How do I write the formula for binary compounds

Determine the basic oxidation<br>number for elements in s and p<br>orbitals

Andrews [41]

Answer:

Explanation:

The oxidation state, sometimes referred to as oxidation number, describes the degree of oxidation (loss of electrons) of an atom in a chemical compound. Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds.

The term oxidation was first used by Antoine Lavoisier to signify reaction of a substance with oxygen. Much later, it was realized that the substance, upon being oxidized, loses electrons, and the meaning was extended to include other reactions in which electrons are lost, regardless of whether oxygen was involved.

Helped?

Brainliest?

7 0

2 years ago

The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant,

Genrish500 [490]

Answer: (E) 300 bq

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is represented by $t_{\frac{1}{2}$

Half life of Thallium-208 = 3.053 min

Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. $\frac{2400}{2}=1200$ , after second half life, the activity would be reduced to half of 1200 i.e. $\frac{1200}{2}=600$ , and after third half life, the activity would be reduced to half of 600 i.e. $\frac{600}{2}=300$ ,

Thus the activity 9 minutes later is 300 bq.

7 0

2 years ago

Direct electron transfer from a singlet reduced species to a triplet oxidizing species is quantum-mechanically forbidden

posledela

Direct electron transfer from a a singlet reduced species to a triplet oxidizing species is quantum-mechanically forbidden.

<h3><u>Transfer from singlet to triplet:</u></h3>

Either an excited singlet state or an excited triplet state will occur when an electron in a molecule with a singlet ground state is stimulated (through radiation absorption) to a higher energy level.
All electron spins in a molecule electronic state known as a singlet are coupled.
In other words, the ground state electron and the stimulated electron's spin are still coupled (a pair of electrons in the same energy level must have opposite spins, per the Pauli exclusion principle).
The excited electron and ground state electron are parallel in a triplet state because they are no longer coupled (same spin).
It is less likely that a triplet state would arise when the molecule absorbs radiation since excitation to a triplet state necessitates an additional "forbidden" spin transfer.

To view more questions on quantum mechanism, refer to:

brainly.com/question/13639384

#SPJ4

8 0

2 years ago

(Please help, ASAP) How many grams are in 3.45x10^23 atoms of P?

Daniel [21]

one mole of P weights about 31 grams

in one mole there are 6.022*10^23 atoms

we use the rule of threes

6.022*10^23atoms......weight..........31 grams

3.45*10^23 atoms.........weight...........x grams

x=(3.45*10^23*31)/6.022*10^23

x=106.95/6.022=<u><em>17.76 grams</em></u>

5 0

2 years ago

2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to

evablogger [386]

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

3 0

2 years ago