM) × number of moles (n)According to this question, a chemist determines by measurements that 0.0550 moles of nitrogen gas (N2) participate in a chemical reaction.
The molecular mass of nitrogen gas (N2) = 14.01(2)= 28.02g/mol
Hence, the mass of the nitrogen gas that participated in the chemical reaction is calculated as follows:
Mass (g) = 0.0550 mol × 28.02 g/molMass = 1.5411Therefore, the mass of nitrogen gas that participated in the chemical reaction is 1.54g
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what do y’all do when y’all brainly says “locked” and won’t show anymore answers !???
Answers
Answer:
Create a new account as the thing you're experiencing now is due to the fact that Brainly had blocked your account.
what is second explain it briefly
Answers
Explanation:
it is that when an object is moving 25 m per sec to calculate its time we can simply divide 1.5 by 25 and multiply by 1
A chemist sets up a chemical reaction but finds that none of the reactant molecules have the required activation energy. What is the result?
A.
Products will form with less energy input.
B.
No products will be formed.
C.
The products will form too quickly.
D.
Products will convert to reactants.
Answers
Answer:
B.
No products will be formed.
I forgot to add the answers i had to my question (While a balloon is being filled, if the temperature of the air in the balloon increases, what happens to it’s volume)
A) It stays the same
B) It deflates
C) It decreases
D) It increases
Answers
The correct answer is D ... and if the answer was added to the word "explode" .. it would have been more correct .
5+64
this is TOTALLY SOOOO hard so help me
^thats what she said
Answers
Answer:
69
Explanation:
I thought that I answer is correct
Answer:
The answer is 69
Explanation:
Because if you add one its 66, and so on, hope this helps! but if you add five it will be 69, if you need help with easy equations let me know
describe how thin layer chromatography is used in the isolation and extraction of lipids
Answers
Thin layer chromatography(TLC) works with the principle of separation through adsorption.
It is used in the isolation and extraction of lipids through the following steps:
apply the lipid samples spots in the bottom of the plate.also apply sample solution to the marked spotpour the mobile phase into the TLC chamber and use a moist filter paper to cover it. this is done to maintain equal humidity.then place the plate in the TLC chamber and close it with a lid.the plate is immersed into the solvent (mobile phase) for its development. this is done, keeping in mind that the sample spot should be above the solvent.once the sample spots are developed, they are removed and dried.this is later viewed using the UV light chamber to see the isolation of the lipid sample.Learn more here:
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hii pls help me to balance the equation and state the symbols thanksss
Answers
Answer:
I hope that the link above will be helpful
▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓
[tex]\boxed{\pmb{\green{\sf{2Mg +H_{2}SO_{4}\dashrightarrow H _{2}+Mg_{2}SO_{4}}}}}[/tex]
▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓▓
At the beginning of the reaction: N2 + 3H2 -> 2NH there is 1 mol of N. and 3 moles of H, and at the end of it is present a mixture formed by 2.5 moles in total, what is the performance for the reaction?
5.- Al comenzar la reacción: N2+ 3H2 -> 2NH existe 1 mol de N. y 3 moles de H, y al finalizarla está presente una mezcla formada por 2,5 moles en total, ¿cuál es el rendimiento para la reacción?
Answers
Answer:
75%
Explanation:
call realistic moles of N2 in the reaction is a
moles of H2 :3a
moles of NH3: 2a
the moles of residual N2 after reaction: 1-a
the moles of residual H2 after react: 3- 3a
the moles of mixture: 2a+1-a+3-3a=2.5
-> a=0.75
H=0.75/1=75%
how many atoms of gold are present in 0.6 gm of 18 caral gold? ( 24 caral gold in 100% pure) atomic weight of gold = 197?
Answers
Answer:
HERE'S YOUR ANSWERWe are given that the ring is made up of 20 carrat gold. 20 carrat gold means 20 parts of gold and 4 parts of other metal (usually copper or silver). Thus, the percentage of gold in 20 carrat gold will be
= (20 / 24) X 100
= 83.333 %
We can now calculate the mass of gold in the ring. We are given that the ring weighs 300 mg. So, the mass of gold in ring will be
= (83.333 / 100) X 300
= 250 mg
Thus, the mass of gold in 300 mg of the ring is 250 mg.
Now the molar mass of gold is 197 g. This means that there are 6.023 X 1023 atoms of gold in 197g of gold. Thus
197g of gold = 6.023 X 1023 atoms of gold
1g of gold = (6.023 X 1023 ) / 197 atoms of gold
250 mg or 0.25g of gold = (6.023 X 1023 X 0.25) / 197 atoms of gold
= 7.643 X 1020 atoms of gold
HOPE IT HELPS MAN ☺️Ribbon diagrams shows secondary structures and appear less detailed than other model types. In one to two sentences, give a reason the chemists would use ribbon diagrams. What type of information do they provide?
Answers
Answer:
They show helixes and proteins
Explanation:
Ribbon diagrams are the Richardson diagrams that represent the protein structures. Chemist prefers using Ribbon diagrams to show the various structure of the proteins.
What are ribbon diagrams?The Ribbon diagram is used to represent the structural organization of the protein in the polypeptide chains or individually. It shows the three-dimensional structure and organization of the protein framework.
The ribbon diagram for the secondary structure of the protein shows the intramolecular and intermolecular structure along with helixes and bonding. It shows the detailed structure of the protein organization including the helix and sheets.
Therefore, the ribbon diagram is used by chemists to determine the structure of the protein as they give the details about the helix and sheet.
Learn more about ribbon diagrams, here:
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water and development are substitute of one another.
explanation with it, please
Answers
what is the reactant(s) in the chemical equation below
3CO(g) + Fe2O3(s) 2Fe(s) + 3CO2(g)
Answers
Answer:
the reactants are carbon dioxide ,iron, oxygen
When 120 g of carbon reacts completely with 320 g of oxygen the mass of carbon dioxide formed will be?
Answers
Which of the following is NOT a physical property?
Viscosity
Reactivity
Malleability
Density
Answers
Answer:
reactivity is NOT a physical property
A student was given a 2.850-g sample of a mixture of potassium nitrate and potassium bromide and was asked to find the percentage of each compound in the mixture. She dissolved the sample and added a solution that contained an excess of silver nitrate, AgNO3. The silver ion precipitated all of the bromide ion in the mixture as AgBr. It was filtered, dried, and weighed. Its mass was 1.740 g. What was the percentage of each compound in the mixture
Answers
Answer:
See explanation
Explanation:
The reaction occurs as follows;
KBr(aq) + AgNO3(aq) ----> AgBr(s) + KNO3(aq)
Number of moles of AgBr formed = mass /molar mass =1.740 g/187.77 g/mol = 0.0093 moles
From the reaction equation;
1 mole of KBr yields 1 mole of AgBr
Hence the number of moles of KBr reacted = 0.0093 moles
Mass of KBr present = 0.0093 moles × 119g/mol = 1.11 g
Mass of KNO3 = 2.850 g - 1.11 g = 1.74 g
Percentage of KBr = 1.11 g/2.850 g × 100 = 38.9%
Percentage of KNO3 = 1.74 g/2.850 g × 100 = 61.1%
two easy uses of mixture
Answers
Explanation:
it helps to make juices.
It helps to make concentrated acid into dilute acid.
Answer:
Explanation:
Here are a few more examples:
1. Smoke and fog (Smog)
2. Dirt and water (Mud)
If the temperature of a reversible reaction in dynamic equilibrium decreases, how will the equilibrium change?
A. It will shift towards the endothermic reaction.
B. It will not change.
C. It will shift towards the products.
D. It will shift towards the exothermic reaction.
Answers
Answer:
D. It will shift towards the exothermic reaction.
Explanation:
PLEASE HELP ASAP
Barney was a very inventive scientist and tried to perform the same experiments as those by famous scientists from long ago. One of the most challenging experiments was to roll out a very thin film of gold and put X-ray film in a circle around it. The difficult part was to get ahold of some radioactivity that he could shoot at the thin film for gold. One day, he ran into The Great Gazoo and told him his goal. The Great Gazoo said he had the perfect substance for him that was radioactive and should work. The element was named after him and had the symbol Gz. Barney quickly went back to his lab to run the experiment. What Barney noticed is that all the radioactive particles went through the gold film and none of them ricocheted back, but all went straight through the gold. Being puzzled, he quickly went to the lab to analyze this unknown radioactive substance. He found that the atomic number was 119, the mass number was 305. After a few more tests, Barney realized what was wrong with Gz. Your task is to figure out what was wrong with Gz by answering the following questions:
Whose experiment was Barney trying to imitate?
Where would Gz be located on Earth’s periodic table, column and row?
Give as many details on Gz that you could predict based on its location on the periodic table: type of chemical (metal, metalloid, or nonmetal), reactivity, ionization energy, electronegativity, and size of element.
What would the elements electron configuration end in?
Why did all the radioactive particles go straight through the gold film?
What would the daughter product be for Gz? Give all details for the atomic number, mass number, number of protons, neutrons and electrons.
If Gz had worked correctly, like Barney intended, what type of decay should Gz have had?
If Gz had worked correctly, what would the daughter product be? Give all details for the atomic number, mass number, number of protons, neutrons and electrons.
If Gz had worked correctly, what element would it become?
Answers
Answer:
a) Barney wants to repeat the rutheford experiment
b) the element will be in row 8 column 1 being an alkali metal
c) it should be a radioactive element, with a lot of mass
Explanation:
n this exercise they indicate that the particle Gz has atomic number 119 and atomic mass 305 amu, when reviewing the element periodicity table with this atomic number it has not yet been discovered, it should be in row 8 column 1 therefore it should be an alkali metal .
Therefore, it has only one electron in its last orbit.
a) Barney wants to repeat the rutheford experiment
b) the element will be in row 8 column 1 being an alkali metal
c) it should be a radioactive element, with a lot of mass
A compound was found to contain 90.6% lead (Pb) and 9.4% oxygen. What is the empirical formula for this compound?
Answers
Answer:
the answer is 47.9 and ik because I just had that question
The empirical formula of the compound is O₄Pb₃.
What is the empirical formula?
An Empirical system is the chemical system of a compound that offers the proportions of the elements gifted within the compound however not the real numbers or arrangement of atoms. This would be the lowest complete variety ratio of the elements within the compound.
Amount of lead (Pb) = 90.6%
⇒and amount of oxygen = 9.4%
taking the whole number ratio
o = 4
Pb = 3
∴ ⇒O: Pb=4:3
O4Pb3 answer.
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Unknown # 29
Flame Test Color: Green
Addition of HCl(aq): Bubbles
Addition of HCl(aq) and BaCl2(aq): NR (No Reaction)
Addition of HNO3(aq) and AgNO3(aq): White precipitate
What is the unknown ionic compound molecular formula?
Answers
Answer:
I think it's a sulphate ion, because when you add hydrochloric acid to it it produces bubbles and when you add barium chloride it then produces a white precipitate.its formula is SO4
I hope this helps
What is the AHsol for LiF → Lit + F-? The lattice energy is -1,036 kJ/
mol, the enthalpy of hydration for Lit is -499
kJ/mol, and the enthalpy of hydration for F-is-431 kJ/mol. Use A Hooi = -A Hat + AHhydr.
0-968 kJ/mol
-106 kJ/mol
106 kJ/mol
1,966 kJ/
mol
Answers
The heat of hydration is defined as the heat absorbed or evolved when one mole of a substance undergoes hydration. The heat of solution is the enthalpy change associated with he dissolution of a solute. The lattice energy is the heat. Lattice energy is the energy released when the components of the lattice are brought together from infinity.
Hence the heat of solution for LiF → Li^+ + F- is -1966 kJ/mol
Given that;
Heat of hydration = ΔH solution – ΔH lattice energy
Where,
ΔH solution = Heat of the solution
ΔH lattice energy = Lattice energy of the solution
The heat of solution or enthalpy of dissolution is defined as the enthalpy change associated with the dissolution of a solute in a solvent
From the formula above;
ΔH solution = Heat of hydration + ΔH lattice energy
Heat of hydration = [(-431) + (-499)] = -930 kJ/mol
ΔH solution = (-930) + (-1,036) = -1966 kJ/mol
Hence the heat of solution for LiF → Li^+ + F- is -1966 kJ/mol
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Give the relation of energy (E) and frequency (v) as given by Planck. Calculate the frequency and energy of a photon of radiation having wavelength 3000°A.
Pls answer correctly,need it urgently...
Answers
Solution:-1
[tex]\\ \sf\longmapsto E=hv[/tex]
E denotes to Energy h denotes to planks constantv denotes to frequency.Solution:-2
Wavelength=3000Å[tex]\boxed{\sf \lambda=\dfrac{c}{v}}[/tex]
[tex]\\ \sf\longmapsto v=\dfrac{c}{\lambda}[/tex]
[tex]\\ \sf\longmapsto v=\dfrac{3\times 10^8m/s}{3000\times 10^{-10}m}[/tex]
[tex]\\ \sf\longmapsto v=0.001\times 10^{18}s^{-1}[/tex]
[tex]\\ \sf\longmapsto v=1.0\times 10^{15}s^{-1}[/tex]
Now
[tex]\\ \sf\longmapsto E=hv[/tex]
[tex]\\ \sf\longmapsto E=6.626\times 10^{-34}Js\times 1.0\times 10^{15}s^{-1}[/tex]
[tex]\\ \sf\longmapsto E=6.6\times 10^{-19}J[/tex]
9. During an experiment the students prepared three mixtures A)Starch in water B) Sodium chloride solution C) Tincture of Iodine. i) Students observed a visible beam of light through mixture A. Why? ii) Tincture of lodTe did not show Tyndall effect. Explain reason. ill) How can you relate particle size to Tyndall effect?
Answers
Answer:
See explanation
Explanation:
Tyndall effect refers to the scattering of light in a solution. Tyndall effect occurs when the size of particles in the solution exceeds 1 nm in diameter. Such solutions are actually called false solutions.
In tincture of iodine, the size of particles in solution is less than 1 nm in diameter hence the solution does not exhibit Tyndall effect. Hence, tincture of iodine is a true solution.
Therefore, if the size of particles in solution exceeded 1nm in diameter, Tyndall effect is observed.
A piece of wood displaces 12.5 mL of water and has a density of 0.97 g/ml. what is the mass of the piece of wood? what will happen if the wood is placed in water?
HELP PLEASE! ASAP YOUR DA BEST! 15 POINTS YALL
Answers
Density=0.97g/mL
Volume=12.5mL
[tex]\\ \sf \longmapsto Density=\dfrac{Mass}{Volume}[/tex]
[tex]\\ \rm\longmapsto Mass=Density(Volume)[/tex]
[tex]\\ \sf \longmapsto Mass=0.97(12.5)[/tex]
[tex]\\ \sf \longmapsto Mass=12.125g[/tex]
